diff --git "a/OE_TO_physics_en_COMP.tsv" "b/OE_TO_physics_en_COMP.tsv"
new file mode 100644--- /dev/null
+++ "b/OE_TO_physics_en_COMP.tsv"
@@ -0,0 +1,13563 @@
+id	question	solution	final_answer	context	image	modality	difficulty	is_multiple_answer	unit	answer_type	error	question_type	subfield	subject	language
+0	"In an old coal factory, a conveyor belt will move at a constant velocity of $20.3 \mathrm{~m} / \mathrm{s}$ and can deliver a maximum power of $15 \mathrm{MW}$. Each wheel in the conveyor belt has a diameter of $2 \mathrm{~m}$. However a changing demand has pushed the coal factory to fill their coal hoppers with a different material with a certain constant specific density. These ""coal"" hoppers have been modified to deliver a constant $18 \mathrm{~m}^{3} \mathrm{~s}^{-1}$ of the new material to the conveyor belt. Assume that the kinetic and static friction are the same and that there is no slippage. What is the maximum density of the material?"	['The maximal force the convey belt can provide to a particle is:\n$$\nF=\\frac{P}{v}\n$$\n\nThe conveyor belt must provide an impulse to the particles to have a momentum of $p=m v$, where $m$ is the mass of the particle and $v$ is the velocity.\n\n$$\nF=\\frac{d p}{d t}\n$$\n\nwhere $\\frac{d p}{d t}$ is:\n\n$$\n\\rho \\dot{V} v\n$$\n\nSolving for for the maximum density we get:\n\n$$\n\\begin{gathered}\n\\rho=\\frac{P}{\\dot{V} v^{2}} \\\\\n\\rho=2022.2 \\frac{\\mathrm{kg}}{\\mathrm{m}^{3}}\n\\end{gathered}\n$$']	['$2022.2$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	$\frac{\mathrm{kg}}{\mathrm{m}^{3}}$	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+1	"Neutrinos are extremely light particles and rarely interact with matter. The Sun emits neutrinos, each with an energy of $8 \times 10^{-14} \mathrm{~J}$ and reaches a flux density of $10^{11}$ neutrinos $/\left(\mathrm{s} \mathrm{cm}^{2}\right)$ at Earth's surface.
+
+In the movie 2012, neutrinos have mutated and now are completely absorbed by the Earth's inner core, heating it up. Model the inner core as a sphere of radius $1200 \mathrm{~km}$, density $12.8 \mathrm{~g} / \mathrm{cm}^{3}$, and a specific heat of $0.400 \mathrm{~J} / \mathrm{g} \mathrm{K}$. The time scale, in seconds, that it will take to heat up the inner core by $1^{\circ} \mathrm{C}$ is $t=1 \times 10^{N}$ where $N$ is an integer. What is the value of $N$ ?"	['The cross sectional area is $\\pi r^{2}$, so the incoming power generated by the neutrinos is:\n$$\nP=\\pi r^{2} E \\Phi\n$$\n\nwhere $E$ is the energy of each neutrino and $\\Phi$ is the flux density. We want to cause a change in energy of:\n\n$$\n\\Delta Q=m c \\Delta T=\\rho \\frac{4}{3} \\pi r^{3} c \\Delta T\n$$\n\nwhich can be accomplished in a time:\n\n$$\nP t=\\Delta Q \\Longrightarrow t=\\frac{\\rho\\left(4 \\pi r^{3}\\right) c \\Delta T}{3 \\pi r^{2} E \\Phi}=\\frac{4 \\rho r c \\Delta T}{3 E \\Phi}=1 \\times 10^{14} \\mathrm{~s}\n$$']	['$1 \\times 10^{14}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	s	Numerical	1e13	Open-ended	Modern Physics	Physics	English
+2	"Eddie is experimenting with his sister's violin. Allow the ""A"" string of his sister's violin have an ultimate tensile strength $\sigma_{1}$. He tunes a string up to its highest possible frequency $f_{1}$ before it breaks. He then builds an exact copy of the violin, where all lengths have been increased by a factor of $\sqrt{2}$ and tunes the same string again to its highest possible frequency $f_{2}$. What is $f_{2} / f_{1}$ ? The density of the string does not change.
+
+Note: The ultimate tensile strength is maximum amount of stress an object can endure without breaking. Stress is defined as $\frac{F}{A}$, or force per unit area."	['We note from a simple dimensional analysis that the angular frequency of the string $\\omega$ will consist\n\n\nof the tension $T$, the length of the string $L$ and the mass of the string $m$.\n\n$$\n\\begin{aligned}\nT & =\\left[M L T^{-2}\\right] \\\\\nL & =[L] \\\\\nm & =[M] \\\\\n\\omega & =\\left[T^{-1}\\right]\n\\end{aligned}\n$$\n\nTherefore, by rearranging, we find that\n\n$$\n\\begin{aligned}\n\\omega & =T^{\\alpha} L^{\\beta} m^{\\gamma} \\\\\n{\\left[T^{-1}\\right] } & =\\left[M L T^{-2}\\right]^{\\alpha}[L]^{\\beta}[M]^{\\gamma}\n\\end{aligned}\n$$\n\nDistributing the exponents, and rearranging gives us\n\n$$\nT^{-1}=M^{\\alpha+\\gamma} L^{\\alpha+\\beta} T^{-2 \\alpha}\n$$\n\nWe now have three equations\n\n$$\n\\begin{aligned}\n\\alpha+\\gamma & =0 \\\\\n\\alpha+\\beta & =0 \\\\\n-2 \\alpha & =-1\n\\end{aligned}\n$$\n\nFrom here, we find that $\\alpha=1 / 2$. Substituting this into the first equation gives us\n\n$$\n1 / 2+\\gamma=0 \\Longrightarrow \\gamma=-1 / 2\n$$\n\nthen substituting $\\alpha$ into the second equation gives us\n\n$$\n1 / 2+\\beta=0 \\Longrightarrow \\beta=-1 / 2\n$$\n\nWe now find that the angular frequency is given by\n\n$$\n\\omega=A \\sqrt{\\frac{T}{L m}}\n$$\n\nwhere $A$ is an arbritary constant. Noting that $\\omega=2 \\pi f$, we find that\n\n$$\nf=\\frac{A}{2 \\pi} \\sqrt{\\frac{T}{L m}}\n$$\n\nFrom this analysis, we can then see that $f_{2} / f_{1}=\\sqrt{2} / 2 \\approx 0.707$.']	['$\\frac{\\sqrt{2}}{2}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-3	Open-ended	Mechanics	Physics	English
+3	"A one horsepower propeller powered by a battery and is used to propel a small boat initially at rest. You have two options:
+
+1. Put the propeller on top of the boat and push on the air with an initial force $F_{1}$
+2. Put the propeller underwater and push on the water with an initial force $F_{2}$.
+
+The density of water is $997 \mathrm{~kg} / \mathrm{m}^{3}$ while the density of air is $1.23 \mathrm{~kg} / \mathrm{m}^{3}$. Assume that the force is both cases is dependent upon only the density of the medium, the surface area of the propeller, and the power delivered by the battery. What is $F_{2} / F_{1}$ ? You may assume (unrealistically) the efficiency of the propeller does not change. Round to the nearest tenths."	['The force exerted on the fluid is roughly proportional to the change in momentum with respect to time:\n$$\nF=\\frac{d p}{d t}=v \\frac{d m}{d t}=v \\frac{d}{d t}(\\rho A x)=\\rho A v^{2}\n$$\n\nIt is kept at a constant power $P=F v$, which can allow us to solve for the speed $v$ of the propellers.\n\n$$\nP=\\rho A v^{3} \\Longrightarrow v=\\left(\\frac{P}{\\rho A}\\right)^{1 / 3}\n$$\n\nso the force is given by:\n\n$$\nF=\\rho A\\left(\\frac{P}{\\rho A}\\right)^{2 / 3} \\Longrightarrow F \\propto \\rho^{1 / 3}\n$$\n\nTherefore:\n\n$$\nF_{2} / F_{1}=(997 / 1.23)^{1 / 3}=9.26 \\text { times }\n$$']	['9.26']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	times	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+4	"A professional pastry chef is making a sweet which consists of 3 sheets of chocolate. The chef leaves a gap with width $d_{1}=0.1 \mathrm{~m}$ between the top and middle layers and fills it with a chocolate syrup with uniform viscosity $\eta_{1}=10 \mathrm{~Pa} \cdot \mathrm{s}$ and a gap with width $d_{2}=0.2 \mathrm{~m}$ between the middle and bottom sheet and fills it with caramel with uniform viscosity $\eta_{2}=15 \mathrm{~Pa} \cdot \mathrm{s}$. If the chef pulls the top sheet with a velocity $2 \mathrm{~m} / \mathrm{s}$ horizontally, at what speed must he push the bottom sheet horizontally such that the middle sheet remains stationary initially? Ignore the weight of the pastry sheets throughout the problem and the assume the sheets are equally sized.
+
+Note: Shear stress is governed by the equation $\tau=\eta \times$ rate of strain."	['The plates are equal sizes so all we have to do is simply balance the shear stresses which act in opposing directions on the middle plate:\n$$\n\\begin{gathered}\n\\tau_{1}=\\tau_{2} \\\\\n\\eta_{1} \\cdot \\frac{v_{1}}{d_{1}}=\\eta_{2} \\cdot \\frac{v_{2}}{d_{2}} \\\\\n10 \\cdot \\frac{2}{0.1}=15 \\frac{v}{0.2} \\\\\nv=2.667 \\mathrm{~m} / \\mathrm{s}\n\\end{gathered}\n$$']	['$2.667$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	$\mathrm{~m} / \mathrm{s}$	Numerical	1e-2	Open-ended	Mechanics	Physics	English
+5	"A magnetic field is located within a region enclosed by an elliptical island with semi-minor axis of $a=100 \mathrm{~m}$ and semi-major axis of $b=200 \mathrm{~m}$. A car carrying charge $+Q=1.5 \mathrm{C}$ drives on the boundary of the island at a constant speed of $v=5 \mathrm{~m} / \mathrm{s}$ and has mass $m=2000 \mathrm{~kg}$. Any dimensions of the car can be assumed to be much smaller than the dimensions of the island. Ignore any contributions to the magnetic field from the moving car and assume that the car has enough traction to continue driving in its elliptical path.
+
+Let the center of the island be located at the point $(0,0)$ while the semi major and semi minor axes lie on the $x$ and $y$-axes, respectively.
+
+On this island, the magnetic field varies as a function of $x$ and $y: B(x, y)=k_{b} e^{c_{b} x y} \hat{z}$ (pointing in the upward direction, perpendicular to the island plane in the positive $z$-direction). The constant $c_{b}=10^{-4} \mathrm{~m}^{-2}$ and the constant $k_{b}=2.1 \mu \mathrm{T}$
+
+At what point on the island is the force from the magnetic field a maximum? Write the distance of this point from the $x$-axis in metres."	['To find a minimum or maximum, the gradient of the constraint function $f(x, y)=\\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}-1$ and the gradient of the $B$ field function should be scalar multiples of each other.\n\n$$\n\\begin{aligned}\n& \\frac{2 x}{b^{2}} \\mu=c_{b} y e^{x y} \\\\\n& \\frac{2 y}{a^{2}} \\mu=c_{b} x e^{x y}\n\\end{aligned}\n$$\n\nSolving the two equations, we get that a maximum point $(x, y)$ is of the form $\\left(\\frac{b}{\\sqrt{2}}, \\frac{a}{\\sqrt{2}}\\right)$ or $\\left(-\\frac{b}{\\sqrt{2}},-\\frac{a}{\\sqrt{2}}\\right)$. The distance from the $\\mathrm{y}$-axis is thus $\\frac{a}{\\sqrt{2}}=70.7 \\mathrm{~m}$.']	['70.7']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	1e-1	Open-ended	Electromagnetism	Physics	English
+6	"Inside a laboratory at room temperature, a steel tuning fork in the shape of a $\mathrm{U}$ is struck and begins to vibrate at $f=426 \mathrm{~Hz}$. The tuning fork is then brought outside where it is $10^{\circ} \mathrm{C}$ hotter and the experiment is performed again. What is the change in frequency, $\Delta f$ of the tuning fork? (A positive value will indicate an increase in frequency, and a negative value will indicate a decrease.)
+
+Note: The linear thermal coefficient of expansion for steel is $\alpha=1.5 \times 10^{-5} \mathrm{~K}^{-1}$ and you may assume the expansion is isotropic and linear. When the steel bends, there is a restoring torque $\tau=-\kappa \theta$ such that $\kappa \equiv G J$ where $G=77 \mathrm{GPa}$ is constant and $J$ depends on the geometry and dimensions of the cross-sectional area."	['Note that $\\kappa$ has units of torque so dimensionally, $J$ must be proportional to $L^{3}$. Therefore, we have:\n$$\n\\beta M L^{2} \\alpha \\propto-L^{3} \\theta \\Longrightarrow f \\propto \\sqrt{L}\n$$\n\nTherefore, we have:\n\n$$\n\\frac{\\Delta f}{f}=\\frac{\\Delta L}{2 L}\n$$\n\nSince $\\frac{\\Delta L}{L}=\\alpha \\Delta T$, this gives us:\n\n$$\n\\Delta f=\\frac{1}{2} f \\alpha \\Delta T=0.0320 \\mathrm{~Hz}\n$$']	['0.0320']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	Hz	Numerical	1e-3	Open-ended	Thermodynamics	Physics	English
+7	A large metal conducting sphere with radius $10 \mathrm{~m}$ at an initial potential of 0 and an infinite supply of smaller conducting spheres of radius $1 \mathrm{~m}$ and potential $10 \mathrm{~V}$ are placed into contact in such a way: the large metal conducting sphere is contacted with each smaller sphere one at a time. You may also assume the spheres are touched using a thin conducting wire that places the two spheres sufficiently far away from each other such that their own spherical charge symmetry is maintained. What is the least number of smaller spheres required to be touched with the larger sphere such that the potential of the larger sphere reaches $9 \mathrm{~V}$ ? Assume that the charges distribute slowly and that the point of contact between the rod and the spheres is not a sharp point.	['Let each sphere with radius $1 \\mathrm{~m}$ have charge $q$. Note that each time the large metal conducting sphere is contacted with each of the smaller spheres, the potential is equalized between the two objects. The potential on a sphere is proportional to $\\frac{q}{r}$, so the large conducting sphere must retain $\\frac{10}{11}$ of the total charge after it is contacted with a smaller sphere. Furthermore, to reach $9 \\mathrm{~V}$, the required end charge on the sphere of radius $10 \\mathrm{~m}$ is at least $9 q$. Thus, we get a recursion for the charge of the large square $Q$ in terms of the number of small spheres touched $n$.\n$$\nQ(n+1)=(Q(n)+q) \\cdot \\frac{10}{11}\n$$\n\nInductively applying this recursion, we obtain\n\n$$\nQ(n)=q\\left[\\left(\\frac{10}{11}\\right)^{n}+\\cdots+\\frac{10}{11}\\right]\n$$\n\nWe can now sum this geometric series:\n\n$$\nQ(n)=q\\left(\\frac{10}{11} \\cdot \\frac{\\left(\\frac{10}{11}\\right)^{n}-1}{\\frac{10}{11}-1}\\right) .\n$$\n\nThus, using $Q(n) \\geq 9 q$, we find that $\\left(\\frac{10}{11}\\right)^{n} \\leq 0.1$, which provides $n \\geq 24.1588$, or $n=25$ as the answer.']	['25']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-3	Open-ended	Electromagnetism	Physics	English
+8	"During high speed motion in a strong electric field, a charged particle can ionize air molecules it collides with.
+
+A charged particle of mass $m=0.1 \mathrm{~kg}$ and charge $q=0.5 \mu \mathrm{C}$ is located in the center of a cubical box. Each vertex of the box is fixed in space and has a charge of $Q=-4 \mu \mathrm{C}$. If the side length of the box is $l=1.5 \mathrm{~m}$ what minimum speed (parallel to an edge) should be given to the particle for it to exit the box (even if it's just momentarily)? Let the energy loss from Corona discharge and other radiation effects be $E=0.00250 \mathrm{~J}$."	['Conservation of energy gives:\n$$\nT_{i}+U_{i}=T_{f}+U_{f}+E\n$$\n\nSolving for the initial potential energy gives\n\n$$\nU_{i}=-8 \\frac{k q Q}{l \\sqrt{3} / 2}=-\\frac{16 k q Q}{\\sqrt{3} l}\n$$\n\nAnd since the final kinetic energy is zero, the final potential energy is\n\n$$\nU_{f}=-4 \\frac{k q Q}{l / \\sqrt{2}}-4 \\frac{k q Q}{l \\sqrt{\\frac{3}{2}}}=-\\left(4 \\sqrt{2}+4 \\sqrt{\\frac{2}{3}}\\right) \\frac{k q Q}{l}\n$$\n\nand thus solving for the initial kinetic energy:\n\n$$\n\\frac{1}{2} m v^{2}=\\left(-4 \\sqrt{2}-4 \\sqrt{\\frac{2}{3}}+\\frac{16}{\\sqrt{3}}\\right) \\frac{k q Q}{l}+E\n$$\n\nThe final answer is $v=0.354 \\mathrm{~m} / \\mathrm{s}$.']	['0.354']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m/s	Numerical	1e-2	Open-ended	Electromagnetism	Physics	English
+9	"Max finds himself trapped in the center of a mirror walled equilateral triangular room. What minimum beam angle must his flashlight have so that any point of illumination in the room can be traced back to his flashlight with at most 1 bounce? (Answer in degrees.) Since the room is large, assume the person is a point does not block light. Visualize the questions in a 2D setup. The floor/ceiling is irrelevant.
+
+The point of illumination refers to any point in the room that is lit."	['Each time light hits a mirror, we can reflect the entire equilateral triangle about that mirror and continue to trace the straight-line path of the light. For a maximum of 1 bounce, we can reflect our triangle about each of its initial sides. In order for the light to hit every part of the triangle (or an image of that part), by symmetry we require a $120^{\\circ}$ angle. In this case, we would just shine the flashlight so that light directly reaches the entirety of one side, and reflections of light will fully reach the other two sides of the triangle.']	['$120$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	$^{\circ}$	Numerical	0	Open-ended	Optics	Physics	English
+10	"Kushal finds himself trapped in a large room with mirrors as walls. Being scared of the dark, he has a powerful flashlight to light the room. All references to ""percent"" refer to area. Since the room is large, assume the person is a point does not block light. Visualize the questions in a 2D setup. The floor/ceiling is irrelevant. The point of illumination refers to any point in the room that is lit.
+
+What percent of a large circular room can be lit up using a flashlight with a 20 degree beam angle if Kushal stands in the center?"	['Each ray emitted follows a straight line, even when it is reflected since it originates from the center of a circle. Thus, the light rays in total trace out two circular sectors with an angle of $\\theta=20^{\\circ}$ each. Thus, the total percent of the room illuminated is:\n\n$$\nf=\\frac{2 \\theta}{360}=11.1 \\% \\text {. }\n$$']	['11.1']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	%	Numerical	1e-1	Open-ended	Optics	Physics	English
+11	"Two identical neutron stars with mass $m=4 \times 10^{30} \mathrm{~kg}$ and radius $15 \mathrm{~km}$ are orbiting each other a distance $d=700 \mathrm{~km}$ away from each other ( $d$ refers to the initial distance between the cores of the neutron stars). Assume that they orbit as predicted by classical mechanics, except that they generate gravitational waves. The power dissipated through these waves is given by:
+
+$$
+P=\frac{32 G^{4}}{5}\left(\frac{m}{d c}\right)^{5}
+$$
+
+How long does it take for the two stars to collide? Answer in seconds. Note: $d$ is the distance between the cores of the stars."	['Due to Virial theorem, we have:\n$$\nK=-\\frac{1}{2} U\n$$\n\nso the total energy is:\n\n$$\nE=U-\\frac{1}{2} U=-\\frac{G m^{2}}{2 R}\n$$\n\nWe know that the power dissipated gives the change in energy, or:\n\n$$\nP=\\frac{32 G^{4}}{5}\\left(\\frac{m}{R c}\\right)^{5}=\\frac{d}{d t} \\frac{G m^{2}}{2 R}\n$$\n\nor:\n\n$$\n\\frac{32 G^{4}}{5}\\left(\\frac{m}{R c}\\right)^{5} d t=-\\frac{G m^{2}}{2 R^{2}} d R \\Longrightarrow \\int_{0}^{t} \\frac{64 G^{3}}{5} \\frac{m^{3}}{c^{5}} d t=\\int_{d}^{2 r}-R^{3} d R\n$$\n\nSolving this leads us to:\n\n$$\n\\frac{64 G^{3} m^{3}}{5 c^{5}} t=\\frac{d^{4}-r^{4}}{4} \\Longrightarrow t=\\frac{5 c^{5}\\left(d^{4}-16 r^{4}\\right)}{256 G^{3} m^{3}}\n$$\n\nPlugging in the numbers gives:\n\n$$\nt=590 \\mathrm{sec}\n$$\n\nNote that we can also assume that $d^{4} \\gg(2 r)^{4}$ which will simplify calculations, and not introduce any noticeable error.']	['590']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	seconds	Numerical	1e1	Open-ended	Modern Physics	Physics	English
+12	"In the cosmic galaxy, the Sun is a mainsequence star, generating its energy mainly by nuclear fusion of hydrogen nuclei into helium. In its core, the Sun fuses hydrogen to produce deuterium $(2 \mathrm{H})$ and tritium $(3 \mathrm{H})$, then makes about 600 million metric tons of helium (4He) per second. Of course, there are also some relatively smaller portions of fission reactions in the Sun's core, e.g. a nuclear fission reaction with Uranium-235 (235U). The Fusion reaction:
+
+$$
+{ }^{2} \mathrm{H}+{ }^{3} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+n+\text { Released Energy }
+$$
+
+The Fission reaction:
+
+$$
+{ }^{235} U+n+(\text { Initial Energy }) \rightarrow{ }^{144} \mathrm{Ba}+{ }^{90} \mathrm{Kr}+2 n+\text { Released Energy }
+$$
+
+Isotope Mass (at rest) 
+
+| Isotope Names | Mass (at rest) $(\mathrm{u})$ |
+| :--- | :--- |
+| Deuterium $\left({ }^{2} \mathrm{H}\right)$ | 2.0141 |
+| Tritium $\left({ }^{3} \mathrm{H}\right)$ | 3.0160 |
+| Helium $\left({ }^{4} \mathrm{He}\right)$ | 4.0026 |
+| Neutron $(\mathrm{n})$ | 1.0087 |
+| Uranium-235 $\left({ }^{235} \mathrm{U}\right)$ | 235.1180 |
+| Barium-144 $\left({ }^{144} \mathrm{Ba}\right)$ | 143.8812 |
+| Krypton-90 $\left({ }^{90} \mathrm{Kr}\right)$ | 89.9471 |
+
+Calculate the kinetic energy (in $\mathrm{MeV}$ ) released by the products in one fusion reaction."	['Let the kinetic energy released be\n\n$$\nK E_{\\text {released }}=-\\Delta m c^{2}\n$$\n\nLet the mass of helium be $m_{h}$, deuterium be $m_{d}$, tritium $m_{t}$, and mass of neutron $m_{n}$ Therefore.\n\n$$\n-\\Delta m=m_{d}+m_{t}-m_{h}-m_{n}=3.0160+2.0141-4.0026-1.0087=0.0188 \\mathrm{u}\n$$\n\nwhich gives\n\n$$\nK E_{\\text {released }}=(0.0188 \\mathrm{u}) \\cdot\\left(931.494 \\frac{\\mathrm{MeV}}{\\mathrm{u}}\\right)=17.51 \\mathrm{MeV}\n$$']	['17.51']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	MeV	Numerical	1e-1	Open-ended	Modern Physics	Physics	English
+13	A particle of rest mass $m$ moving at a speed $v=0.7 c$ decomposes into two photons which fly off at a separated angle $\theta$. What is the minimum value of the angle of separation assuming that the two photons have equal wavelength. (Answer in degrees)	['Conservation of momentum and energy gives:\n$$\n\\begin{gathered}\np_{m}=2 E_{\\gamma} \\cos (\\theta / 2) \\\\\nE_{m}=2 E_{\\gamma}\n\\end{gathered}\n$$\n\nRelativity demands that:\n\n$$\nE_{m}^{2}=m^{2}+p_{m}^{2}\n$$\n\nSolving this system of three equations gives:\n\n$$\n\\begin{aligned}\n4 E_{\\gamma}^{2} & =m^{2}+p_{m}^{2} \\\\\n\\frac{\\gamma^{2} m^{2} v^{2}}{\\cos ^{2}(\\theta / 2)} & =m^{2}+\\gamma^{2} m^{2} v^{2} \\\\\n\\gamma^{2} v^{2}\\left(-1+\\frac{1}{\\cos ^{2}(\\theta / 2)}\\right) & =1 \\\\\n\\frac{1}{\\cos ^{2}(\\theta / 2)} & =\\frac{1}{\\gamma^{2} v^{2}}+1 \\\\\n\\cos (\\theta / 2) & =v \\\\\n\\theta & =91.1^{\\circ}\n\\end{aligned}\n$$']	['$91.1$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	$^{\circ}$	Numerical	1e0	Open-ended	Modern Physics	Physics	English
+14	"At an amusement park, there is a ride with three ""teacups"" that are circular with identical dimensions. Three friends, Ethan, Rishab, and Kushal, all pick a teacup and sit at the edge. Each teacup rotates about its own axis clockwise at an angular speed $\omega=1 \mathrm{rad} / \mathrm{s}$ and can also move linearly at the same time.
+
+The teacup Ethan is sitting on (as always) is malfunctional and can only rotate about its own axis. Rishab's teacup is moving linearly at a constant velocity $2 \mathrm{~m} / \mathrm{s}[\mathrm{N}]$ and Kushal's teacup is also moving linearly at a constant velocity of $4 \mathrm{~m} / \mathrm{s}\left[\mathrm{N} 60^{\circ} \mathrm{E}\right]$. All three teacups are rotating as described above. Interestingly, they observe that at some point, all three of them are moving at the same velocity. What is the radius of each teacup?
+
+Note: $\left[\mathrm{N} 60^{\circ} \mathrm{E}\right]$ means $60^{\circ}$ clockwise from north e.g. $60^{\circ}$ east of north."	['We can plot the motion on a $v_{y}-v_{z}$ graph instead of carrying out calculations. We have three points at locations $(0,0),(0,2)$, and $(2 \\sqrt{3}, 2)$ which represent the velocity of the center of mass of the teacups. The velocity that they are moving at can be traced as a circle with radius $r \\omega$, centered at these points.\nThe problem now becomes, at what value $r$ will the three circles intersect. Drawing a diagram, or carrying out trigonometric calculations gives $r=2 \\mathrm{~m}$.']	['2']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+15	"Life on Earth would not exist as we know it without the atmosphere. There are many reasons for this, but one of which is temperature. Let's explore how the atmosphere affects the temperature on Earth. Assume that all thermal energy striking the earth uniformly and ideally distributes itself across the Earth's surface.
+
+- Assume that the Earth is a perfect black body with no atmospheric effects. Let the equilibrium temperature of Earth be $T_{0}$. (The sun outputs around $3.846 \times 10^{2^{6}} \mathrm{~W}$, and is $1.496 \times 10^{8} \mathrm{~km}$ away.)
+- Now assume the Earth's atmosphere is isothermal. The short wavelengths from the sun are nearly unaffected and pass straight through the atmosphere. However, they mostly convert into heat when they strike the ground. This generates longer wavelengths that do interact with the atmosphere. Assume that the albedo of the ground is 0.3 and $e$, the emissivity and absorptivity of the atmosphere, is 0.8 . Let the equilibrium average temperature of the planet now be $T_{1}$.
+
+What is the percentage increase from $T_{0}$ to $T_{1}$ ?
+
+Note: The emissivity is the degree to which an object can emit longer wavelengths (infrared) and the absorptivity is the degree to which an object can absorb energy. Specifically, the emissivity is the ratio between the energy emitted by an object and the energy emitted by a perfect black body at the same temperature. On the other hand, the absorptivity is the ratio of the amount of energy absorbed to the amount of incident energy."	['Let us solve this problem in the case of the Earth being a graybody first and then substitute values for when it is a blackbody. The portion of energy that reaches the Earth is given by the ratio between the cross-sectional area of the satellite and the area of an imaginary sphere centered around the sun with a radius of $L$. Thus, the incoming radiation is multiplied by a factor of $\\gamma=(R / 2 L)^{2}$. The energy from the sun that the surface absorbs is $\\gamma(1-\\alpha) E$, where $E$ is the energy output of the sun. Here $\\gamma=1 / 4$ as the sphere encompassing will be 4 times the area of its intercept.\nWe can now write two systems of equations at the atmosphere and the ground of the Earth. At the top of the atmosphere, we require equilibrium meaning that zero net radiation leaves the atmosphere or:\n\n$$\n-\\frac{1}{4} S_{0}(1-\\alpha)+\\varepsilon \\sigma T_{a}^{4}+(1-\\varepsilon) \\sigma T_{s}^{4}=0\n$$\n\nSimilarly, at the ground, we write another equilibrium equation of:\n\n$$\n\\frac{1}{4} S_{0}(1-\\alpha)+\\varepsilon \\sigma T_{a}^{4}-\\sigma T_{s}^{4}=0\n$$\n\nThus, solving the ground equilibrium equation yields us $T_{a}=2^{-1 / 4} T_{s}$ and plugging back into the atmosphere equilibrium equation tells us:\n\n$$\n\\frac{1}{4} S_{0}(1-\\alpha)=\\left(1-\\frac{\\varepsilon}{2}\\right) \\sigma T_{s}^{4} \\Longrightarrow T_{s}=289.601 \\mathrm{~K}\n$$']	['289.601']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	K	Numerical	1e0	Open-ended	Thermodynamics	Physics	English
+16	"Mountains have two sides: windward and leeward. The windward side faces the wind and typically receives warm, moist air, often from an ocean. As wind hits a mountain, it is forced upward and begins to move towards the leeward side. During social distancing, Rishab decides to cross a mountain from the windward side to the leeward side of the mountain. What he finds is that the air around him has warmed when he is on the leeward side of the mountain.
+
+Let us investigate this effect. Consider the warm, moist air mass colliding with the mountain and moving upwards on the mountain. Disregard heat exchange with the air mass and the mountain. Let the humidity of the air on the windward side correspond to a partial vapor pressure $0.5 \mathrm{kPa}$ at $100.2 \mathrm{kPa}$ and have a molar mass of $\mu_{a}=28 \mathrm{~g} / \mathrm{mole}$. The air predominantly consists of diatomic molecules of oxygen and nitrogen. Assume the mountain to be very high which means that at the very top of the mountain, all of the moisture in the air condenses and falls as precipitation. Let the precipitation have a heat of vaporization $L=2.4 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}$ and molar mass $\mu_{p}=18.01 \mathrm{~g} / \mathrm{mole}$. Calculate the total change in temperature from the windward side to the leeward side in degrees Celsius."	['We use the first law of thermodynamics to solve this problem. For diatomic molecules, the internal energy per mole is given by $\\frac{5}{2} R T$. If the molar mass of the air is $\\mu_{a}$, then we have that the change in internal energy of the air is given by\n$$\n\\Delta U=\\frac{5}{2} \\frac{M}{\\mu_{a}} R \\Delta T\n$$\n\nWe also note that the total work performed by the gas is\n\n$$\nW=P_{2} V_{2}-P_{1} V_{1}\n$$\n\nsince the process is adiabatic, we can use the ideal gas equation $P V=\\nu R T=(M / \\mu) R T$ to express the total work as\n\n$$\nW=\\frac{M}{\\mu_{a}} R \\Delta T\n$$\n\nThe heat that is taken away during condensation at the top of the mountain is given by $Q=L \\Delta m$ where $\\Delta m$ is the total mass of the precipitation. According to the ideal gas law, we have that\n\n$$\nP V_{1}=\\frac{\\Delta m}{\\mu_{p}} R T_{1}, \\quad P_{1} V_{1}=\\frac{M}{\\mu_{a}} R T_{1}\n$$\n\nrecombining these equations and equating them gives us\n\n$$\n\\begin{aligned}\n\\frac{\\Delta m}{\\mu_{p} P} R T_{1} & =\\frac{M}{\\mu_{a} P_{1}} R T_{1} \\\\\n\\Delta m & =M \\frac{\\mu_{p} P}{\\mu_{a} P_{1}}\n\\end{aligned}\n$$\n\n\n\nTherefore,\n\n$$\nQ=L M \\frac{\\mu_{a} P}{\\mu_{p} P_{1}}\n$$\n\nWe finally can now use the first law of thermodynamics\n\n$$\nQ=\\Delta U+W \\Longrightarrow L M \\frac{\\mu_{p} P}{\\mu_{a} P_{1}}=\\frac{5}{2} \\frac{M}{\\mu_{a}} R \\Delta T+\\frac{M}{\\mu_{a}} R \\Delta T\n$$\n\nWe then simplify this equation to get\n\n$$\n\\begin{aligned}\nL M \\frac{\\mu_{p} P}{\\mu_{a} P_{1}} & =\\frac{7}{2} \\frac{M}{\\mu_{a}} R \\Delta T \\\\\n\\Delta T & =\\frac{2}{7} \\frac{L \\mu_{p} P}{R P_{1}}=7.41 \\mathrm{~K}\n\\end{aligned}\n$$\n\nA simpler approach could be to assume that the number of moles of water vapour in the atmosphere is equal to number of moles of water condensed. Then, the mass of precipitated water is $\\mu_{p} n \\frac{P}{P_{1}}$, where $n$ is the number of moles of air. Thus,\n\n$$\n\\mu_{p} n \\frac{P}{P_{1}} L=\\frac{7}{2} n R \\Delta T\n$$']	['7.41']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	K	Numerical	1e-1	Open-ended	Thermodynamics	Physics	English
+17	A planet has a radius of $10 \mathrm{~km}$ and a uniform density of $5 \mathrm{~g} / \mathrm{cm}^{3}$. A powerful bomb detonates at the center of the planet, releasing $8.93 \times 10^{17} \mathrm{~J}$ of energy, causing the planet to separate into three large sections each with equal masses. You may model each section as a perfect sphere of radius $r^{\prime}$. The initial and final distances between the centers of any two given sections is $2 r^{\prime}$. How long does it take for the three sections to collide again?	['Due to conservation of momentum, the three masses must form an equilateral triangle at all times. Let us determine the force as a function of $r$, the distance between each mass and the center.\n\n\n<img_4307>\n\nThe vector sum of the net force on any individual mass is\n\n$$\nF=\\frac{G m^{2}}{d^{2}} \\sqrt{2-2 \\cos 120^{\\circ}}=\\frac{\\sqrt{3} G m^{2}}{d^{2}}\n$$\n\nwhere $d$ is the distance between the mass and the center.\n\n$$\nd^{2}=3 r^{2}\n$$\n\nThe net force is thus\n\n$$\nF=\\frac{G m^{2}}{\\sqrt{3} r^{2}}\n$$\n\nThe system behaves as if there was a stationary mass $m^{\\prime}=m / \\sqrt{3}$ at the center, simplifying the problem greatly into a restricted two body system. Next, we need to figure out the height of the apoapsis. This can be done via conservation of energy.\n\n$$\nE_{\\text {binding,initial }}+E=3 E_{\\text {binding,final }}-\\frac{G m^{2}}{\\sqrt{3} \\ell} \\Longrightarrow \\ell=-\\frac{G m^{2}}{\\sqrt{3}\\left(-\\frac{3 G M^{2}}{5 R}+8.93 \\cdot 10^{17}-\\left(-3 \\frac{3 G m^{2}}{5 r_{f}}\\right)\\right)}=101,000 \\mathrm{~m}\n$$\n\nIf you have a stationary mass $M$ at the center. The time it takes for an object to fall into it is:\n\n$$\nT=\\pi \\sqrt{\\frac{\\ell^{3}}{8 G M}}\n$$\n\nour time will be double this, and the mass in the center will be $M=m / \\sqrt{3}$. So plugging in numbers gives:\n\n$$\nt=2 \\pi \\sqrt{\\frac{l^{3}}{8 G\\left(\\frac{m}{\\sqrt{3}}\\right)}}=138,000\n$$\n\n\n\nIf we take into account a nonzero radius so final separation is $10 / \\cos \\left(30^{\\circ}\\right) \\mathrm{km}$, then the answer should be:\n\n$$\n-2 \\int_{l}^{\\frac{10000}{\\cos \\left(\\frac{\\pi}{6}\\right)}}\\left(\\sqrt{\\frac{x l}{2 G\\left(\\frac{m}{\\sqrt{3}}\\right)(l-x)}}\\right) d x=136,000 \\mathrm{~s}\n$$\n\nNote if you set the upper bound to zero, you get the same answer as before. Both these answers will be accepted.']	['136000']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	s	Numerical	1e3	Open-ended	Mechanics	Physics	English
+18	"A point charge $+q$ is placed a distance $a$ away from an infinitely large conducting plate. The force of the electrostatic interaction is $F_{0}$. Then, an identical conducting plate is placed a distance $3 a$ from the charge, parallel to the first one such that the charge is ""sandwiched in."" The new electrostatic force the particle feels is $F^{\prime}$. What is $F^{\prime} / F_{0}$ ? Round to the nearest hundredths."	"[""We solve this via the method of image charges. Let us first reflect the charge $+q$ across the closest wall. The force of this interaction will be:\n$$\nF_{0}=\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}} \\frac{1}{2^{2}}\n$$\n\nand this will be the only image charge we need to place if there were only one conducting plane. Since there is another conducting plane, another charge will be reflected to a distance $a+4 a$ past the other conducting plane, and thus will be $a+4 a+3 a=8 a$ away from the original charge. All these reflections cause a force that points in the same direction, which we will label as the positive + direction. Therefore:\n\n$$\nF_{+}=\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}}\\left(\\frac{1}{2^{2}}+\\frac{1}{8^{2}}+\\frac{1}{10^{2}}+\\frac{1}{16^{2}} \\frac{1}{18^{2}}+\\frac{1}{24^{2}}+\\cdots\\right)\n$$\n\nNow let us look at what happens if we originally reflect the charge $+q$ across the other wall. Repeating the steps above, we see that through subsequent reflections, each force will point in the negative - direction. Therefore:\n\n$$\nF_{-}=\\frac{-q^{2}}{4 \\pi \\epsilon_{0} a^{2}}\\left(\\frac{1}{6^{2}}+\\frac{1}{8^{2}}+\\frac{1}{14^{2}}+\\frac{1}{16^{2}}+\\frac{1}{22^{2}}+\\frac{1}{24^{2}}+\\cdots\\right)\n$$\n\nThe net force is a result of the superposition of these two forces, giving us:\n\n$$\n\\begin{aligned}\nF^{\\prime}=\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}} & \\left(\\frac{1}{2^{2}}+\\frac{1}{8^{2}}+\\frac{1}{10^{2}}+\\frac{1}{16^{2}} \\frac{1}{18^{2}}+\\frac{1}{24^{2}}+\\cdots\\right. \\\\\n& \\left.-\\frac{1}{6^{2}}-\\frac{1}{8^{2}}-\\frac{1}{14^{2}}-\\frac{1}{16^{2}}-\\frac{1}{22^{2}}-\\frac{1}{24^{2}}-\\cdots\\right)\n\\end{aligned}\n$$\n\nEven terms can be cancelled out to give:\n\n$$\n\\begin{aligned}\nF^{\\prime} & =\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}}\\left(\\frac{1}{2^{2}}-\\frac{1}{6^{2}}+\\frac{1}{10^{2}}-\\frac{1}{14^{2}}+\\frac{1}{18^{2}}-\\frac{1}{22^{2}}+\\cdots\\right) \\\\\n& =\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}} \\frac{1}{4}\\left(\\frac{1}{1^{2}}-\\frac{1}{3^{2}}+\\frac{1}{5^{2}}-\\frac{1}{7^{2}}+\\frac{1}{9^{2}}-\\frac{1}{11^{2}}+\\cdots\\right)\n\\end{aligned}\n$$\n\nYou may recognize the infinite series inside the parentheses to be Catalan's constant $G \\approx 0.916$. Alternatively, you can use a calculator and evaluate the first seven terms to get a rough answer (but will still be correct since we asked for it to be rounded). Therefore:\n\n$$\nF^{\\prime} / F=\\frac{1}{1^{2}}-\\frac{1}{3^{2}}+\\frac{1}{5^{2}}-\\frac{1}{7^{2}}+\\frac{1}{9^{2}}-\\frac{1}{11^{2}}+\\cdots=G \\approx 0.916\n$$""]"	['0.916']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	5e-2	Open-ended	Electromagnetism	Physics	English
+19	"Jerry spots a truckload of his favourite golden yellow Swiss cheese being transported on a cart moving at a constant velocity $v_{0}=5 \mathrm{~m} / \mathrm{s} \hat{i}$ along the x-axis, which is initially placed at $(0,0)$. Jerry, driven by desire immediately starts pursuing the cheese-truck in such a way that his velocity vector always points towards the cheese-truck; however, Jerry is smart and knows that he must maintain a constant distance $\ell=10 \mathrm{~m}$ from the truck to avoid being caught by anyone, no matter what. Note that Jerry starts at coordinates $(0, \ell)$.
+
+
+Let the magnitude of velocity (in $\mathrm{m} / \mathrm{s}$ ) and acceleration (in $\mathrm{m} / \mathrm{s}^{2}$ ) of Jerry at the moment when the (acute) angle between the two velocity vectors is $\theta=60^{\circ}$ be $\alpha$ and $\beta$ respectively. Compute $\alpha^{2}+\beta^{2}$."	"[""If the distance between Jerry and the cheese truck is constant, then Jerry moves in circle of radius $\\ell$ in the reference frame of the cheese truck. There is no radial component of Jerry's velocity in this reference frame, so we must have $\\alpha=v_{0} \\cos \\theta=\\frac{5}{2}$. In this case, the tangential velocity is $v_{0} \\sin \\theta$. Furthermore, the radial acceleration in this frame is given by the centripetal acceleration which is $\\frac{\\left(v_{0} \\sin \\theta\\right)^{2}}{\\ell}=\\frac{v_{0}^{2} \\sin ^{2} \\theta}{\\ell}$. The tangential acceleration is\n\n$$\n\\frac{d}{d t}\\left(v_{0} \\sin \\theta\\right)=v_{0} \\cos \\theta \\cdot \\frac{d \\theta}{d t}=v_{0} \\cos \\theta \\cdot \\frac{-v_{0} \\sin \\theta}{\\ell}=-\\frac{v_{0}^{2} \\sin \\theta \\cos \\theta}{\\ell}\n$$\n\nThe vector sum of these accelerations has magnitude\n\n$$\n\\beta=\\sqrt{\\left(\\frac{v_{0}^{2} \\sin ^{2} \\theta}{\\ell}\\right)^{2}+\\left(\\frac{v_{0}^{2} \\sin \\theta \\cos \\theta}{\\ell}\\right)^{2}}=\\frac{v_{0}^{2} \\sin \\theta}{\\ell}=\\frac{5 \\sqrt{3}}{4} .\n$$\n\nThe final answer is $\\alpha^{2}+\\beta^{2}=10.9375$.""]"	['10.9375']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-1	Open-ended	Mechanics	Physics	English
+20	Consider an LC circuit with one inductor and one capacitor. The amplitude of the charge on the plates of the capacitor is $Q=10 \mathrm{C}$ and the two plates are initially at a distance $d=1 \mathrm{~cm}$ away from each other. The plates are then slowly pushed together to a distance $0.5 \mathrm{~cm}$ from each other. Find the resultant amplitude of charge on the parallel plates of the capacitor after this process is completed. Note that the initial current in the circuit is zero and assume that the plates are grounded.	"[""In slow steady periodic processes (when the time for the change in parameters $\\tau$ is much less than the total systems frequency $f$ ), a quantity called the adiabatic invariant $I$ is conserved ${ }^{a}$. The adiabatic invariant corresponds to the area of a closed contour in phase space (a graph with momentum $p$ and position $x$ as its axes). Note the we can electrostatically map this problem to a mechanics one as the charge corresponds to position, while the momentum would correspond to $L I$ where $I$ is the current and $L$ is the inductance. Thus, in phase space, we have an elliptical contour corresponding to the equation: $\\frac{Q^{2}}{2 C}+\\frac{(L I)^{2}}{2 L}=C$ where $C$ is a constant in the system. As the area under the curve is conserved, then it can be written that $\\pi Q_{0} L I_{0}=\\pi Q_{f} L I_{f}$. It is also easy to conserve energy such that $L I^{2}=\\frac{Q^{2}}{C}$ which tells us $I=\\frac{Q}{\\sqrt{L C}}$. As $C \\propto 1 / x$, we then can write the adiabatic invariant as $x q^{4}$ which tells us $Q_{f}=\\sqrt[4]{2} Q$.\nWe can also solve this regularly by looking at the changes analytically. From Gauss's law, the electric field between the plates of the capacitators initially can be estimated as\n\n$$\nE=\\frac{Q}{2 \\varepsilon_{0} A}\n$$\n\nwhere $A$ is the area of the plate. The plates of the capacitator is attracted to the other one with a force of\n\n$$\nF=Q E=\\frac{Q^{2}}{2 \\varepsilon_{0} A}\n$$\n\nThe charges of the plates as a function of time can be approximated as\n\n$$\nQ_{c}= \\pm Q \\sin (\\omega t+\\phi)\n$$\n\nwhere $\\omega=\\frac{1}{\\sqrt{L C}}$. Using this equation, we estimate the average force $\\langle F\\rangle$ applied on the plate after a period of oscillations to be\n\n$$\n\\langle F\\rangle=\\frac{\\left\\langle Q^{2}\\right\\rangle}{2 \\varepsilon_{0} A}=\\frac{Q^{2}}{2 \\varepsilon_{0} A}\\left\\langle\\sin ^{2}(\\omega t+\\phi)\\right\\rangle=\\frac{Q^{2}}{2 \\varepsilon_{0} A} \\cdot\\left(\\frac{1}{2 \\pi} \\int_{0}^{2 \\pi} \\sin ^{2}(x) d x\\right)=\\frac{Q^{2}}{4 \\varepsilon_{0} A}\n$$\n\nthis means that after one period, the amount of work done to push the plates closer together is given by\n\n$$\nW_{F}=\\langle F\\rangle d x=\\frac{Q^{2}}{4 \\varepsilon_{0} A} d x\n$$\n\nIn this cycle, the amount of incremental work done by the $\\mathrm{LC}$ circuit will be given by\n\n$$\nd W_{\\mathrm{LC}}=\\Delta(F x)=\\Delta\\left(\\frac{Q^{2} x}{2 \\varepsilon_{0} A}\\right)=\\frac{Q x}{\\varepsilon_{0} A} d Q+\\frac{Q^{2}}{2 \\varepsilon_{0} A} d x\n$$\n\n\n\nFrom conservation of energy, $W_{F}=W_{L C}$. Or in other words,\n\n$$\n\\frac{Q^{2}}{4 \\varepsilon_{0} A} d x=\\frac{Q x}{\\varepsilon_{0} A} d Q+\\frac{Q^{2}}{2 \\varepsilon_{0} A} d x\n$$\n\nsimplifying gives us\n\n$$\n\\begin{aligned}\n\\frac{Q x}{\\varepsilon_{0} A} d Q & =-\\frac{Q^{2}}{4 \\varepsilon_{0} A} d x \\\\\n\\frac{1}{4} \\int \\frac{d x}{x} & =-\\int \\frac{d Q}{Q} \\\\\n\\frac{1}{4} \\ln x+\\ln Q & =\\text { const. }\n\\end{aligned}\n$$\n\nWe now find our adiabatic invariant to be\n\n$$\nx Q^{4}=\\text { const. }\n$$\n\nSubstituting values into our equation, we find that\n\n$$\nd Q_{i}^{4}=\\frac{d}{2} Q_{f}^{4} \\Longrightarrow Q_{f}=\\sqrt[4]{2} Q=11.892 \\mathrm{C}\n$$[^0]""]"	['$11.892$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	C	Numerical	2e-1	Open-ended	Electromagnetism	Physics	English
+21	A child attaches a small rock of mass $M=0.800 \mathrm{~kg}$ to one end of a uniform elastic string of mass $m=0.100 \mathrm{~kg}$ and natural length $L=0.650 \mathrm{~m}$. He grabs the other end and swings the rock in uniform circular motion around his hand, with angular velocity $\omega=6.30 \mathrm{rad} / \mathrm{s}$. Assume his hand is stationary, and that the elastic string behaves like a spring with spring constant $k=40.0 \mathrm{~N} / \mathrm{m}$. After that, at time $t=0$, a small longitudinal perturbation starts from the child's hand, traveling towards the rock. At time $t=T_{0}$, the perturbation reaches the rock. How far was the perturbation from the child's hand at time $t=\frac{T_{0}}{2}$ ? Ignore gravity.	"[""Let $x$ be the distance from a point on the unstretched elastic string to the center of rotation (child's hand). Note that $x$ varies from 0 to $L$. However, the string stretches, so let $r$ be the distance from a point on the stretched string (in steady state) to the center of rotation. Let $T$ be the tension in the string as a function of position. Let $\\lambda=\\frac{m}{L}$. Consider a portion of the string $d x$. We know that the portion as spring constant $k \\frac{L}{d x}$ and it is stretched by $d r-d x$, so by Hooke's Law, we have $T=k \\frac{L}{d x}(d r-d x)=k L\\left(\\frac{d r}{d x}-1\\right)$. Also, by applying Newton's Second Law on the portion, we get $d T=-\\lambda d x \\cdot \\omega^{2} r$, which implies $\\frac{d T}{d x}=-\\lambda \\omega^{2} r$. Combining the two equations, we obtain\n$$\nT=-k L\\left(\\frac{1}{\\lambda \\omega^{2}} T^{\\prime \\prime}+1\\right)\n$$\nWe know that\n$$\nT^{\\prime}(x=0)=0,\n$$\nsince $r=0$ when $x=0$. The general solution is\n$$\nT=A \\cos \\left(\\frac{\\omega}{L} \\sqrt{\\frac{m}{k}} x\\right)-k L\n$$\nfor some constant $A$. Thus, we have\n$$\nr=-\\frac{1}{\\lambda \\omega^{2}} T^{\\prime}=\\frac{A}{\\omega \\sqrt{k m}} \\sin \\left(\\frac{\\omega}{L} \\sqrt{\\frac{m}{k}} x\\right)\n$$\nAlso, we have that\n$$\nT(x=L)=M \\omega^{2} \\int_{0}^{L} r x d x=M \\omega^{2} \\int_{0}^{L}\\left(\\frac{T}{k L}+1\\right) x .\n$$\nPlugging in our general solution, we can get $A \\cos \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)-k L=M \\omega^{2} \\cdot \\frac{A}{k L} \\frac{L}{\\omega} \\sqrt{\\frac{k}{m}} \\sin \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)$. Solving for $A$, we obtain\n$$\nA=\\frac{k L}{\\cos \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)-\\frac{M \\omega}{\\sqrt{k m}} \\sin \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)}\n$$\nWe now introduce a claim:\n\n\n\nClaim. The speed of a longitudinal wave on a spring with spring constant $k$, length $L$, and mass $m$ is given by $v=L \\sqrt{\\frac{k}{m}}$\n\nProof. Let a spring with spring constant $k$ and mass $m$ be stretched to length $L$. The spring constant of a small portion $d x$ of the spring is $k \\frac{L}{d x}$, and the excess tension is $\\delta T=k \\frac{L}{d x} d s=k L \\frac{d s}{d x}$, where $s$ is the displacement from equilibrium. By Newton's second law on the portion, we get $d T=\\frac{m}{L} d x \\cdot \\frac{d^{2} s}{d t^{2}}$, or $\\frac{d T}{d x}=\\frac{m}{L} \\frac{d^{2} s}{d t^{2}}$. Thus, $L^{2} \\frac{k}{m} \\frac{d^{2} s}{d x^{2}}=\\frac{d^{2} s}{d t^{2}}$, which we recognize as the wave equation with speed $v=L \\sqrt{\\frac{k}{m}}$ and the time it takes to traverse the spring is $\\sqrt{\\frac{m}{k}}$.\n\nThus, we have\n$$\nt=\\int d t=\\int_{0}^{x} \\sqrt{\\frac{\\lambda d x}{k \\cdot \\frac{L}{d x}}}=\\int_{0}^{x} \\sqrt{\\frac{\\lambda}{k L}} d x=\\sqrt{\\frac{m}{k}} \\frac{x}{L}\n$$\nSince we know $x=L$ when $t=T_{0}$, we have $x=\\frac{L}{2}$ when $t=\\frac{T_{0}}{2}$. Therefore, our answer is\n$$\nr=\\frac{A}{\\omega \\sqrt{k m}} \\sin \\left(\\frac{\\omega}{L} \\sqrt{\\frac{m}{k}} x\\right)=\\frac{1}{\\omega} \\sqrt{\\frac{k}{m}} \\frac{L \\sin \\left(\\frac{1}{2} \\omega \\sqrt{\\frac{m}{k}}\\right)}{\\cos \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)-\\frac{M \\omega}{\\sqrt{k m}} \\sin \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)}=1.903 \\mathrm{~m}\n$$""]"	['1.903']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+22	What is the smallest number of $1 \Omega$ resistors needed such that when arranged in a certain arrangement involving only series and parallel connections, that the equivalent resistance is $\frac{7}{6} \Omega$ ?	['We can write:\n$$\n\\frac{7}{6}=\\frac{1}{2}+\\frac{2}{3}\n\\tag{2}\n$$\n\nIt takes two resistors (connected in parallel) to create a $\\frac{1}{2} \\Omega$ resistor. If we write $\\frac{2}{3}=\\frac{2 \\cdot 1}{2+1}$, then it takes three resistors to create a $\\frac{2}{3} \\Omega$ (this is accomplished by connecting a 2 resistor in parallel with a 1 resistor.\n\nCombining the $\\frac{1}{2} \\Omega$ element in series with the $\\frac{2}{3} \\Omega$ element gives us our desired amount. To prove this is the minimum, we can easily check all possible combinations using 4 or fewer resistors.']	['5']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	0	Open-ended	Electromagnetism	Physics	English
+23	A coaxial cable is cylindrically symmetric and consists of a solid inner cylinder of radius $a=2 \mathrm{~cm}$ and an outer cylindrical shell of inner radius $b=5 \mathrm{~cm}$ and outer radius $c=7 \mathrm{~cm}$. A uniformly distributed current of total magnitude $I=5 \mathrm{~A}$ is flowing in the inner cylinder and a uniformly distributed current of the same magnitude but opposite direction flows in the outer shell. Find the magnitude $B(r)$ of the magnetic field $B$ as a function of distance $r$ from the axis of the cable. As the final result, submit $\int_{0}^{\infty} B(r) \mathrm{d} r$. In case this is infinite, submit 42 .	"[""Ampere's law $\\int B \\cdot \\mathrm{d} l=\\mu_{0} I$ is all we need. For every point on the wire, we can write the magnetic field as a function of the distance from its center $r$. Thus,\n$$\nB(r)= \\begin{cases}\\frac{5 \\mu_{0} r}{8 \\pi} & r \\leq 2 \\\\ \\frac{5 \\mu_{0}}{2 \\pi r} & 2<r<5 \\\\ \\frac{5 \\mu_{0}\\left(-r^{2}+49\\right)}{48 \\pi r} & 5 \\leq r \\leq 7 \\\\ 0 & r>7\\end{cases}\n$$\n\nNow we just sum each integral from each interval, or in othe words\n\n$$\n\\int_{0}^{\\infty} B(r) \\mathrm{d} r=\\int_{0}^{2} B(r) \\mathrm{d} r+\\int_{2}^{5} B(r) \\mathrm{d} r+\\int_{5}^{7} B(r) \\mathrm{d} r\n$$\n\nThis is now straightforward integration.""]"	['$1.6 \\times 10^{-8}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-9	Open-ended	Electromagnetism	Physics	English
+24	"A train of length $100 \mathrm{~m}$ and mass $10^{5} \mathrm{~kg}$ is travelling at $20 \mathrm{~m} / \mathrm{s}$ along a straight track. The driver engages the brakes and the train starts deccelerating at a constant rate, coming to a stop after travelling a distance $d=2000 \mathrm{~m}$. As the train decelerates, energy released as heat from the brakes goes into the tracks, which have a linear heat capacity of $5000 \mathrm{~J} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$. Assume the rate of heat generation and transfer is uniform across the length of the train at any given moment.
+If the tracks start at an ambient temperature of $20^{\circ} \mathrm{C}$, there is a function $T(x)$ that describes the temperature (in Celsius) of the tracks at each point $x$, where the rear of where the train starts is at $x=0$. Assume (unrealistically) that $100 \%$ of the original kinetic energy of the train is transferred to the tracks (the train does not absorb any energy), that there is no conduction of heat along the tracks, and that heat transfer between the tracks and the surroundings is negligible.
+
+Compute $T(20)+T(500)+T(2021)$ in degrees celsius."	"[""Consider a small element of the tracks at position $x$ with width $\\mathrm{d} x$. Since the rate of heat generation is uniform along the length of the train $L$, we have that the rate of heat given to the track element is mav $\\frac{\\mathrm{d} x}{L}$, where $m$ is the train's mass, $a$ is the train's deceleration, and $v$ is the trains speed. Integrating over time gives the total heat given to the track element: $\\mathrm{d} Q=m a \\frac{\\mathrm{d} x}{L} \\Delta x$, where $\\Delta x$ is the total distance the train slips on the track element. Combining with $\\mathrm{d} Q=c \\mathrm{~d} x \\cdot \\Delta T$, we get $T(x)=T_{0}+\\frac{m a}{c L} \\Delta x$, where $c$ is the linear heat capacity. Now we split into 3 cases:\n- $0<x<L$,\n- $L<x<d$,\n- $d<x<d+L$\n\nwhere $d$ is the total distance. If $0<x<L$, the train slips a distance $\\Delta x=x$ on the track element, so we have $T(x)=T_{0}+\\frac{m a}{c L} x$ (linear). If $L<x<d$, the train slips a distance $\\Delta x=L$ on the track element, so we have $T(x)=T_{0}+\\frac{m a}{c}$ (constant). If $d<x<d+L$, the train slips a distance $\\Delta x=d+L-x$, so we have $T(x)=T 0+\\frac{m a}{c L}(d+L-x)$ (linear). Note that $a=\\frac{v_{0}^{2}}{2 d}$ where $v_{0}$ is the train's initial velocity.\n\n$63.98^{\\circ} \\mathrm{C}$""]"	['$63.98$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	$^{\circ} \mathrm{C}$	Numerical	1e-1	Open-ended	Thermodynamics	Physics	English
+25	"Consider a gas of mysterious particles called nieons that all travel at the same speed, $v$. They are enclosed in a cubical box, and there are $\rho$ nieons per unit volume. A very small hole of area $A$ is punched in the side of the box. The number of nieons that escape the box per unit time is given by
+$$
+\alpha v^{\beta} A^{\gamma} \rho^{\delta}
+\tag{10}
+$$
+
+where $\alpha, \beta, \gamma$, and $\delta$ are all dimensionless constants. Calculate $\alpha+\beta+\gamma+\delta$."	"['The main idea is this: if a nieon is to escape in time $\\Delta t$, then it must be traveling towards the hole and be within a hemisphere of radius $v \\Delta t$, centered on the hole.\n(Note that we will ignore all collisions between nieons. As with many calculations in these kinds of problems, collisions will not affect our answer. Even if they did, we can assume that the hole is so small that we can take $v \\Delta t \\rightarrow 0$, making the radius of the hemisphere much much smaller than the mean free path.)\n\nFirst, we will break up this hemisphere into volume elements. We will calculate how many nieons are inside each volume element. Then we will find out how many of these nieons are traveling in the direction of the hole.\n\nWe define a spherical coordinate system $(r, \\theta, \\phi)$. The volume of each differential element is $r^{2} \\sin \\theta d r d \\theta d \\phi$, which means that there are $\\rho r^{2} \\sin \\theta d r d \\theta d \\phi$ nieons inside this volume element.\n\nWithin this volume element, how many of these nieons are going towards the hole? We must think about what each nieon ""sees."" Visualize a sphere of radius $r$ around the volume element. The surface of the sphere will pass through the small hole in the wall. Each nieon is equally likely to go in any direction and is equally likely to end up going through any patch of this spherical surface. Thus, each nieon within this volume element has a probability $A^{\\prime} / 4 \\pi r^{2}$ of going through the hole, where $A^{\\prime}$ is the perceived area of the hole from where the nieon is situated.\n\n$A^{\\prime}$ can be calculated as follows. $\\hat{\\mathbf{j}}$ is the unit vector pointing perpendicular to the hole. Let $\\hat{\\mathbf{r}}$ be the\n\n\n\nunit vector pointing from the hole to the volume element. Then it can be verified that $A^{\\prime}=A \\cos \\psi$, where $\\psi$ is the angle between $\\hat{\\mathbf{j}}$ and $\\hat{\\mathbf{r}}$.\n\nExpressing $\\hat{\\mathbf{r}}$ in terms of the other unit vectors, it can be shown that $A^{\\prime}=A \\sin \\theta \\sin \\phi$. Hence, the fraction of nieons inside the volume element that are going towards the hole is\n\n$$\n\\frac{A \\sin \\theta \\sin \\phi}{4 \\pi r^{2}}\n$$\n\nIntegrating over the entire hemisphere, the number of nieons escaping through the hole in a time $\\Delta t$ is\n\n$$\n\\int_{0}^{v \\Delta t} d r \\int_{0}^{\\pi} d \\phi \\int_{0}^{\\pi} d \\theta\\left(\\frac{A}{4 \\pi r^{2}} \\sin \\theta \\sin \\phi\\right)\\left(\\rho r^{2} \\sin \\theta\\right)=\\frac{1}{4} v A \\rho \\cdot \\Delta t\n$$']"	['3.25']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-1	Open-ended	Modern Physics	Physics	English
+26	Now, Poncho has encountered a different Pico-Pico game that uses the same shaped frictionless track, but lays it horizontally on a table with friction and coefficient of friction $\mu=0.8$. In addition, the ball, which can once again be considered a point mass, is placed on the other side of the track as the ball in part 1. Finally, a buzzer on the other side of the track requires the mass to hit with at least velocity $v_{f}=2 \mathrm{~m} / \mathrm{s}$ in order to trigger the buzzer and win the game. Find the minimum velocity $v_{0}$ required for the ball to reach the end of the track with a velocity of at least $v_{f}$. The initial velocity must be directed along the track.	['We simply need to find the work done by friction and we can finish with conservation of energy. To get the work done by friction, since the force is constant, we just need to find the arc length of the track. That is\n$$\n\\ell=\\int_{-\\sqrt{5 / 2}}^{\\sqrt{5 / 2}} \\sqrt{1+h^{\\prime}(x)^{2}} \\mathrm{~d} x=\\int_{-\\sqrt{5 / 2}}^{\\sqrt{5 / 2}} \\sqrt{1+16 x^{2}} \\mathrm{~d} x=10.76 \\mathrm{~m}\n$$\n\nThen by conservation of energy,\n\n$$\n\\frac{1}{2} v_{0}^{2}-\\mu g \\ell=\\frac{1}{2} v_{f}^{2} \\Longrightarrow v_{0}=\\sqrt{v_{f}^{2}+2 \\mu g \\ell}\n$$\n\nWe simply plug in the numbers to finish.']	['$13.1$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m/s	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+27	Let $A B C$ be a solid right triangle $(A B=5 s, A C=12 s$, and $B C=13 s$ ) with uniform charge density $\sigma$. Let $D$ be the midpoint of $B C$. We denote the electric potential of a point $P$ by $\phi(P)$. The electric potential at infinity is 0 . If $\phi(B)+\phi(C)+\phi(D)=\frac{k \sigma s}{\epsilon_{0}}$ where $k$ is a dimensionless constant, determine $k$.	['If we put two of these right triangles together, we can form a rectangle with side lengths $5 s$ and $12 s$. Let $V$ be the potential at the center of this rectangle. By superposition, $\\phi(D)=\\frac{V}{2}$. Consider the potential at the corner. It can be decomposed into the potential from each of the right triangles, which is precisely $\\phi(B)+\\phi(C)$. Now, also note that if we put 4 of these rectangles side by side, to make a larger rectangle with side lengths $10 s$ and $24 s$, the potential is scaled by a factor of 2 due to dimensional analysis arguments. Thus, the potential at the center of this larger rectangle is $2 \\mathrm{~V}$, but this can be decomposed into the sum of the potentials at the\n\n\ncorners of the 4 smaller rectangles. Thus, the potential in the corner of the smaller rectangle is $\\frac{V}{2}=\\phi(B)+\\phi(C)$. Thus, we obtain $\\phi(B)+\\phi(C)+\\phi(D)=V$.\n\nNow, we will find the potential at the center of the rectangle. Note that it suffices to find the potential at the vertex of an isosceles triangle because we can connect the center of the rectangle to the 4 corners to create 4 isosceles triangles. Suppose we have an isosceles triangle with base $2 x$ and height $y$. The potential at the vertex is\n\n$$\n\\iint \\frac{\\sigma}{4 \\pi \\epsilon_{0} r}(r d r d \\theta)=\\frac{\\sigma}{4 \\pi \\epsilon_{0}} \\int_{-\\tan ^{-1}\\left(\\frac{x}{y}\\right)}^{\\tan ^{-1}\\left(\\frac{x}{y}\\right)} \\int_{0}^{\\frac{y}{\\cos \\theta}} d r d \\theta=\\frac{\\sigma y}{2 \\pi \\epsilon_{0}} \\log \\left(\\frac{x+\\sqrt{x^{2}+y^{2}}}{y}\\right)\n$$\n\nIf the sides of the rectangle are $a$ and $b$, we then obtain the potential at the center is\n\n$$\nV=\\frac{\\sigma}{2 \\pi \\epsilon_{0}}\\left(a \\log \\left(\\frac{b+\\sqrt{a^{2}+b^{2}}}{a}\\right)+b \\log \\left(\\frac{a+\\sqrt{a^{2}+b^{2}}}{b}\\right)\\right)\n$$\n\nIn this case,\n\n$$\nk=\\frac{5}{2 \\pi} \\ln 5+\\frac{6}{\\pi} \\ln \\frac{3}{2}\n$$']	['2.055']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	6e-2	Open-ended	Electromagnetism	Physics	English
+28	A straight ladder $A B$ of mass $m=1 \mathrm{~kg}$ is positioned almost vertically such that point $B$ is in contact with the ground with a coefficient of friction $\mu=0.15$. It is given an infinitesimal kick at the point $A$ so that the ladder begins rotating about point $B$. Find the value $\phi_{m}$ of angle $\phi$ of the ladder with the vertical at which the lower end $B$ starts slipping on the ground.	['By conservation of energy, we have $\\frac{1}{2} I \\omega_{m}^{2}=m g \\frac{L}{2}\\left(1-\\cos \\phi_{m}\\right)$ where $I=\\frac{1}{3} m L^{2}$. Thus,\n$$\n\\omega_{m}=\\sqrt{\\frac{3 g\\left(1-\\cos \\phi_{m}\\right)}{L}}\n$$\n\nAlso, by torque analysis about B, we have $\\tau=m g \\frac{L}{2} \\sin \\phi_{m}=I \\alpha_{m}$ which means\n\n$$\n\\alpha_{m}=\\frac{3 g}{2 L} \\sin \\phi_{m}\n$$\n\nThus, the centripetal and tangential accelerations of the ladder are $a_{c}=\\omega_{m}^{2} \\frac{L}{2}=\\frac{3}{2} g\\left(1-\\cos \\phi_{m}\\right)$ and $a_{t}=\\alpha_{m} \\frac{L}{2}=\\frac{3}{4} g \\sin \\phi_{m}$ respectively. The normal force is thus $N=m g-m a_{c} \\cos \\phi-m a_{t} \\sin \\phi$, so\n\n$$\n\\frac{N}{m g}=1-\\frac{3}{2} \\cos \\phi_{m}\\left(1-\\cos \\phi_{m}\\right)-\\frac{3}{4} \\sin ^{2} \\phi_{m}\n$$\n\nThe frictional force is thus $f=m a_{t} \\cos \\phi_{m}-m a_{c} \\sin \\phi_{m}$ so\n\n$$\n\\frac{f}{m g}=\\frac{3}{4} \\sin \\phi_{m} \\cos \\phi_{m}-\\frac{3}{2} \\sin \\phi_{m}\\left(1-\\cos \\phi_{m}\\right)\n$$\n\nSetting $\\frac{f}{N}=\\mu$, we have $6 \\sin \\phi_{m}\\left(1-\\cos \\phi_{m}\\right)-3 \\sin \\phi_{m} \\cos \\phi_{m}=-4 \\mu+6 \\mu \\cos \\phi_{m}\\left(1-\\cos \\phi_{m}\\right)+$ $3 \\mu \\sin ^{2} \\phi_{m}$. Simplifying\n\n$$\n6 \\sin \\phi_{m}-9 \\sin \\phi_{m} \\cos \\phi_{m}+9 \\mu \\cos ^{2} \\phi_{m}-6 \\mu \\cos \\phi_{m}=-\\mu\n$$\n\nWe then can solve for $\\phi_{m}$ numerically.']	['$11.5$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	$^{\circ}$	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+29	Two Ladders Two straight ladders $A B$ and $C D$, each with length $1 \mathrm{~m}$, are symmetrically placed on smooth ground, leaning on each other, such that they are touching with their ends $B$ and $C$, ends $A$ and $D$ are touching the floor. The friction at any two surfaces is negligible. Initially both ladders are almost parallel and vertical. Find the distance $A D$ when the points $B$ and $C$ lose contact.	['The center of mass of both of the ladders moves in a circle, centered at the point on the ground directly beneath $B / C$. So we find when the required normal force between the two ladders is 0 . That is, when the total net force on one of the ladders is when the two ladders lose contact. Let $2 r=\\ell$. Now by conservation of energy,\n$$\n\\frac{1}{2} m v^{2}+\\frac{1}{2} \\frac{m r^{2}}{3} \\frac{v^{2}}{r^{2}}=m g r(1-\\cos \\theta)\n$$\n\nwhere $\\theta$ is defined as the angle the ladder makes witht he vertical. So we have\n\n$$\nv^{2}\\left(1+\\frac{1}{3}\\right)=2 g r(1-\\cos \\theta) \\Longrightarrow v^{2}=\\frac{3}{2} g r(1-\\cos \\theta)\n$$\n\nSo the centripetal acceleration is\n\n$$\na_{c}=\\frac{v^{2}}{r}=\\frac{3}{2} g(1-\\cos \\theta)\n$$\n\n\n\nAnd the tangential acceleration is\n\n$$\n\\frac{d v}{d t}=\\sqrt{\\frac{3 g r}{2}} \\frac{\\sin \\theta}{2 \\sqrt{1-\\cos \\theta}} \\frac{d \\theta}{d t}\n$$\n\nAnd $\\frac{d \\theta}{d t}=\\frac{v}{r}$, so\n\n$$\na_{\\theta}=\\frac{d v}{d t}=\\frac{3 g}{2} \\frac{\\sin \\theta}{2}\n$$\n\nNow for the total acceleration to be vertical, we need\n\n$$\na_{c} \\tan \\theta=a_{\\theta} \\text {, }\n$$\n\nso\n\n$$\n1-\\cos \\theta=\\frac{\\cos \\theta}{2}\n$$\n\nThis simplifies to\n\n$$\n2=3 \\cos \\theta\n$$\n\nSo the distance between the two ends is $4 r \\sin \\theta=4 r \\sqrt{1-\\frac{4}{9}}=\\frac{4 r \\sqrt{5}}{3}$.']	['$\\frac{2\\sqrt{5}}{3}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+30	An evil gamma photon of energy $E_{\gamma 1}=200 \mathrm{keV}$ is heading towards a spaceship. The commander's only choice is shooting another photon in the direction of the gamma photon such that they 'collide' head on and produce an electron-positron pair (both have mass $m_{e}$ ). Find the lower bound on the energy $E_{\gamma 2}$ of the photon as imposed by the principles of special relativity such that this occurs. Answer in keV.	['The key claim is that energy is minimized when both particles are moving at the same velocity after the collision. This can be proved by transforming into the frame where the total momentum is 0 .\nThis idea is sufficient because it implies both the electron and positron have the same momentum and energy after the collision. Let them both have momentum $p$. We then have\n\n$$\n2 p=\\frac{E_{\\gamma_{1}}-E_{\\gamma_{2}}}{c} \\Longrightarrow p c=\\frac{E_{\\gamma_{1}}-E_{\\gamma_{2}}}{2}\n\\tag{21}\n$$\n\nBy energy conservation, both the electron and positron have energy $\\frac{E_{\\gamma_{1}}+E_{\\gamma_{2}}}{2}$. Using the result $E=(p c)^{2}+\\left(m c^{2}\\right)^{2}$, we obtain,\n\n$$\n\\left(\\frac{E_{\\gamma_{1}}+E_{\\gamma_{2}}}{2}\\right)^{2}=(p c)^{2}+\\left(m_{e} c^{2}\\right)^{2}\n\\tag{22}\n$$\n\nCombining equation 21 and equation 22 , we get\n\n$$\n\\left(\\frac{E_{\\gamma_{1}}+E_{\\gamma_{2}}}{2}\\right)^{2}=\\left(\\frac{E_{\\gamma_{1}}-E_{\\gamma_{2}}}{2}\\right)^{2}+\\left(m_{e} c^{2}\\right)^{2} \\Longrightarrow E_{\\gamma_{2}}=\\frac{m_{e}^{2} c^{4}}{E_{\\gamma_{1}}}\n$$']	['1306']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	keV	Numerical	1e0	Open-ended	Modern Physics	Physics	English
+31	Adithya is launching a package from New York City $\left(40^{\circ} 43^{\prime} \mathrm{N}\right.$ and $\left.73^{\circ} 56^{\prime} \mathrm{W}\right)$ to Guam $\left(13^{\circ} 27^{\prime} \mathrm{N}\right.$ and $\left.144^{\circ} 48^{\prime} \mathrm{E}\right)$. Find the minimal launch velocity $v_{0}$ from New York City to Guam. Ignore the rotation of the earth, effects due to the atmosphere, and the gravitational force from the sun. Additionally, assume the Earth is a perfect sphere with radius $R_{\oplus}=6.37 \times 10^{6} \mathrm{~m}$ and mass $M_{\oplus}=5.97 \times 10^{24} \mathrm{~kg}$.	['We first want to find the angular distance between New York City and Guam. Let this be $\\theta$. Let New York City be point $A$ and Guam be point $B$. Consider the north pole $P$ and the spherical triangle $P A B$. By the spherical law of cosines,\n$$\n\\cos \\theta=\\cos \\left(90^{\\circ}-\\phi_{A}\\right) \\cos \\left(90^{\\circ}-\\phi_{B}\\right)+\\sin \\left(90^{\\circ}-\\phi_{A}\\right) \\sin \\left(90^{\\circ}-\\phi_{B}\\right) \\cos \\left(\\ell_{B}-\\ell_{A}\\right) .\n\\tag{23}\n$$\n\n\n\nEquation 23 simplifies to\n\n$$\n\\cos \\theta=\\sin \\phi_{A} \\sin \\phi_{B}+\\cos \\phi_{A} \\cos \\phi_{B} \\cos \\left(\\ell_{B}-\\ell_{A}\\right)\n\\tag{24}\n$$\n\nfrom which we find $\\theta=115.05^{\\circ}$.\n\nNow that we have determined the angular distance, we will proceed with the orbital mechanics problem. By the vis-viva equation, the speed at the launch point is\n\n$$\nv_{0}=\\sqrt{G M_{\\oplus}\\left(\\frac{2}{R_{\\oplus}}-\\frac{1}{a}\\right)}\n\\tag{25}\n$$\n\nwhere $a$ is the semimajor axis for the orbit. It is clear that in order to minimize $v_{0}$, we must minimize $a$. The orbit is an ellipse with focii $F_{1}$ and $F_{2}$, where $F_{1}$ is the center of the earth. By the definition of an ellipse,\n\n$$\nA F_{1}+A F_{2}=2 a\n\\tag{26}\n$$\n\nSince $A F_{1}=R_{\\oplus}$, is suffices to minimize $A F_{2}$. By symmetry, line $F_{1} F_{2}$ is the perpendicular bisector of $A B$. Since $F_{2}$ is on a fixed line, to minimize $F_{2}$, we place it at the foot of the perpendicular from $A$ to this line. Then, we obtain $A F_{2}=R_{\\oplus} \\sin \\left(\\frac{\\theta}{2}\\right)$. Thus,\n\n$$\na=\\frac{1+\\sin \\left(\\frac{\\theta}{2}\\right)}{2} R_{\\oplus}\n$$\n\nand from Equation 25 we find\n\n$$\nv_{0}=\\sqrt{\\frac{G M_{\\oplus}}{R_{\\oplus}} \\frac{2 \\sin \\frac{\\theta}{2}}{1+\\sin \\frac{\\theta}{2}}} .\n$$']	['$7564$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m/s	Numerical	1e2	Open-ended	Mechanics	Physics	English
+32	Consider a container filled with argon, with molar mass $39.9 \mathrm{~g} \mathrm{~mol}^{-1}$ whose pressure is much smaller than that of atmospheric pressure. Suppose there is a plate of area $A=10 \mathrm{~mm}^{2}$ moving with a speed $v$ perpendicular to its plane. If the gas has density $\rho=4.8 \times 10^{-7} \mathrm{~g} \mathrm{~cm}^{-3}$, and temperature $T=100 \mathrm{~K}$, find an approximate value for the drag force acting on the plate. Suppose that the speed of the plate is $v=100 \mathrm{~m} \mathrm{~s}^{-1}$.	['Let $N$ be the number of particles colliding with the plate in a unit time, and let all the molecules arrive at the plate with the same speed $u$. The force on the plate will be the momentum imparted over a unit time, i.e. $F=2 N m(v-u) / t$ where the coefficient off two reflects on the fact that the average speed of the molecules hitting the wall is the same as the molecules departing. Note that Maxwells distribution dictates that the average speed of particles in all direction $\\langle u\\rangle=0$ which means that the average force acting on the plate is simply $\\langle F\\rangle=2 N m v / t$. During a time period $t$, these molecules arrive a wall that is of thickness $v_{R} t$ where $v_{R}$ is the relative velocity between the molecules and the plate (which is not negligible as the mean free path and velocity of the plate are comparable.) This means the number of collisions found in this layer will be $N=\\frac{1}{2} n V \\approx A v_{R} t$ where the factor of $1 / 2$ reflects on how half the molecules go to the plate and the other half go the other direction. Thus the force acting on the plate will become $\\left\\langle F_{D}\\right\\rangle=2 m v N / t=2 n m v \\cdot v_{R} A$. To find the relative velocity, consider the vector diagram:\n\n\n<img_4277>\n\nBy law of cosines, the magnitude of $v_{R}$ will simply be\n\n$$\nv_{R}^{2}=v^{2}+u^{2}-2 \\boldsymbol{v} \\cdot \\boldsymbol{u}\n$$\n\nThe direction of $\\boldsymbol{u}$ points homogeneously in all directions as the orientation of molecules changes with each individual one. Therefore, $2 \\boldsymbol{v} \\cdot \\boldsymbol{u}$ will point in all directions averaging to 0 . Thus, the magnitude of $v_{R}$ will simply be $\\left\\langle v_{R}\\right\\rangle=\\sqrt{v^{2}+\\langle u\\rangle^{2}}$ where $\\langle u\\rangle^{2}$ is the average thermal velocity of molecules. In terms of density, we can then express the drag force to be $2 \\rho A v \\sqrt{v^{2}+\\langle u\\rangle^{2}}$. $2.41 \\times 10^{-4} \\mathrm{~N}$.']	['$2.41\\times 10^{-4}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	N	Numerical	1e-5	Open-ended	Mechanics	Physics	English
+33	"Consider a $1 \mathrm{~cm}$ long slit with negligible height. First, we divide the slit into thirds and cover the middle third. Then, we perform the same steps on the two shorter slits. Again, we perform the same steps on the four even shorter slits and continue for a very long time.
+Then, we shine a monochromatic, coherent light source of wavelength $500 \mathrm{~nm}$ on our slits, which creates an interference pattern on a wall 10 meters away. On the wall, what is the distance between the central maximum and the first side maximum? Assume the distance to the wall is much greater than the width of the slit. Answer in millimeters."	['This problem is essentially an infinite convolution. In detail, consider two separate amplitude functions $f(x)$ and $g(x)$ that correspond to two different interference patterns. One can think of a convolution of $f$ and $g$ as taking $f$ and placing the same function in every position that exists in $g$. To be more specific, the amplitude of one interference pattern can be thought of as the product of the amplitude patterns of two separate amplitude functions. For example, the diffraction pattern of two slits is the same as the product of the amplitudes for one slit and two light sources (this will be given as an exercise to prove). To make more sense of this, let us consider that very example and let $f$ pertain to the amplitude function of a single slit and let $g$ pertain to the amplitude function of two light sources. One can then write in phase space with a generalized coordinate $x^{\\prime}$ that $\\{f * g\\}(x)=\\int_{-\\infty}^{\\infty} f\\left(x-x^{\\prime}\\right) g\\left(x^{\\prime}\\right) \\mathrm{d} x^{\\prime}$ according to the convolution theorem. The integral then runs through all values and places a copy of $f\\left(x-x^{\\prime}\\right)$ at the peaks of $g\\left(x^{\\prime}\\right)$.\nWith this in hand, we can use the convolution theorem to our advantage. Let us designate the amplitude function of the entire Cantor slit to be $F(\\theta)$. Since $\\theta$ is small, we can designate the phase angle as $\\phi=k x \\theta$ where $x$ is a variable moving through all of the slits. If we consider a single slit that is cut into a third, the angle produced will be a third of its original as well, or the new amplitude function will simply be a function of $\\theta / 3$ or $F(\\theta / 3)$. The distance between the midpoints of the two slits will simply be $d=2 / 3 \\mathrm{~cm}$ which then becomes $2 / 9$, then $2 / 27$ and so on. In terms of the original function, we can decompose it into the function of $F(\\theta / 3)$ and the amplitude function of a single light source. In other words, $F(\\theta)=F(\\theta / 3) \\cos (k d \\theta / 2)$ where $k=\\frac{2 \\pi}{\\lambda}$. To the limit of infinity, this can be otherwise written as\n\n$$\nF(\\theta)=\\prod_{i=0}^{\\infty} \\cos \\left(\\frac{k d \\theta}{2 \\cdot 3^{i}}\\right)=\\cos \\left(\\frac{k d \\theta}{2}\\right) \\cos \\left(\\frac{k d \\theta}{6}\\right) \\cos \\left(\\frac{k d \\theta}{18}\\right) \\ldots\n$$\n\nFinding an exact mathematical solution would be difficult, thus it is simple enough to just multiply the first 4 or 5 terms to achieve the final answer.']	['$0.647$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	mm	Numerical	1e-2	Open-ended	Mechanics	Physics	English
+34	"A certain planet with radius $R=$ $3 \times 10^{4} \mathrm{~km}$ is made of a liquid with constant density $\rho=1.5 \mathrm{~g} / \mathrm{cm}^{3}$ with the exception of a homogeneous solid core of radius $r=10 \mathrm{~km}$ and mass $m=2.4 \times 10^{16} \mathrm{~kg}$. Normally, the core is situated at the geometric center of the planet. However, a small disturbance has moved the center of the core $x=1 \mathrm{~km}$ away from the geometric center of the planet. The core is released from rest, and the fluid is inviscid and incompressible.
+
+Calculate the magnitude of the force due to gravity that now acts on the core. Work under the assumption that $R \gg r$."	['We solve by simplifying the configuration into progressively simpler but mathematically equivalent formulations of the same problem.\nBecause the core is spherical and homogeneous, all the gravitational forces on the core are equivalent to the gravitational forces on a point mass located at the center of the core. Thus, the problem is now the force of gravity on a point mass of mass $m$ located $x$ distance away from the center of planet, with a sphere of radius $r$ evacuated around the point mass.\n\nBut if we filled this evacuated sphere, would the force on the point mass be any different? No! If we filled up the sphere, all the added liquid (of the same density as the rest of the planet) would add no additional force on the point mass because it is symmetrically distributed around the point mass. Thus, the problem is now the force of gravity on a point mass of mass $m$ located $x$ distance away from the center of the planet.\n\nBy shell theorem, we can ignore all the fluid that is more than a distance $x$ away from the center of the planet. Thus, the problem is now the force of gravity on a point mass $m$ situated on the surface of a liquid planet of radius $x$. This force is not difficult to calculate at all:\n\n$$\n\\begin{aligned}\nF & =\\frac{G M m}{x^{2}} \\\\\n& =\\frac{G m}{x^{2}}\\left(\\frac{4}{3} \\pi x^{3} \\rho\\right) \\\\\n& =\\frac{4}{3} \\pi G m \\rho x\n\\end{aligned}\n$$\n\n$1.0058 \\times 10^{13} \\mathrm{~N}$']	['$1.0058 \\times 10^{13}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	N	Numerical	1e11	Open-ended	Mechanics	Physics	English
+35	"A scientist is doing an experiment with a setup consisting of 2 ideal solenoids that share the same axis. The lengths of the solenoids are both $\ell$, the radii of the solenoids are $r$ and $2 r$, and the smaller solenoid is completely inside the larger one. Suppose that the solenoids share the same (constant) current $I$, but the inner solenoid has $4 N$ loops while the outer one has $N$, and they have opposite polarities (meaning the current is clockwise in one solenoid but counterclockwise in the other).
+Model the Earth's magnetic field as one produced by a magnetic dipole centered in the Earth's core.
+
+
+
+Let $F$ be the magnitude of the total magnetic force the whole setup feels due to Earth's magnetic field. Now the scientist replaces the setup with a similar one: the only differences are that the the radii of the solenoids are $2 r$ (inner) and $3 r$ (outer), the length of each solenoid is $7 \ell$, and the number of loops each solenoid is $27 N$ (inner) and $12 N$ (outer). The scientist now drives a constant current $2 I$ through the setup (the solenoids still have opposite polarities), and the whole setup feels a total force of magnitude $F^{\prime}$ due to the Earth's magnetic field. Assuming the new setup was in the same location on Earth and had the same orientation as the old one, find $F^{\prime} / F$.
+
+Assume the dimensions of the solenoids are much smaller than the radius of the Earth."	"[""We can solve the problem by assuming that the location of the setup is at the North Pole and that the solenoids are oriented so that their axis intersects the Earth's core. Note that if we had some other location or orientation, then both $F$ and $F^{\\prime}$ would be multiplied by the same factor, so their ratio remains the same.\nSuppose the radii of the solenoids are $r$ and $\\alpha r$, where the number of inner and outer loops are $N$ and $\\frac{N}{\\alpha^{2}}$, respectively. To find the force the Earth's dipole exerts on the solenoids, we can calculate the force the solenoids exert on the dipole. To do this, we need to find the gradient of the magnetic field produced by the solenoids at the dipole's location. Let the radius of the Earth be $R$.\n\nConsider the field produced by 2 concentric, coaxial, ring currents, the inner ring with current $I$ radius $r$ and the outer one with current $\\frac{I}{\\alpha^{2}}$ and radius $\\alpha r$. The currents are in opposite directions. At a distance $R$ away from the center of the rings, along their axis, the magnetic field is given by\n\n$$\n\\begin{aligned}\nB & =\\frac{\\mu_{0} I r^{2}}{2\\left(R^{2}+r^{2}\\right)^{\\frac{3}{2}}}-\\frac{\\mu_{0} I r^{2}}{2\\left(R^{2}+(\\alpha r)^{2}\\right)^{\\frac{3}{2}}} \\\\\n& =\\frac{\\mu_{0} I^{2}}{2 R^{3}}\\left(\\left(1+\\frac{r^{2}}{R^{2}}\\right)^{-\\frac{3}{2}}-\\left(1+\\frac{\\alpha^{2} r^{2}}{R^{2}}\\right)^{-\\frac{3}{2}}\\right) \\\\\n& \\approx \\frac{\\mu_{0} I^{2}}{2 R^{3}}\\left(\\frac{3}{2}\\left(\\alpha^{2}-1\\right) \\frac{r^{2}}{R^{2}}\\right) \\\\\n& =\\frac{3 \\mu_{0} I r^{4}}{4 R^{5}}\\left(\\alpha^{2}-1\\right)\n\\end{aligned}\n$$\n\nThus, the gradient of the magnetic field is proportional to $\\operatorname{Ir}^{4}\\left(\\alpha^{2}-1\\right)$. Now we consider the actual setup. The new setup multiplies the effective current $\\frac{27}{4} \\cdot \\frac{2}{1}=\\frac{27}{2}$ times, while multiplying $r$ by 2 . The factor $\\alpha^{2}-1$ changed from 3 to $\\frac{5}{4}$. Combining, we get $\\frac{F^{\\prime}}{F}=\\frac{27}{2} \\cdot 2^{4} \\cdot \\frac{5}{12}=90$.""]"	['90']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-1	Open-ended	Mechanics	Physics	English
+36	"Adithya is in a rocket with proper acceleration $a_{0}=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}^{2}$ to the right, and Eddie is in a rocket with proper acceleration $\frac{a_{0}}{2}$ to the left. Let the frame of Adithya's rocket be $S_{1}$, and the frame of Eddie's rocket be $S_{2}$. Initially, both rockets are at rest with respect to each other, they are at the same location, and Adithya's clock and Eddie's clock are both set to 0 .
+
+
+At the moment Adithya's clock reaches $0.75 \mathrm{~s}$ in $S_{2}$, what is the velocity of Adithya's rocket in $S_{2}$ ?"	"[""Throughout this solution, we will let $c=a_{0}=1$, for simplicity. We will work in the inertial frame that is initially at rest with both rockets at $t_{1}=t_{2}=0$. First, we determine the velocity of Adithya's rocket in this frame as a function of the proper time that has elapsed $t_{1}$. In frame. $S_{1}$, in a time $d t_{1}$, the rocket's velocity increases by $d t_{1}$, so by velocity addition, the new velocity in the inertial frame is\n$$\n\\frac{v_{1}+d t_{1}}{1+v_{1} d t_{1}} \\approx v_{1}+\\left(1-v_{1}^{2}\\right) d t_{1}\n$$\n\nTherefore, we have\n\n$$\n\\frac{d v_{1}}{d t_{1}}=1-v_{1}^{2}\n$$\n\nUpon separating and integrating, we find $v_{1}=\\tanh \\left(t_{1}\\right)$. Similarly, the velocity of Eddie's rocket in the inertial frame is $v_{2}=\\tanh \\left(t_{2} / 2\\right)$. Now, in this inertial frame, let the time between events $A$ and $B$ be $t$, and let the distance between rockets $A$ and $B$ be $x$. By a Lorentz transformation, the time between the events in frame $S_{2}$ is\n\n$$\nt^{\\prime}=\\gamma\\left(t-v_{2} x\\right)=0\n$$\n\nsince the events are simultaneous in $S_{2}$. Therefore, we must have $t=v_{2} x$. Note that $t=t_{A}-t_{B}$ where $t_{A}$ and $t_{B}$ denote the times of events $A$ and $B$ in the inertial frame, respectively. Also, note that $x=x_{A}+x_{B}$ where $x_{A}$ and $x_{B}$ are the respective displacements of the two rockets in the inertial frame. By the effects of time dilation, we have\n\n$$\nt_{A}=\\int \\gamma d t_{1}=\\int \\frac{1}{\\sqrt{1-v_{1}^{2}}} d t_{1}=\\int_{0}^{t_{1}} \\cosh \\left(t_{1}^{\\prime}\\right) d t_{1}^{\\prime}=\\sinh \\left(t_{1}\\right)\n$$\n\nSimilarly, $t_{B}=2 \\sinh \\left(t_{2} / 2\\right)$, and we obtain $t=2 \\sinh \\left(t_{2} / 2\\right)-\\sinh \\left(t_{1}\\right)$. Additionally, from the above result,\n\n$$\nx_{A}=\\int v_{1} d t=\\int v_{1} \\cosh \\left(t_{1}\\right) d t_{1}=\\int_{0}^{t_{1}} \\sinh \\left(t_{1}\\right) d t_{1}^{\\prime}=\\cosh \\left(t_{1}\\right)-1\n$$\n\nSimilarly, $x_{B}=2 \\cosh \\left(t_{2} / 2\\right)-2$, and $x=\\cosh \\left(t_{1}\\right)+2 \\cosh \\left(t_{2} / 2\\right)-3$. Thus, from $t=v_{2} x$,\n\n$$\n\\begin{gathered}\n2 \\sinh \\left(t_{2} / 2\\right)-\\sinh \\left(t_{1}\\right)=\\tanh \\left(t_{2} / 2\\right)\\left(\\cosh \\left(t_{1}\\right)+2 \\cosh \\left(t_{2} / 2\\right)-3\\right) . \\\\\n-\\sinh \\left(t_{1}\\right)=\\tanh \\left(t_{2} / 2\\right)\\left(\\cosh \\left(t_{1}\\right)-3\\right) . \\\\\n\\tanh \\left(t_{2} / 2\\right)=\\frac{\\sinh \\left(t_{1}\\right)}{3-\\cosh \\left(t_{1}\\right)} .\n\\end{gathered}\n$$\n\nNow, this is the velocity of Eddie's rocket as measured from the inertial frame, so by velocity addition, the velocity of Adithya's rocket as seen by Eddie is\n\n$$\nv=\\frac{\\tanh \\left(t_{1}\\right)+\\frac{\\sinh \\left(t_{1}\\right)}{3-\\cosh \\left(t_{1}\\right)}}{1+\\frac{\\sinh \\left(t_{1}\\right) \\tanh \\left(t_{1}\\right)}{3-\\cosh \\left(t_{1}\\right)}}=\\frac{3 \\tanh \\left(t_{1}\\right)}{3-\\cosh \\left(t_{1}\\right)+\\sinh \\left(t_{1}\\right) \\tanh \\left(t_{1}\\right)} .\n$$""]"	['$2.564 \\times 10^{8}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m/s	Numerical	5e6	Open-ended	Mechanics	Physics	English
+37	"Suppose a ping pong ball of radius $R$, thickness $t$, made out of a material with density $\rho_{b}$, and Young's modulus $Y$, is hit so that it resonates in mid-air with small amplitude oscillations. Assume $t \ll R$. The density of air around (and inside) the ball is $\rho_{a}$, and the air pressure is $p$, where $\rho_{a} \ll \rho_{b} \frac{t}{R}$ and $p \ll Y \frac{t^{3}}{R^{3}}$.
+
+
+
+An estimate for the resonance frequency is $\omega \sim R^{a} t^{b} \rho_{b}^{c} Y^{d}$. Find the value of $4 a^{2}+3 b^{2}+2 c^{2}+d^{2}$.
+Hint: The surface of the ball will oscillate by ""bending"" instead of ""stretching"", since the former takes much less energy than the latter."	"['Throughout the problem, we will work to order of magnitude and ignore prefactors.\nThe hint says the surface of the ball will bend instead of stretch, so we need to develop a theory of bending. First, we consider the simplified scenario of a long beam with thickness $t$, width $w$, length $L$, made out of a material with Young\'s modulus $Y$ and density $\\rho$. When the beam bends, the top part of the beam will be in tension and the bottom part will be in compression. Thus, this is how the potential energy is stored in the beam. Suppose the beam is bent with curvature $\\kappa$. Then the top part of the beam will stretch by $L t \\kappa$, and the bottom part will compress by the same amount. Using Hooke\'s law, we can approximate the total potential energy stored in the beam as $U \\sim \\frac{Y t w}{L}(L t \\kappa)^{2} \\sim Y t^{3} w L \\kappa^{2}$. Note that if the relaxed state of the beam was already curved, we simply replace $\\kappa$ with the change in curvature.\n\nTo find the oscillation frequency of the beam, we need to find the kinetic energy in terms of $\\dot{\\kappa}$. Since curvature is on the order of second derivative of displacement, we can multiply $\\kappa$ by $L^{2}$ to get an estimate for displacement. Then $\\dot{\\kappa} L^{2}$ gives an estimate for speed, so the kinetic energy is $K \\sim \\rho t w L\\left(\\dot{\\kappa} L^{2}\\right)^{2} \\sim \\rho t w L^{5} \\dot{\\kappa}^{2}$. Thus, the frequency of oscillations is $\\omega \\sim \\sqrt{\\frac{Y t^{2}}{\\rho L^{4}}}$. Again, if the beam\n\n\n\nwas already curved, we can replace $\\kappa$ everywhere with the change in curvature.\n\nWe can model the ping pong ball as a ""beam"" of length order $R$, width order $R$, and thickness $t$. This is a very crude approximation, but will give a dimensionally correct answer (since we are ignoring prefactors). The angular frequency is thus $\\omega \\sim \\frac{t}{R^{2}} \\sqrt{\\frac{Y}{\\rho_{b}}}$.']"	['19.75']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	3e-1	Open-ended	Thermodynamics	Physics	English
+38	"A player throws two tennis balls on a level ground at $v=20 \mathrm{~m} / \mathrm{s}$ in the same direction, once at an angle of $\alpha=35^{\circ}$ and once at an angle $\beta=55^{\circ}$ to the horizontal. The distance between the landing spots of the two balls is $d$. Find $d$ in meters.
+Assume the height of the player is negligble and ignore air resistance."	['The range of a projectile is proportional as $R \\propto \\sin 2 \\theta$, or $R \\propto \\cos \\theta \\sin \\theta$. As $\\cos (90-\\theta)=\\sin \\theta$, and $\\alpha+\\beta=90$, the distance travelled by both projectiles are the same.\n$0 \\mathrm{~m}$']	['0']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	0	Open-ended	Mechanics	Physics	English
+39	"For this problem, assume the Earth moves in a perfect circle around the sun in the $x y$ plane, with a radius of $r=1.496 \times 10^{11} \mathrm{~m}$, and the Earth has a mass $m=5.972 \times 10^{24} \mathrm{~kg}$. An alien stands far away from our solar system on the $x$ axis such that it appears the Earth is moving along a one dimensional line, as if there was a zero-length spring connecting the Earth and the Sun.
+For the alien at this location, it is impossible to tell just from the motion if it's 2D motion via gravity or 1D motion via a spring. Let $U_{g}$ be the gravitational potential energy ignoring its self energy if Earth moves via gravity, taking potential energy at infinity to be 0 and $U_{s}$ be the maximum spring potential energy if Earth moves in $1 \mathrm{D}$ via a spring. Compute $U_{g} / U_{s}$."	['One naive idea is to directly compute $U_{g}$ and $U_{s}$, but we can use the fact that their frequencies are the same, or:\n$$\n\\omega^{2}=\\frac{k}{m}=\\frac{G M}{r^{3}} \\Longrightarrow k r^{2}=\\frac{G M}{r}\n$$\n\nThen,\n\n$$\nU_{g}=-\\frac{G M m}{r}=-k r^{2}\n$$\n\nand\n\n$$\nU_{s}=\\frac{1}{2} k r^{2}\n$$\n\nTherefore,\n\n$$\nU_{g} / U_{s}=-2\n$$']	['-2']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-1	Open-ended	Mechanics	Physics	English
+40	"Battle ropes can be used as a full body workout (see photo). It consists of a long piece of thick rope (ranging from $35 \mathrm{~mm}$ to $50 \mathrm{~mm}$ in diameter), wrapped around a stationary pole. The athlete grabs on to both ends, leans back, and moves their arms up and down in order to create waves, as shown in the photo.
+
+The athlete wishes to upgrade from using a $35 \mathrm{~mm}$ diameter rope to a $50 \mathrm{~mm}$ diameter rope, while keeping everything else the same (rope material, rope tension, amplitude, and speed at which her arms move back and forth). By doing so, the power she needs to exert changes from $P_{0}$ to $P_{1}$. Compute $P_{1} / P_{0}$."	['The power transmitted by a wave is given by\n$$\nP=\\frac{1}{2} \\mu \\omega^{2} A^{2} v\n$$\n\nwhere $\\mu=\\frac{m}{L}$ is the linear mass density, $A$ is the amplitude, and $v$ is the speed of the wave. The speed of a wave on a rope is given by\n\n$$\nv=\\sqrt{\\frac{T}{\\mu}}\n\\tag{2}\n$$\n\nwhere $T$ is the tension. Note that $\\omega, A, T$ will all remain constant when changing the radius. Thus, $P \\propto \\sqrt{\\mu} \\propto \\sqrt{m}$. As we increase the radius by a factor of $f=\\frac{50}{35}$, we change the mass by $f^{2}$, so the power changes by a factor of $f$, giving us\n\n$$\nP_{1} / P_{0}=f=1.43\n$$']	['1.43']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-1	Open-ended	Mechanics	Physics	English
+41	Given vertically polarized light, you're given the task of changing it to horizontally polarized light by passing it through a series of $N=5$ linear polarizers. What is the maximum possible efficiency of this process? (Here, efficiency is defined as the ratio between output light intensity and input light intensity.)	['Let $\\theta_{0}=0$ be the original direction of polarization and $\\theta_{5}=\\pi / 2$ the final direction of polarization. The 5 polarizers are directed along $\\theta_{1}, \\theta_{2}, \\ldots, \\theta_{5}$. Let $\\delta_{k}=\\theta_{k}-\\theta_{k-1}$, so that the efficiency is\n\n$$\n\\eta=\\prod_{k=1}^{5} \\cos ^{2} \\delta_{k}\n$$\n\nWe wish to maximize $\\eta$ subject to the constraint that $\\sum_{k} \\delta_{k}=\\pi / 2$. Clearly, the $\\delta_{k}^{\\prime} s$ should be non-negative, implying that $0 \\leq \\delta_{k} \\leq \\pi / 2$ and thus $\\cos \\delta_{k} \\geq 0$ for all $k$.\n\n\n\nWe claim that the maximum is achieved when all $\\delta_{k}$ are equal. If not, let $\\delta_{i} \\neq \\delta_{i+1}$. Then\n\n$$\n\\begin{aligned}\n\\cos \\delta_{i} \\cos \\delta_{i+1} & =\\frac{1}{2}\\left[\\cos \\left(\\delta_{i}+\\delta_{i+1}\\right)+\\cos \\left(\\delta_{i}-\\delta_{i+1}\\right)\\right] \\\\\n& <\\frac{1}{2}\\left[\\cos \\left(\\delta_{i}+\\delta_{i+1}\\right)+1\\right] \\\\\n& =\\frac{1}{2}\\left[\\cos \\left(\\delta_{i}^{\\prime}+\\delta_{i+1}^{\\prime}\\right)+\\cos \\left(\\delta_{i}^{\\prime}-\\delta_{i+1}^{\\prime}\\right)\\right]\n\\end{aligned}\n$$\n\nwhere $\\delta_{i}^{\\prime}=\\delta_{i+1}^{\\prime}=\\frac{\\delta_{i}+\\delta_{i+1}}{2}$. So replacing $\\delta_{i}, \\delta_{i+1}$ with $\\delta_{i}^{\\prime}, \\delta_{i+1}^{\\prime}$ increases $\\eta$.\n\nSo $\\eta$ is maximized when all $\\delta_{k}$ are equal, i.e., $\\delta_{k}^{*}=\\frac{\\pi}{10}$ for all $k$. Then\n\n$$\n\\eta^{*}=\\cos ^{10}\\left(\\frac{\\pi}{10}\\right) \\approx 0.6054\n$$']	['$\\cos ^{10}(\\frac{\\pi}{10})$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	1e-8	Open-ended	Optics	Physics	English
+42	"In this problem, we explore how fast an iceberg can melt, through the dominant mode of forced convection. For simplicity, consider a very thin iceberg in the form of a square with side lengths $L=100 \mathrm{~m}$ and a height of $1 \mathrm{~m}$, moving in the arctic ocean at a speed of $0.2 \mathrm{~m} / \mathrm{s}$ with one pair of edges parallel to the direction of motion (Other than the height, these numbers are typical of an average iceberg). The temperature of the surrounding water and air is $2^{\circ} \mathrm{C}$, and the temperature of the iceberg is $0^{\circ} \mathrm{C}$. The density of ice is $917 \mathrm{~kg} / \mathrm{m}^{3}$ and the latent heat of melting is $L_{w}=334 \times 10^{3} \mathrm{~J} / \mathrm{kg}$.
+The heat transfer rate $\dot{Q}$ between a surface and the surrounding fluid is dependent on the heat transfer coefficient $h$, the surface area in contact with the fluid $A$, and the temperature difference between the surface and the fluid $\Delta T$, via $\dot{Q}=h A \Delta T$.
+
+In heat transfer, three useful quantities are the Reynold's number, the Nusselt number, and the Prandtl number. Assume they are constant through and given by (assuming laminar flow):
+
+$$
+\operatorname{Re}=\frac{\rho v_{\infty} L}{\mu}, \quad \mathrm{Nu}=\frac{h L}{k}, \quad \operatorname{Pr}=\frac{c_{p} \mu}{k}
+$$
+
+where:
+
+- $\rho$ : density of the fluid
+- $v_{\infty}$ : speed of the fluid with respect to the object (at a very far distance)
+- $L$ : length of the object in the direction of motion
+
+
+
+- $\mu$ : dynamic viscosity of the fluid
+- $k$ : thermal conductivity of the fluid
+- $c_{p}$ : the specific heat capacity of the fluid
+
+Through experiments, the relationship between the three dimensionless numbers is, for a flat plate:
+
+$$
+\mathrm{Nu}=0.664 \operatorname{Re}^{1 / 2} \operatorname{Pr}^{1 / 3}
+$$
+
+Use the following values for calculations:
+
+|  | Air | Water |
+| :--- | :--- | :--- |
+| $\rho\left(\mathrm{kg} / \mathrm{m}^{3}\right)$ | 1.29 | 1000 |
+| $\mu(\mathrm{kg} /(\mathrm{m} \cdot \mathrm{s}))$ | $1.729 \times 10^{-5}$ | $1.792 \times 10^{-3}$ |
+| $c_{p}(\mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K}))$ | 1004 | 4220 |
+| $k(\mathrm{~W} /(\mathrm{m} \cdot \mathrm{K}))$ | 0.025 | 0.556 |
+
+The initial rate of heat transfer is $\dot{Q}$. Assuming this rate is constant (this is not true, but will allow us to obtain an estimate), how long (in days) would it take for the ice to melt completely? Assume convection is only happening on the top and bottom faces. Round to the nearest day."	['The heat transfer coefficient for water-ice and air-ice contact can be figured out with the relationship between the three dimensionless numbers\n$$\n\\mathrm{Nu}=0.664 \\operatorname{Re}^{1 / 2} \\operatorname{Pr}^{1 / 3} \\Longrightarrow h=0.664 \\frac{k}{L}\\left(\\frac{\\rho v_{\\infty} L}{\\mu}\\right)^{1 / 2}\\left(\\frac{c_{p} \\mu}{k}\\right)^{1 / 3}\n$$\n\nAs\n\n$$\n\\frac{\\mathrm{d} Q}{\\mathrm{~d} t}=h A \\Delta T\n$$\n\nwe then have\n\n$$\nt\\left(\\dot{Q}_{a}+\\dot{Q}_{w}\\right)=\\left(h_{w} A_{w}+h_{a} A_{a}\\right) \\Delta T \\Longrightarrow t=\\frac{\\rho L^{2} H L_{w}}{\\Delta T} \\frac{1}{h_{w} L^{2}+h_{a} L^{2}}=59.84 \\approx 60 \\text { days. }\n$$']	['60']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	days	Numerical	1e0	Open-ended	Thermodynamics	Physics	English
+43	"In a galaxy far, far away, there is a planet of mass $M=6 \cdot 10^{27} \mathrm{~kg}$ which is a sphere of radius $R$ and charge $Q=10^{3} \mathrm{C}$ uniformly distributed. Aliens on this planet have devised a device for transportation, which is an insulating rectangular plate with mass $m=1 \mathrm{~kg}$ and charge $q=10^{4} \mathrm{C}$. This transportation device moves in a circular orbit at a distance $r=8 \cdot 10^{6} \mathrm{~m}$ from the center of the planet. The aliens have designated this precise elevation for the device, and do not want the device to deviate at all. In order to maintain its orbit, the device contains a relatively small energy supply. Find the power (in Watts) that the energy supply must release in order to sustain this orbit.
+The velocity of the device can be assumed to be much smaller than the speed of light, so that relativistic effects can be ignored. The device can also be assumed to be small in comparison to the size of the planet."	['The centripetal force is given by\n$$\n\\frac{m v^{2}}{r}=\\frac{G M m}{r^{2}}-\\frac{q Q}{4 \\pi \\epsilon_{0} r^{2}}\n$$\n\n\n\nwhich implies the centripetal acceleration is\n\n$$\na=\\frac{v^{2}}{r}=\\frac{G M}{r^{2}}-\\frac{q Q}{4 \\pi \\epsilon_{0} m r^{2}}\n$$\n\nNow, the device loses energy due to its acceleration, as given by the Larmor formula. The power needed to sustain motion is\n\n$$\nP=\\frac{q^{2} a^{2}}{6 \\pi \\epsilon_{0} c^{3}}=0.522 \\mathrm{~W}\n$$']	['$0.522$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	W	Numerical	5e-2	Open-ended	Electromagnetism	Physics	English
+44	A raindrop of mass $M=0.035 \mathrm{~g}$ is at height $H=2 \mathrm{~km}$ above a large lake. The raindrop then falls down (without initial velocity), mixing and coming to equilibrium with the lake. Assume that the raindrop, lake, air, and surrounding environment are at the same temperature $T=300 \mathrm{~K}$. Determine the magnitude of entropy change associated with this process (in $J / K$ ).	"[""The total heat gain is equal to the change in potential energy of the raindrop, which spreads through out the whole environment at thermally equilibrium temperature $T$ (the environment is very large so any change in $T$ is negligible). The entropy gain $\\Delta S$ is thus generated by the dissipation of this potential energy $M g H$ to internal energy $\\Delta U$ in the environment (given that the specific volume of water doesn't change much, $\\Delta U \\approx M g H$ ). Hence the entropy change associated with this process can be estimated by:\n$$\nS=\\frac{\\Delta U}{T} \\approx \\frac{M g H}{T} \\approx 2.29 \\times 10^{-3} \\mathrm{~J} / \\mathrm{K}\n\\tag{8}\n$$""]"	['$2.29 \\times 10^{-3}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	J/K	Numerical	1e-4	Open-ended	Thermodynamics	Physics	English
+45	"A rocket with mass of 563.17 (not including the mass of fuel) metric tons sits on the launchpad of the Kennedy Space Center (latitude $28^{\circ} 31^{\prime} 27^{\prime \prime} \mathrm{N}$, longitude $80^{\circ} 39^{\prime} 03^{\prime \prime} \mathrm{W}$ ), pointing directly upwards. Two solid fuel boosters, each with a mass of $68415 \mathrm{~kg}$ and providing $3421 \mathrm{kN}$ of thrust are pointed directly downwards.
+The rocket also has a liquid fuel engine, that can be throttled to produce different amounts of thrust and gimbaled to point in various directions. What is the minimum amount of thrust, in $\mathrm{kN}$, that this engine needs to provide for the rocket to lift vertically (to accelerate directly upwards) off the launchpad?
+
+Assume $G=6.674 \times 10^{-11} \frac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{3}}$, and that the Earth is a perfect sphere of radius $6370 \mathrm{~km}$ and mass $5.972 \times 10^{24} \mathrm{~kg}$ that completes one revolution every $86164 \mathrm{~s}$ and that the rocket is negligibly small compared to the Earth. Ignore buoyancy forces."	"[""Note the additional information provided in the problem (latitude, Earth radius, revolution period), which makes it clear that the effect of the rotation of the Earth must also be considered.\nWe first compute local gravitational acceleration:\n\n$$\ng=\\frac{G M}{R^{2}}=9.823 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}\n$$\n\n\n\nAnd also acceleration due to the Earth's rotation:\n\n$$\na=(R \\cos \\theta) \\omega^{2}=0.02976 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}\n$$\n\nThen vertical acceleration is:\n\n$$\ng-a \\cos \\theta=9.7965 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}\n$$\n\nAnd horizontal acceleration:\n\n$$\na \\sin \\theta=0.0142 \\frac{m}{s^{2}}\n$$\n\nThe total mass of the craft is 700 metric tons, so the needed forces for liftoff are:\n\n$$\n\\left(F_{x}, F_{y}\\right)=(6857527,9948) \\mathrm{N}\n$$\n\nSubtracting out the effect of the solid fuel boosters, the force the liquid fuel engine needs to provide is:\n\n$$\n\\left(F_{x}, F_{y}\\right)=(15527,9948) \\mathrm{N}\n$$""]"	['$18.44$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	kN	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+46	"A spacecraft is orbiting in a very low circular orbit at a velocity $v_{0}$ over the equator of a perfectly spherical moon with uniform density. Relative to a stationary frame, the spacecraft completes a revolution of the moon every 90 minutes, while the moon revolves in the same direction once every 24 hours. The pilot of the spacecraft would like to land on the moon using the following process:So the final answer is 
+
+1. Start by firing the engine directly against the direction of motion.So the final answer is 
+
+2. Orient the engine over time such that the vertical velocity of the craft remains 0 , while the horizontal speed continues to decrease.
+
+3. Once the velocity of the craft relative to the ground is also 0 , turn off the engine.
+
+Assume that the engine of the craft can be oriented instantly in any direction, and the craft has a TWR (thrust-to-weight ratio, where weight refers to the weight at the moon's surface) of 2, which remains constant throughout the burn. If the craft starts at $v_{0}=500 \mathrm{~m} / \mathrm{s}$, compute the delta-v expended to land, minus the initial velocity, i.e. $\Delta v-v_{0}$."	"[""The trick in this question is to work in dimensionless units. Let $v$ be the ratio of the craft's velocity to orbital velocity. Then, if the craft has horizontal velocity $v$, the acceleration downwards is the following:\n\n$$\na_{V}=\\frac{v_{0}^{2}}{r}-\\frac{\\left(v_{0} v\\right)^{2}}{r}=\\frac{v_{0}^{2}}{r}\\left(1-v^{2}\\right)=g_{m}\\left(1-v^{2}\\right)\n$$\n\nAs the TWR is 2 , the total acceleration the engine provides is $a_{0}=2 g_{m}$, where $g_{m}$ is the surface gravity of the moon. As this total acceleration is the sum of horizontal and vertical components, and the vertical component cancels out the downwards acceleration:\n\n$$\na_{H}=\\sqrt{\\left(2 g_{m}\\right)^{2}-\\left(g_{m}\\left(1-v^{2}\\right)\\right)^{2}}=g_{m} \\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}}\n$$\n\nAnd $a_{H}$ is related to $v$ (which is dimensionless!) by the following relation:\n\n$$\n\\begin{gathered}\n\\frac{\\mathrm{d}\\left(v_{0} v\\right)}{\\mathrm{d} t}=-a_{H} \\\\\n\\frac{\\mathrm{d} v}{\\mathrm{~d} t}=-\\frac{g_{m}}{v_{0}} \\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}} \\\\\n\\frac{\\mathrm{d} v}{\\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}}}=-\\frac{g_{m}}{v_{0}} \\mathrm{~d} t\n\\end{gathered}\n$$\n\nAt the start, $v=1$. However, at landing, velocity of the craft is 0 relative to the surface, not a stationary frame! Therefore, we use the orbital periods data to determine the final $v$ to be $\\frac{1.5}{24}=\\frac{1}{16}$. Then integrating:\n\n$$\n\\int_{v=1}^{v=1 / 16} \\frac{\\mathrm{d} v}{\\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}}}=-\\frac{g_{m}}{v_{0}} t\n$$\n\nAnd as delta-v is related to time by $\\Delta v=a_{0} t$ :\n\n$$\n\\Delta v=a_{0} t=a_{0} \\frac{v_{0}}{g_{m}} \\int_{1 / 16}^{1} \\frac{\\mathrm{d} v}{\\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}}}=a_{0} \\frac{v_{0}}{a_{0}} 2(0.503)=500 \\cdot 1.006=503.06 \\frac{\\mathrm{m}}{\\mathrm{s}}\n$$\n\nAnd subtracting:\n\n$$\n\\Delta v-v_{0}=3.06 \\frac{\\mathrm{m}}{\\mathrm{s}}\n$$""]"	['$3.06$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m/s	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+47	"Consider two points $S$ and $S^{\prime}$ randomly placed inside a $D$-dimensional hyper-rectangular room with walls that are perfect-reflecting $(D-1)$-dimensional hyper-plane mirrors. How many different
+
+light-rays that start from $S$, reflect $N$ times on one of the walls and $N-1$ times on each of the rest, then go to $S^{\prime}$ ? Use $D=7$ and $N=3$."	"[""Using the hyper-rectangular room as the fundamental unit-cell of an infinite hyper-grid in space, then find all possible positions of $S^{\\prime}$-images through reflections: we can realize that there is only one position of $S^{\\prime}$-image inside every unit-cell.\n\n(A)\n<img_4257>\n<img_4259>\n\nConsider two opposite hyper-plane mirrors, since the rest of the mirrors are perpendicular to them, the numbers of reflections on them for any light-path traveled from point $S$ and point $S^{\\prime}$ can only different by 1 or less. If the numbers are both equal and non-zero, then the available positions $S^{\\prime}$ image the light-path from $S$ should reach are two unit-cell hyper-rows that parallel to the mirrors. If the numbers are not equal, then the available positions $S^{\\prime}$-image the light-path from $S$ should reach are one unit-cell hyper-rows that that parallel to the mirrors.\n\nSay, without loss of generality, pick one mirror to be reflected $N$ times and the rest to be reflected $N-1$ times each, then the number of light-rays for that pick should equal to the number of unit-cell intersections between all relevant unit-cell hyper-rows, which is half the total number of vertices a hyper-rectangular has, thus $2^{D-1}$. There are $2 D$ walls, thus the total number of light-rays that satisfies the task given is $2 D \\times 2^{D-1}=D 2^{D}$, independent of $N$ for all values $N>1$.\n\n\n\nTo illustrate the above explanation, let's take a look at the simple case of $D=2$ and $N=2>1$. Consider a rectangular room, with random points $S, S^{\\prime}$ and $2 D=4$ walls $W_{1}, W_{2}, W_{3}, W_{4}$ (see Fig. A). Without loss of generality, we want to find light-rays that go from $S$, reflect $N=2$ times on $W_{1}$ and $N-1=1$ times on $W_{2}, W_{3}, W_{4}$ then come to $S^{\\prime}$. Each image of $S^{\\prime}$ is an unique point in a unit-cell generated by the room (see Fig. B). Note that every light-ray from $S$ that reach $S^{\\prime}$-images in the $\\rightarrow W_{2} \\rightarrow W_{4} \\rightarrow$ and $\\rightarrow W_{4} \\rightarrow W_{2} \\rightarrow$ unit-cell green-rows will satisfy the requirement of one reflection on each of $W_{2}$ and $W_{4}$, every light-ray from $S$ that reach $S^{\\prime}$-images in the $\\rightarrow W_{1} \\rightarrow W_{3} \\rightarrow W_{1} \\rightarrow$ unit-cell blue-rows will satisfy the requirement of two reflection on $W_{1}$ and one reflection on $W_{3}$ (see Fig. C). The intersection of these rows are two unit-cells, corresponds two possible images thuss two possible light-rays that satisfies the requirement (see Fig. D).\n\nFor $D=7$ and $N=3>1$, we get 896 light-rays.""]"	['895']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	0	Open-ended	Optics	Physics	English
+48	Two concentric isolated rings of radius $a=1 \mathrm{~m}$ and $b=2 \mathrm{~m}$ of mass $m_{a}=1 \mathrm{~kg}$ and $m_{b}=2 \mathrm{~kg}$ are kept in a gravity free region. A soap film of surface tension $\sigma=0.05 \mathrm{Nm}^{-1}$ with negligible mass is spread over the rings such that it occupies the region between the rings. The smaller ring is pulled slightly along the axis of the rings. Find the time period of small oscillation in seconds.	['The force on the two rings when they are a distance $L$ apart follows as\n$$\nF=4 \\pi r \\sigma \\sin \\theta\n$$\n\nIn small displacements, the change in $\\theta$ is small. Therefore,\n\n$$\n\\begin{aligned}\nF & =4 \\pi r \\sigma \\theta \\\\\n\\frac{F}{4 \\pi r \\sigma} & =\\frac{\\mathrm{d} y}{\\mathrm{~d} r} \\\\\n\\int_{a}^{b} \\frac{F}{4 \\pi \\sigma r} \\mathrm{~d} r & =\\int_{0}^{L} \\mathrm{~d} y \\\\\n\\frac{F}{4 \\pi \\sigma} \\ln (b / a) & =L\n\\end{aligned}\n$$\n\nNow let $a_{1}$ and $a_{2}$ be acceleration of a and $\\mathrm{b}$ respectively. We have that\n\n$$\n\na_{\\text {net }}=a_{1}+a_{2}\n\\tag{12}\n$$\n$$\na_{\\text {net }}=F\\left(\\frac{m_{1}+m_{2}}{m_{1} m_{2}}\\right)\n\\tag{13}\n$$\n$$\nL \\omega^{2}=\\frac{4 \\pi \\sigma}{\\ln (b / a)} L\\left(\\frac{m_{1}+m_{2}}{m_{1} m_{2}}\\right)\n\\tag{14}\n$$\n$$\nT=2 \\pi \\sqrt{\\frac{\\ln (b / a) m_{1} m_{2}}{4 \\pi \\sigma\\left(m_{1}+m_{2}\\right)}}\n\\tag{15}\n$$\n$$\nT=2 \\pi \\sqrt{\\frac{10 \\ln (2)}{3 \\pi}}=5.388 \\mathrm{~s}\n\\tag{16}\n$$']	['$2 \\pi \\sqrt{\\frac{10 \\ln (2)}{3 \\pi}}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	s	Numerical	1e-2	Open-ended	Mechanics	Physics	English
+49	"A table of unknown material has a mass $M=100 \mathrm{~kg}$, width $w=4 \mathrm{~m}$, length $\ell=3 \mathrm{~m}$, and 4 legs of length $L=0.5 \mathrm{~m}$ with a Young's modulus of $Y=1.02 \mathrm{MPa}$ at each of the corners. The cross-sectional area of a table leg is approximately $A=1 \mathrm{~cm}^{2}$. The surface of the table has a coefficient of friction of $\mu=0.1$. A point body with the same mass as the table is put at some position from the geometric center of the table. What is the minimum distance the body must be placed from the center such that it slips on the table surface immediately after? Report your answer in centimeters.
+The table surface and floor are non-deformable."	['This problem requires some 3 dimensional reasoning. Suppose $\\mathbf{s}=\\left(s_{x}, s_{y}\\right)$ is the gradient of the table. We can use this to calculate the additional torque from the displacement of the mass. The forces from each table leg are\n$$\nF_{i}=\\frac{Y A}{L}\\left( \\pm s_{x} \\frac{\\ell}{2} \\pm s_{y} \\frac{w}{2}\\right)\n$$\n\nTaking the cross product as $\\boldsymbol{\\tau}=\\mathbf{r} \\times \\mathbf{F}_{\\mathbf{i}}$ shows that torque is given as\n\n$$\n\\boldsymbol{\\tau}=\\frac{Y A}{L}\\left(\\begin{array}{c}\n-s_{y} w^{2} \\\\\ns_{x} \\ell^{2}\n\\end{array}\\right)\n$$\n\nwhich must balance out the torque $M g d$ from a point mass. Hence, rewriting yields\n\n$$\nd=\\frac{Y A}{M g L} \\sqrt{s_{y}^{2} w^{4}+s_{x}^{2} \\ell^{4}}\n$$\n\nFurthermore, note that the angle required from slipping is given from a force analysis as\n\n$$\nm g \\sin \\theta=\\mu m g \\cos \\theta \\Longrightarrow \\mu=\\tan \\theta=|\\mathbf{s}|=\\sqrt{s_{x}^{2}+s_{y}^{2}}\n$$\n\nWhen $w>\\ell$, we can rewrite\n\n$$\ns_{y}^{2} w^{4}+s_{x}^{2} \\ell^{4}=s_{y}^{2}\\left(w^{4}-\\ell^{4}\\right)+\\ell^{4} \\mu^{2}\n$$\n\nwhich is minimized to $\\ell^{4} \\mu^{2}$ when $s_{y}=0$. Hence, we obtain\n\n$$\nd=\\frac{\\mu \\ell^{2} Y A}{M g L}\n$$']	['$18.71$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	1e-1	Open-ended	Mechanics	Physics	English
+50	Dipole Conductor An (ideal) electric dipole of magnitude $p=1 \times 10^{-6} \mathrm{C} \cdot \mathrm{m}$ is placed at a distance $a=0.05 \mathrm{~m}$ away from the center of an uncharged, isolated spherical conductor of radius $R=0.02 \mathrm{~m}$. Suppose the angle formed by the dipole vector and the radial vector (the vector pointing from the sphere's center to the dipole's position) is $\theta=20^{\circ}$. Find the (electrostatic) interaction energy between the dipole and the charge induced on the spherical conductor.	"[""We can use the fact that if a charge $Q$ is placed at a distance $a$ from a grounded, conducting sphere of radius $R$, as far as the field outside the sphere is concerned, there is an image charge of magnitude $-Q \\frac{R}{a}$ at a position $\\frac{R^{2}}{a}$ from the origin, on the line segment connecting the origin and charge $Q$. It is straightforward to check that indeed, the potential on the sphere due to this image charge prescription is 0 . If the point charge is instead a dipole $p$, we can think of this as a superposition of 2 point charges, and use the fact above. In particular, there is one charge $-Q$ at point $(a, 0,0)$ and another charge $Q$ at point $(a+s \\cos \\theta, s \\sin \\theta, 0)$, where $s$ is small and $Q s=p$. Note that the dipole points in the direction $\\theta$ above the $\\mathrm{x}$-axis. Consequently, there will be an image charge at $\\left(\\frac{R^{2}}{a}, 0,0\\right)$ with magnitude $Q \\frac{R}{a}$ and an image charge at $\\left(\\frac{R^{2}}{a+s \\cos \\theta}, \\frac{R^{2} s \\sin \\theta}{a(a+s \\cos \\theta)}, 0\\right)$ with magnitude $-Q \\frac{R}{a+s \\cos \\theta}$. The image charges are close to each other but do not cancel out exactly,\n\n\n\nso they can be represented as a superposition of an image point charge $Q^{\\prime}$ and an image dipole $p^{\\prime}$. The image point charge has magnitude $Q^{\\prime}=-Q R\\left(\\frac{1}{a+s \\cos \\theta}-\\frac{1}{a}\\right)=\\frac{Q R s \\cos \\theta}{a^{2}}$. The image dipole has magnitude $p^{\\prime}=Q \\frac{R}{a} * \\frac{R^{2} s}{a^{2}}=\\frac{Q R^{3} s}{a^{3}}$ and points towards the direction $\\theta$ below the positive x-axis. Finally, since the sphere in the problem is uncharged instead of grounded, to ensure the net charge in the sphere is 0 , we place another image charge $-Q^{\\prime}$ at the origin.\n\nNow we can calculate the desired interaction energy, which is simply the interaction energy between the image charges and the real dipole. Using the dipole-dipole interaction formula, the interaction between the image dipole and the real dipole is given by:\n\n$$\nU_{1}=\\frac{k p p^{\\prime}}{\\left(a-\\frac{R^{2}}{a}\\right)^{3}}\\left(\\cos (2 \\theta)-3 \\cos ^{2} \\theta\\right)\n$$\n\nThe interaction between the image charge at the image dipole's position and the real dipole is given by:\n\n$$\nU_{2}=-\\frac{k p Q^{\\prime} \\cos \\theta}{\\left(a-\\frac{R^{2}}{a}\\right)^{2}}\n$$\n\nThe interaction between the image charge at the center and the real dipole is given by:\n\n$$\nU_{3}=\\frac{k p Q^{\\prime} \\cos \\theta}{a^{2}}\n$$\n\nThe final answer is $U=U_{1}+U_{2}+U_{3}=\\frac{p^{2} R}{4 \\pi \\epsilon_{0}}\\left(\\frac{\\cos ^{2} \\theta}{a^{4}}-\\frac{a^{2} \\cos ^{2} \\theta+R^{2}}{\\left(a^{2}-R^{2}\\right)^{3}}\\right)=-25.22 \\mathrm{~J} .$""]"	['$-25.22$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	J	Numerical	1e-1	Open-ended	Electromagnetism	Physics	English
+51	"A uniform spherical metallic ball of mass $m$, resistivity $\rho$, and radius $R$ is kept on a smooth friction-less horizontal ground. A horizontal uniform, constant magnetic field $B$ exists in the space parallel to the surface of ground. The ball was suddenly given an impulse perpendicular to magnetic field such that ball begin to move with velocity $v$ without losing the contact with ground. Find the time in seconds required to reduce its velocity by half.
+Numerical Quantities: $m=2 \mathrm{~kg}, 4 \pi \epsilon_{0} R^{3} B^{2}=3 \mathrm{~kg}, \rho=10^{9} \Omega \mathrm{m}, v=\pi \mathrm{m} / \mathrm{s}$."	['WLOG assuming magnetic field to be into the plane (negative z-axis) and velocity of the block along x-axis. Let any time $\\mathrm{t}$ ball has velocity $\\mathrm{v}$ and surface charge $\\sigma \\cos \\theta$, where $\\theta$ is measured from $\\mathrm{Y}$-axis. As the ball is moving it will also generate and electric field $\\mathrm{E}=\\mathrm{vB}$ along positive Y-axis. Which will be opposed by the electric field of ball. Also we know the electric filed generated by the ball is $\\frac{\\sigma}{3 \\epsilon_{0}}$ in negative $\\mathrm{Z}$-axis. As this charge distribution will arise form a vertical electric field and the subsequent induced charges will also produce only a vertical electric field only thus our assumption about charge distribution and net electric field must be true.\nNow at any point on the surface of the sphere rate of increase in surface charge density is given by-\n\n\n\n$$\n\\begin{gathered}\nJ d A \\cos \\theta=\\frac{d(\\sigma \\cos \\theta)}{d t} d A \\\\\nJ=\\frac{d \\sigma}{d t} \\\\\nE-\\frac{\\sigma}{3 \\epsilon_{0}}=\\rho \\frac{d \\sigma}{d t} \\\\\nv B-\\frac{\\sigma}{3 \\epsilon_{0}}=\\rho \\frac{d \\sigma}{d t}\n\\end{gathered}\n\\tag{44}\n$$\n\nAs the magnetic field is uniform to calculate force path of the current will not matter. Hence assuming it to be straight line between two points located at $\\theta$ and $-\\theta$. So the force can be written as -\n\n$$\n\\begin{gathered}\nd F=B \\times d I \\times l \\\\\nd F=B(2 \\pi(R \\sin \\theta) \\times R d \\theta \\times J)(2 R \\cos (\\theta)) \\\\\nF=4 \\pi R^{3} B \\frac{d \\sigma}{d t} \\int_{0}^{\\pi / 2} \\sin \\theta \\cos ^{2} \\theta d \\theta \\\\\nF=\\frac{4}{3} \\pi R^{3} B \\frac{d \\sigma}{d t}\n\\end{gathered}\n\\tag{45}\n$$\n\nNow writing force equation on the sphere we have\n\n$$\n\\begin{gathered}\nF=-m \\frac{d v}{d t} \\\\\n\\frac{4}{3} \\pi R^{3} B \\frac{d \\sigma}{d t}=-m \\frac{d v}{d t} \\\\\n\\frac{4}{3} \\pi R^{3} B \\int_{0}^{\\sigma} d \\sigma=-m \\int_{v_{0}}^{v} d v \\\\\n\\frac{4}{3} \\pi R^{3} B \\sigma=m v_{0}-m v\n\\end{gathered}\n\\tag{46}\n$$\n\nSolving equation 1 and 3 gives us\n\n$$\n\\left(m+4 \\pi R^{3} B^{2} \\epsilon_{0}\\right) v-m v_{0}=-3 m \\rho \\epsilon_{0} \\frac{d v}{d t}\n$$\n\nIntegrating it from $v_{0}$ to $\\frac{v_{0}}{2}$ gives\n\n$$\n\nt=\\frac{3 m \\rho \\epsilon_{0}}{\\left(m+4 \\pi R^{3} B^{2} \\epsilon_{0}\\right)} \\ln \\left(\\frac{8 \\pi R^{3} B^{2} \\epsilon_{0}}{4 \\pi R^{3} B^{2} \\epsilon_{0}-m}\\right)\n\\tag{47}\n$$\n$$\nt=\\frac{6 \\ln (6)}{5} \\rho \\epsilon_{0}=0.019 \\mathrm{~s}\n\\tag{48}\n$$']	['$0.019$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	s	Numerical	1e-3	Open-ended	Electromagnetism	Physics	English
+52	"In quantum mechanics, when calculating the interaction between the electron with the proton in a hydrogen atom, it is necessary to compute the following volume integral (over all space):
+$$
+\mathbf{I}=\int \mathbf{B}(\mathbf{r})|\Psi(\mathbf{r})|^{2} d V
+$$
+
+where $\Psi(\mathbf{r})$ is the spatial wavefunction of the electron as a function of position $\mathbf{r}$ and $\mathbf{B}(\mathbf{r})$ is the (boldface denotes vector) magnetic field produced by the proton at position $\mathbf{r}$. Suppose the proton is located at the origin and it acts like a finite-sized magnetic dipole (but much smaller than $a_{0}$ ) with dipole moment
+
+$\mu_{p}=1.41 \times 10^{-26} \mathrm{~J} / \mathrm{T}$. Let the hydrogen atom be in the ground state, meaning $\Psi(\mathbf{r})=\frac{e^{-r / a_{0}}}{\sqrt{\pi a_{0}^{3}}}$, where $a_{0}=5.29 \times 10^{-11} \mathrm{~m}$ is the Bohr radius. Evaluate the magnitude of the integral $|\mathbf{I}|$ (in SI units)."	"[""First, note that the result of the integral will be a vector in the direction the dipole is pointing, call it the z-direction. Thus we can replace $\\mathbf{B}$ in the integral with $B_{z}$. Note that for any $R>0$, the integral over the space outside the sphere of radius $R$ is 0 . To show this, since $|\\Psi|$ is exponentially decaying, we only need to show that the integral over a spherical shell is 0 . To show this, we can show that the integral of $\\mathbf{B}$ inside a sphere of radius $R$ is independent of $R$. Indeed, this quickly follows from dimensional analysis (the only relevant quantities are $\\mu_{0}, \\mu_{p}$, and $R$, and one can check that $\\mu_{0} \\mu_{p}$ already gives the right dimensions, and there is no dimensionless combination of these 3 quantities. In fact, we will actually compute this integral at the end.)\n\nNow, it suffices to compute the integral of $\\mathbf{B}|\\Psi|^{2}$ inside the sphere. Since $R$ was arbitrary, we can make it very small, much smaller than $a_{0}$. Then we can replace $|\\Psi(\\mathbf{r})|^{2}$ with $|\\Psi(0)|^{2}=\\frac{1}{\\pi a_{0}^{3}}$, a constant that can be factored out. The problem reduces to computing the integral of $\\mathbf{B}$ inside a sphere of radius $R$.\n\nWe can compute this integral by splitting the sphere up into many thin discs, all perpendicular to the $z$ axis. We have to add up the $\\mathbf{B}$ field integrated over the volume of each disc, which is equivalent to the magnetic flux through the disc times the thickness of the disc. The magnetic flux through each disc can be calculated using the mutual inductance reciprocity theorem. Suppose a current $I$ goes around the boundary of the disc (a ring) with radius $r$. Then the mutual inductance $M$ between the ring and the dipole is given by the flux through the dipole divided by $I$ :\n\n$$\nM=\\frac{B * A}{I}\n$$\n\nwhere $B$ is the magnetic field produced by the ring's current at the dipole's position, and $A$ is the area of the dipole. The dipole itself carries current $i=\\frac{\\mu_{p}}{A}$, so the flux through the ring is given by\n\n$$\n\\Phi=M * i=\\frac{B i A}{I}=\\frac{\\mu_{p} B}{I}\n$$\n\nwhere $B=\\frac{\\mu_{0} I}{4 \\pi} * \\frac{2 \\pi r^{2}}{\\left(r^{2}+z^{2}\\right)^{\\frac{3}{2}}}=\\frac{\\mu_{0} I r^{2}}{2\\left(r^{2}+z^{2}\\right)^{\\frac{3}{2}}}$, where $z$ is the $z$ coordinate of the ring. Using $r=R \\sin \\theta$ and $z=R \\cos \\theta$, we obtain\n\n\n\n$$\n\\Phi(\\theta)=\\frac{\\mu_{0} \\mu_{p} r^{2}}{2\\left(r^{2}+z^{2}\\right)^{\\frac{3}{2}}}=\\frac{\\mu_{0} \\mu_{p} \\sin ^{2} \\theta}{2 R}\n$$\n\nFinally, we integrate over the thickness of the $\\operatorname{disc} R \\sin \\theta d \\theta$ to get :\n\n$$\n\\int_{0}^{\\pi} \\Phi(\\theta) R \\sin \\theta d \\theta=\\frac{1}{2} \\mu_{0} \\mu_{p} \\int_{0}^{\\pi} \\sin ^{3} \\theta d \\theta=\\frac{2}{3} \\mu_{0} \\mu_{p}\n$$\n\nThus, $|\\mathbf{I}|=\\frac{2}{3} \\mu_{0} \\mu_{p} * \\frac{1}{\\pi a_{0}^{3}}=\\frac{2 \\mu_{0} \\mu_{p}}{3 \\pi a_{0}^{3}}=0.0254 \\mathrm{~T}$.""]"	['$0.0254$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	T	Numerical	5e-3	Open-ended	Modern Physics	Physics	English
+53	"Zed is trying to model the repulsive interaction between 2 objects, $A$ and $B$ (with masses $m_{A}$ and $m_{B}$, respectively), in a relativistic setting. He knows that in relativity, forces cannot act at a distance, so he models the repulsive force with a small particle of mass $m$ that bounces elastically between $A$ and $B$. Throughout this problem, assume everything moves on the x-axis. Suppose that initially, $A$ and $B$ have positions and velocities $x_{A}, v_{A}$ and $x_{B}, v_{B}$, respectively, where $x_{A}<x_{B}$ and $v_{A}>v_{B}$. The particle has an initial (relativistic) speed $v$.
+For simplicity, assume that the system has no total momentum. You may also assume that $v_{A}, v_{B} \ll v$, and that $p_{m} \ll p_{A}, p_{B}$, where $p_{m}, p_{A}, p_{B}$ are the momenta of the particle, $A$, and $B$, respectively. Do NOT assume $v \ll c$, where $c$ is the speed of light.
+
+Find the position (in $\mathrm{m}$ ) of $A$ when its velocity is 0 , given that $m_{A}=1 \mathrm{~kg}, m_{B}=2 \mathrm{~kg}, v_{A}=0.001 c$, $m=1 \times 10^{-6} \mathrm{~kg}, v=0.6 c, x_{A}=0 \mathrm{~m}, x_{B}=1000 \mathrm{~m}$.
+
+Note: Answers will be tolerated within $0.5 \%$, unlike other problems."	"[""Since total momentum is 0 , we have $m_{A} v_{A}+m_{B} v_{B}=0$, so $v_{B}=-0.0005 c$ By conservation of energy:\n\n$$\n\\frac{1}{2} m_{A} v_{A}^{2}+\\frac{1}{2} m_{B} v_{B}^{2}+\\gamma_{0} m c^{2}=\\gamma m c^{2}\n$$\n\nwhere we define $\\gamma=\\frac{1}{\\sqrt{1-\\frac{v^{2}}{c^{2}}}}$ to correspond to the final state of the particle and $\\gamma_{0}$ to the initial state. This equation allows us to solve for $\\gamma$. Since the speed of the particle is much larger than the speed of the masses $A$ and $B$, we can imagine the particle moving in a infinite well potential where the walls are slowly moving. Applying the adiabatic theorem, we get that the adiabatic invariant $p x$ is conserved, where $p$ is the particle's momentum, and $x$ is the distance between $A$ and $B$. Thus, $\\gamma v x$ is conserved, so\n\n$$\n\\gamma v x=\\gamma_{0} v_{0}\\left(x_{B}-x_{A}\\right)\n$$\n\nWe can solve for $x$, since we know $\\gamma$ and $v$. Finally, we realize that the center of mass stays at $x_{c} m=\\frac{m_{A} x_{A}+m_{B} x_{B}}{m_{A}+m_{B}}$, so the final position of $A$ is simply\n\n$$\nx_{c} m-\\frac{m_{B}}{m_{A}+m_{B}} x=378 \\mathrm{~m}\n$$""]"	['378']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	2e0	Open-ended	Modern Physics	Physics	English
+54	"Consider an optical system consisting of two thin lenses sharing the same optical axis. When a cuboid with a side parallel to the optical axis is placed to the left of the left lens, its final image formed by the optical system is also a cuboid but with 500 times the original volume. Assume the two
+
+lenses are $10 \mathrm{~cm}$ apart and such a cuboid of volume $1 \mathrm{~cm}^{3}$ is placed such that its right face is $2 \mathrm{~cm}$ to the left of the left lens. What's the maximum possible volume of the intermediate image (i.e., image formed by just the left lens) of the cuboid? Answer in $\mathrm{cm}^{3}$."	"[""First, note that the two lenses share a focal point. Here's why. For any cuboid with four edges parallel to the optical axis, consider the four parallel rays of light that these four edges lie on. The intermediate images formed by the left lens of these four edges lie on these same light rays after they've passed through the left lens, and the final images of the edges (images formed by the right lens of the intermediate images) lie on these same light rays after they've also passed through the right lens. Since the initial rays were parallel and the same goes for the final rays, the intermediate rays intersect at a focal point of both the left lens and the right lens at the same time.\n\nNow, let $f, f^{\\prime}$ be the focal lengths of the left and right lenses, respectively. Although they can be any nonzero real number, we will WLOG assume that they are positive. (The following derivation will still hold with either $f$ or $f^{\\prime}$ negative, as long as we have the right sign conventions for all the variables.)\n\nFor a point at a distance $x_{1}$ to the left of $F$, its image is at a distance $x_{2}^{\\prime}=-\\left(\\frac{f^{\\prime}}{f}\\right)^{2} x_{1}$ to the right of $F^{\\prime}$. This follows from applying Newton's formula twice:\n\n$$\nx_{2}^{\\prime}=\\frac{f^{\\prime 2}}{x_{1}^{\\prime}}=-\\frac{f^{\\prime 2}}{x_{2}}=-\\frac{f^{\\prime 2}}{f^{2}} x_{1}\n$$\n\nThus, the optical system magnifies horizontal distances by $\\left(\\frac{f^{\\prime}}{f}\\right)^{2}$.\n\nOn the other hand, for a point at height $h_{1}$ (relative to the optical axis), consider a horizontal light ray through the point. Then the final light ray (after it passes through both lenses) is at a height of\n\n$$\nh_{2}=-\\frac{f^{\\prime}}{f} h_{1}\n$$\n\nwhich is the height of the final image of the point. Hence, the optical system magnifies transverse distances (i.e., distances perpendicular to the optical axis) by $\\frac{f^{\\prime}}{f}$.\n\nThe two results above imply that volumes are magnified by\n\n$$\n\\left(\\frac{f^{\\prime}}{f}\\right)^{2}\\left(\\frac{f^{\\prime}}{f}\\right)^{2}=\\left(\\frac{f^{\\prime}}{f}\\right)^{4}\n$$\n\n(The second factor is squared because there are two transverse dimensions.) Given that volumes are magnified by 500 times, we obtain $\\frac{f^{\\prime}}{f}= \\pm 500^{1 / 4}$.\n\nWe now look at a cuboid with volume $V$ whose right face is at a distance $d$ to the left of $F$. Let it have width $x d$ and transverse cross-sectional area $A=V / x d$. The intermediate image is a frustrum that results from truncating a pyramid with vertex located at $K$.\n\n\n\n<img_4261>\n\nBy Newton's formula, the bases of the frustrum are at distances $\\frac{f^{2}}{d}$ and $\\frac{f^{2}}{d(1+x)}$ to the right of $K$, and they have areas\n\n$$\n\\left(\\frac{\\frac{f^{2}}{d}}{f}\\right)^{2} A=\\frac{V f^{2}}{x d^{3}} \\quad \\text { and } \\quad\\left(\\frac{\\frac{f^{2}}{d(1+x)}}{f}\\right)^{2} A=\\frac{V f^{2}}{x(1+x)^{2} d^{3}}\n$$\n\nThus, the volume of the frustrum is\n\n$$\n\\frac{1}{3}\\left(\\frac{f^{2}}{d} \\frac{V f^{2}}{x d^{3}}-\\frac{f^{2}}{d(1+x)} \\frac{V f^{2}}{x(1+x)^{2} d^{3}}\\right)=\\frac{1}{3} \\frac{V f^{4}}{x d^{4}}\\left(1-\\frac{1}{(1+x)^{3}}\\right) \\leq \\frac{1}{3} \\frac{V f^{4}}{x d^{4}}(1-(1-3 x))=\\frac{V f^{4}}{d^{4}},\n$$\n\nwhere equality is approached as $x \\rightarrow 0$.\n\nSince $f+f^{\\prime}=10 \\mathrm{~cm}$ and $\\frac{f^{\\prime}}{f}= \\pm 500^{1 / 4} \\approx \\pm 4.7287$, either $f=1.7456 \\mathrm{~cm}$ and $d=2 \\mathrm{~cm}-f=$ $0.2544 \\mathrm{~cm}$, which gives $V_{\\max }=\\frac{V f^{4}}{d^{4}}=2216 \\mathrm{~cm}^{3}$, or $f=-2.6819 \\mathrm{~cm}$ and $d=2 \\mathrm{~cm}-f=4.6819 \\mathrm{~cm}$, which gives $V_{\\max }=0.1077 \\mathrm{~cm}^{3}$. The former is larger, so the answer is $2216 \\mathrm{~cm}^{3}$.""]"	['$2216$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	$\mathrm{~cm}^{3}$	Numerical	5e0	Open-ended	Optics	Physics	English
+55	Consider an infinite square grid of equal resistors where the nodes are exactly the lattice points in the 2D Cartesian plane. A current $I=2.7 \mathrm{~A}$ enters the grid at the origin $(0,0)$. Find the current in Amps through the resistor connecting the nodes $(N, 0)$ and $(N, 1)$, where $N=38$ can be assumed to be much larger than 1.	"[""WLOG, let each resistor have unit resistance.\n\nKirchoff's current law says that the total current entering the node $(x, y) \\neq(0,0)$ is zero:\n\n$$\n\\begin{array}{r}\n{[U(x+1, y)-U(x, y)]+[U(x-1, y)-U(x, y)]+[U(x, y+1)-U(x, y)]+[U(x, y-1)-U(x, y)]=0} \\\\\n{[U(x+1, y)-2 U(x, y)+U(x-1, y)]+[U(x, y+1)-2 U(x, y)+U(x, y-1)]=0}\n\\end{array}\n$$\n\nThis is an approximation of the equation\n\n$$\n\\frac{\\partial^{2} U}{\\partial x^{2}}+\\frac{\\partial^{2} U}{\\partial y^{2}}=0\n$$\n\nwhich is Laplace's equation $\\nabla^{2} U=0$ in 2D. Given $U \\rightarrow 0$ as $r \\rightarrow \\infty$, this implies that the potential\n\n\n\nfield approximates that of a point charge at $O$ in 2D. This approximation is valid far from the origin where changes in $U$ over unit length are small.\n\nThe electric field corresponding to this potential field is\n\n$$\n\\mathbf{E}=-\\nabla U=\\left(-\\partial_{x} U,-\\partial_{y} U\\right) \\approx(U(x-1, y)-U(x, y), U(x, y-1)-U(x, y))=\\left(i_{x}(x, y), i_{y}(x, y)\\right)\n$$\n\nfar from the origin, where $i_{x}(x, y), i_{y}(x, y)$ are the horizontal and vertical currents passing through node $(x, y)$. (Note that the current horizontal current is different to the left vs. to the right of the node, and similarly for the vertical current, but the difference is negligible for $N \\gg 1$.)\n\nThe current $i=\\sqrt{i_{x}^{2}+i_{y}^{2}}$ at a distance $r \\gg 1$ away from the origin is given by Gauss's law:\n\n$$\n2 \\pi r i(r) \\approx I\n$$\n\nso at $(N, 0)$ we have\n\n$$\ni_{x}(N, 0)=i(N) \\approx \\frac{I}{2 \\pi N}\n$$\n\nThe difference between the entering and exiting horizontal currents at $(N, 0)$ is approximately\n\n$$\n-\\partial_{x} i_{x}(N, 0)=\\frac{I}{2 \\pi N^{2}}\n$$\n\nThis difference is directed equally into the vertical resistors adjacent to $(N, 0)$, so the final answer is\n\n$$\n\\frac{I}{4 \\pi N^{2}}=1.488 \\times 10^{-4} \\mathrm{~A}\n$$""]"	['$1.488 \\times 10^{-4}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	A	Numerical	2e-2	Open-ended	Electromagnetism	Physics	English
+56	"Suppose we have a non-ideal gas, and in a certain volume range and temperature range, it is found to satisfy the state relation
+$$
+p=A V^{\alpha} T^{\beta}
+$$
+
+where $A$ is a constant, $\alpha=-\frac{4}{5}$ and $\beta=\frac{3}{2}$, and the other variables have their usual meanings. Throughout the problem, we will assume to be always in that volume and temperature range.
+
+Assume that $\gamma=\frac{C_{p}}{C_{V}}$ is found to be constant for this gas ( $\gamma$ is independent of the state of the gas), where $C_{p}$ and $C_{v}$ are the heat capacities at constant pressure and volume, respectively. What is the minimum possible value for $\gamma$ ?"	['We claim that the conditions given uniquely determine $\\gamma$.\n\nThe fundamental thermodynamic relation gives:\n\n$$\n\\mathrm{d} U=T \\mathrm{~d} S-p \\mathrm{~d} V\n$$\n\nSo\n\n$$\n\\left(\\frac{\\partial U}{\\partial V}\\right)_{T}=T\\left(\\frac{\\partial S}{\\partial V}\\right)_{T}-p=T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}-p\n$$\n\nwhere we have used a Maxwell relation.\n\n\n\n$$\n\\mathrm{d} U=\\left(\\frac{\\partial U}{\\partial T}\\right)_{V} \\mathrm{~d} T+\\left(\\frac{\\partial U}{\\partial V}\\right)_{T} \\mathrm{~d} V=C_{V} \\mathrm{~d} T+\\left(T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}-p\\right) \\mathrm{d} V\n$$\n\nWe have\n\n$$\nC_{p}=\\left(\\frac{\\partial U}{\\partial T}\\right)_{p}+p\\left(\\frac{\\partial V}{\\partial T}\\right)_{p}=C_{V}+T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}\\left(\\frac{\\partial V}{\\partial T}\\right)_{p}\n$$\n\nRearranging, gives\n\n$$\nC_{V}=\\frac{T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}\\left(\\frac{\\partial V}{\\partial T}\\right)_{p}}{\\gamma-1}\n$$\n\nFrom the symmetry of mixed second partial derivatives, we know\n\n$$\n\\left(\\frac{\\partial C_{V}}{\\partial V}\\right)_{T}=\\left(\\frac{\\partial^{2} U}{\\partial T \\partial V}\\right)=\\left(\\frac{\\partial}{\\partial T}\\left(\\frac{\\partial U}{\\partial V}\\right)_{T}\\right)_{V}=\\frac{\\partial}{\\partial T}\\left(T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}-p\\right)=\\left(\\frac{\\partial^{2} p}{\\partial T^{2}}\\right)_{V}\n$$\n\nPlugging our expression for $C_{V}$ into here, and plugging in the equation of state, we can solve for $\\gamma$ to get\n\n$$\n\\gamma=\\frac{\\alpha+\\beta}{\\alpha(1-\\beta)}=\\frac{7}{4}\n$$']	['$\\frac{7}{4}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	0	Open-ended	Thermodynamics	Physics	English
+57	"The coin flip has long been recognized as a simple and unbiased method to randomly determine the outcome of an event. In the case of an ideal coin, it is well-established that each flip has an equal $50 \%$ chance of landing as either heads or tails.
+However, coin flips are not entirely random. They appear random to us because we lack sufficient information about the coin's initial conditions. If we possessed this information, we would always be able to predict the outcome without needing to flip the coin. For an intriguing discussion on why this observation is significant, watch this video by Vsauce.
+
+Now, consider a scenario where a coin with uniform density and negligible width is tossed directly upward from a height of $h=0.75 \mathrm{~m}$ above the ground. The coin starts with its heads facing upward and is given an initial vertical velocity of $v_{y}=49 \mathrm{~m} / \mathrm{s}$ and a positive angular velocity of $\omega=\pi \mathrm{rad} / \mathrm{s}$. What face does the coin display upon hitting the ground? Submit $\mathbf{0}$ for heads and $\mathbf{1}$ for tails. You only have one attempt for this problem. Assume the floor is padded and it absorbs all of the coin's energy upon contact. The radius of the coin is negligible."	['We have the following quadratic:\n$$\n\\begin{aligned}\nx & =x_{0}+v_{0} t+\\frac{1}{2} a t^{2} \\\\\n0 & =0.75+49 t-4.9 t^{2} \\\\\nt & =-0.01,10.01\n\\end{aligned}\n$$\n\nThe first solution is extraneous so $t=10.01$ is correct. Now, $\\theta=\\omega t \\approx 10 \\pi$. As one full rotation is $\\phi=2 \\pi$, then the coin performs 5 full rotations before landing on the ground. This means the answer is 0 , or heads.']	['0']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	0	Open-ended	Mechanics	Physics	English
+58	Suppose all cars on a (single-lane) highway are identical. Their length is $l=4 \mathrm{~m}$, their wheels have coefficients of friction $\mu=0.7$, and they all travel at speed $v_{0}$. Find the $v_{0}$ which maximizes the flow rate of cars (i.e. how many cars travel across an imaginary line per minute). Assume that they need to be able to stop in time if the car in front instantaneously stops. Disregard reaction time.	['Suppose the maximum speed is $v^{\\prime}$. Notice that $a=\\mu g$ so it takes a time $t^{\\prime}=\\frac{v^{\\prime}}{\\mu g}$ amount of time to stop. This means the car will travel a distance of\n$$\nd=\\frac{1}{2} \\mu g\\left(\\frac{v^{\\prime}}{\\mu g}\\right)^{2}=\\frac{v^{\\prime 2}}{2 \\mu g}\n$$\n\nThus for every distance $d+l=l+\\frac{v^{\\prime 2}}{2 \\mu g}$ there is a car. This means in unit time $t$, there will be $N=\\frac{d+l}{v}$ cars that passes through the line. To maximize the flow rate, we need to maximize\n\n$$\n\\begin{gathered}\nf\\left(v^{\\prime}\\right)=\\frac{l}{v^{\\prime}}+\\frac{v^{\\prime}}{2 \\mu g} \\\\\nf^{\\prime}=\\frac{-l}{v^{\\prime 2}}+\\frac{1}{2 \\mu g}=0 \\Rightarrow v^{\\prime}=\\sqrt{2 \\mu g l}=7.41 \\mathrm{~m} / \\mathrm{s}\n\\end{gathered}\n$$']	['$7.41$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m/s	Numerical	1e-2	Open-ended	Mechanics	Physics	English
+59	"In a resource-limited ecological system, a population of organisms cannot keep growing forever (such as lab bacteria growing inside culture tube). The effective growth rate $g$ (including
+
+contributions from births and deaths) depends on the instantaneous abundance of resource $R(t)$, which in this problem we will consider the simple case of linear-dependency:
+
+$$
+\frac{\mathrm{d}}{\mathrm{d} t} N=g(R) N=\alpha R N
+$$
+
+where $N(t)$ is the population size at time $t$. The resources is consumed at a constant rate $\beta$ by each organism:
+
+$$
+\frac{\mathrm{d}}{\mathrm{d} t} R=-\beta N
+$$
+
+Initially, the total amount of resources is $R_{0}$ and the population size is $N_{0}$. Given that $\alpha=10^{-9}$ resourceunit $^{-1} \mathrm{~s}^{-1}, \beta=1$ resource-unit/s, $R_{0}=10^{6}$ resource-units and $N_{0}=1$ cell, find the total time it takes from the beginning to when all resources are depleted (in hours)."	['We can find the analytical solution for the following set of two ODEs describing the populationresource dynamics:\n\n$$\n\\frac{d N}{d t}=\\alpha R N,\n\\tag{1}\n$$\n$$\n\\frac{d R}{d t}=-\\beta N .\n\\tag{2}\n$$\n\nDivide Eq.(1) by Eq.(2) on both sides, we get a direct relation between the population size $N(t)$ and the amount of resource $R(t)$ :\n\n$$\n\\frac{d N}{d R}=-\\frac{\\alpha}{\\beta} R \\Longrightarrow N=N_{0}+\\frac{\\alpha}{2 \\beta}\\left(R_{0}^{2}-R^{2}\\right)\n$$\n\nPlug this in Eq.(2), we obtain the total time $T$ it takes from the beginning to when the resource is depleted:\n\n$$\n\\begin{aligned}\n\\frac{d R}{d t}=-\\frac{\\alpha}{2}\\left[\\left(\\frac{2 \\beta}{\\alpha} N_{0}+R_{0}^{2}\\right)-R^{2}\\right] \\Longrightarrow & \\Longrightarrow \\\\\n\\left.\\right|_{R=0} & =\\frac{2}{\\alpha} \\int_{0}^{R_{0}} d R\\left[\\left(\\frac{2 \\beta}{\\alpha} N_{0}+R_{0}^{2}\\right)-R^{2}\\right]^{-1} \\\\\n& =\\frac{2}{\\alpha \\sqrt{\\frac{2 \\beta}{\\alpha} N_{0}+R_{0}^{2}}} \\operatorname{arctanh}\\left(\\frac{R_{0}}{\\sqrt{\\frac{2 \\beta}{\\alpha} N_{0}+R_{0}^{2}}}\\right) .\n\\end{aligned}\n$$\n\nUse the given numerical values, we arrive at $\\left.t\\right|_{R=0} \\approx 7594.3 \\mathrm{~s} \\approx 2.1095 \\mathrm{hrs}$.']	['2.1095']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	h	Numerical	2e-1	Open-ended	Modern Physics	Physics	English
+60	An incandescent lightbulb is connected to a circuit which delivers a maximum power of 10 Watts. The filament of the lightbulb is made of Tungsten and conducts electricity to produce light. The specific heat of Tungsten is $c=235 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{kg})$. If the circuit is alternating such that the temperature inside the lightbulb fluctuates between $T_{0}=3000^{\circ} \mathrm{C}$ and $T_{1}=3200^{\circ} \mathrm{C}$ at a frequency of $\omega=0.02 \mathrm{~s}^{-1}$, estimate the mass of the filament.	['This problem was voided from the test because we gave no method for energy dissipation. Therefore, there was ambiguity and energy would be constantly fed to the lightbulb making it hotter and hotter.\nHowever, the problem can still be solved. Here is a solution written by one of our contestants\n\n\n\nGuangyuan Chen. The circuit continuously delivers sinusoidal power to the lightbulb. In order to maintain a quasi-equilibrium state, there must be some form of heat loss present in the system. We will assume that the main source of heat loss is through radiation, though we will see that the precise form of heat loss is largely inconsequential.\n\nLet the power delivered by the circuit be in the form $P_{d}=P_{0} \\sin ^{2} \\Omega t$. Let the power loss due to the radiation be $P_{l}=-k T^{4}$ where $k$ is some constant and $T$ is the temperature of the filament. We can write:\n\n$$\nm c \\frac{\\mathrm{d} T}{\\mathrm{~d} t}=P_{0} \\sin ^{2} \\Omega t-k T^{4}\n$$\n\nThis is a differential equation that cannot be solved by hand. However, since the deviation in temperature from the equilibrium of roughly $100^{\\circ} \\mathrm{C}$ is much smaller than the equilibrium temperature of roughly $3100^{\\circ} \\mathrm{C}$, we can do a first order approximation of the differential equation. The temperature thus varies sinusoidally with equilibrium temperature $T_{\\text {equi }}=3100^{\\circ} \\mathrm{C}$ and amplitude $T_{a}=100^{\\circ} \\mathrm{C}$. Since the average power delivered is $\\frac{1}{2} P_{0}$, at equilibrium temperature, we have:\n\n$$\n\\begin{gathered}\n\\frac{1}{2} P_{0}=k T_{\\text {equi }}^{4} \\\\\nk=\\frac{P_{0}}{2 T_{e q u i}^{4}}\n\\end{gathered}\n$$\n\nWe rewrite equation 1 with the approximation $T=T_{\\text {equi }}+\\Delta T$ and with the substitution from equation 3.\n\n$$\n\\begin{gathered}\nm c \\frac{\\mathrm{d} \\Delta T}{\\mathrm{~d} t}=P_{0} \\sin ^{2} \\Omega t-\\frac{P_{0}}{2 T_{\\text {equi }}^{4}}\\left(T_{\\text {equi }}+\\Delta T\\right)^{4} \\\\\nm c \\frac{\\mathrm{d} \\Delta T}{\\mathrm{~d} t} \\approx P_{0} \\sin ^{2} \\Omega t-\\frac{P_{0}}{2}\\left(1+4 \\frac{\\Delta T}{T_{\\text {equi }}}\\right) \\\\\nm c \\frac{\\mathrm{d} \\Delta T}{\\mathrm{~d} t} \\approx-\\frac{P_{0}}{2}\\left(\\cos 2 \\Omega t+4 \\frac{\\Delta T}{T_{\\text {equi }}}\\right)\n\\end{gathered}\n$$\n\nAt this point we can see that regardless of the physical mechanism of heat loss, a linear approximation can always be made for sufficiently small variations in temperature which results in an equation of the same form as equation 6 . To solve this equation, we can guess the following solution:\n\n$$\n\\Delta T=A \\sin 2 \\Omega t+B \\cos 2 \\Omega t\n$$\n\nStrictly speaking, we also need to include a term $C e^{-\\lambda t}$. However, for sufficiently long times approaching quasi-equilibrium, this term will go to zero, so we need not include it here. Substituting equation 7 into equation 6 :\n\n$$\n2 m c \\Omega(A \\cos 2 \\Omega t-B \\sin 2 \\Omega t)=-\\frac{P_{0}}{2}\\left(\\cos 2 \\Omega t+\\frac{4}{T_{\\text {equi }}}(A \\sin 2 \\Omega t+B \\cos 2 \\Omega t)\\right)\n$$\n\nEquating the coefficients of the sin and cos:\n\n$$\n\\begin{gathered}\n2 m c \\Omega A=-\\frac{P_{0}}{2}\\left(1+\\frac{4 B}{T_{e q u i}}\\right) \\\\\n-2 m c \\Omega B=-\\frac{2 P_{0} A}{T_{\\text {equi }}}\n\\end{gathered}\n$$\n\n\n\nEquations 9 and 10 can be solved simultaneously to give\n\n$$\n\\begin{aligned}\nA & =-\\frac{m c \\Omega T_{e q u i} P_{0}}{4\\left(\\left(m c \\Omega T_{e q u i}\\right)^{2}+P_{0}^{2}\\right)} T_{e q u i} \\\\\nB & =-\\frac{P_{0}^{2}}{4\\left(\\left(m c \\Omega T_{e q u i}\\right)^{2}+P_{0}^{2}\\right)} T_{e q u i}\n\\end{aligned}\n$$\n\nEquation 7 can be written in the form $T_{a} \\sin (2 \\Omega t+\\phi)$. To find $T_{a}$, we simply have:\n\n$$\nT_{a}=\\sqrt{A^{2}+B^{2}}=\\frac{P_{0}}{4 \\sqrt{\\left(m c \\Omega T_{e q u i}\\right)^{2}+P_{0}^{2}}} T_{e q u i}\n$$\n\nRearranging for $m$, we have:\n\n$$\nm=\\frac{P_{0}}{c \\Omega T_{e q u i}} \\sqrt{\\left(\\frac{T_{e q u i}}{4 T_{a}}\\right)^{2}-1}\n$$\n\nThe final step is to relate $\\Omega$ and $\\omega$. Since $T$ oscillates with angular frequency $2 \\Omega$, its period of oscillation is $\\frac{2 \\pi}{2 \\Omega}$. This is equal to $\\frac{1}{\\omega}$, hence we have:\n\n$$\n\\Omega=\\pi \\omega\n$$\n\nSubstituting into equation 14, we get our final numerical answer.\n\n$$\nm=\\frac{P_{0}}{\\pi c \\omega T_{e q u i}} \\sqrt{\\left(\\frac{T_{e q u i}}{4 T_{a}}\\right)^{2}-1}=1.68 \\times 10^{-3} \\mathrm{~kg}\n$$']	['$1.68 \\times 10^{-3}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	kg	Numerical	2e-5	Open-ended	Thermodynamics	Physics	English
+61	In hyperdrive, Spaceship-0 is relativistically moving at the velocity $\frac{1}{3} c$ with respect to reference frame $R_{1}$, as measured by Spaceship-1. Spaceship-1 is moving at $\frac{1}{2} c$ with respect to reference frame $R_{2}$, as measured by Spaceship-2. Spaceship- $k$ is moving at speed $v_{k}=\frac{k+1}{k+3} c$ with respect to reference frame $R_{k+1}$. The speed of Spaceship-0 with respect to reference frame $R_{20}$ can be expressed as a decimal fraction of the speed of light which has only $x$ number of 9 s following the decimal point (i.e., in the form of $0 . \underbrace{99 \ldots 9}_{x \text { times }} c)$. Find the value of $x$.	"[""Let us define the rapidity as\n$$\n\\tanh \\phi \\equiv \\beta=\\frac{v}{c}\n$$\n\nwhere tanh is the hyperbolic tangent function. Let $\\beta_{1}=\\tanh \\phi_{1}$ and $\\beta_{2}=\\tanh \\phi_{2}$. If we add $\\beta_{1}$ and $\\beta_{2}$ using the relativistic velocity addition formula, we find that\n\n$$\n\\beta=\\frac{\\beta_{1}+\\beta_{2}}{1-\\beta_{1} \\beta_{2}}=\\frac{\\tanh \\phi_{1}+\\tanh \\phi_{2}}{1+\\tanh \\phi_{1} \\tanh \\phi_{2}}=\\tanh \\left(\\phi_{1}+\\phi_{2}\\right)\n$$\n\nWe can then rewrite the problem as\n\n$$\nv_{f}=\\tanh \\left(\\operatorname{arctanh} \\frac{1}{3}+\\operatorname{arctanh} \\frac{2}{4}+\\cdots+\\operatorname{arctanh} \\frac{20}{22}\\right)\n$$\n\n\n\nUsing the fact that $\\operatorname{arctanh}(\\phi)=\\frac{1}{2} \\ln \\left(\\frac{1+\\phi}{1-\\phi}\\right)$, we can find that\n\n$$\n\\begin{aligned}\nv_{f} & =\\tanh \\left(\\frac{1}{2} \\sum_{k=0}^{19} \\ln \\left(\\frac{1+\\frac{k+1}{k+3}}{1-\\frac{k+1}{k+3}}\\right)\\right)=\\tanh \\left(\\frac{1}{2} \\sum_{k=0}^{19} \\ln (k+2)\\right) \\\\\n& =\\tanh (\\ln \\sqrt{2 \\cdot 3 \\cdot 4 \\cdots 21})=\\tanh (\\ln \\sqrt{21 !})\n\\end{aligned}\n$$\n\nAs $\\tanh \\phi=\\left(e^{\\phi}-e^{-\\phi}\\right) /\\left(e^{\\phi}+e^{-\\phi}\\right)$, then\n\n$$\n\\tanh (\\ln (\\phi))=\\frac{\\phi-\\frac{1}{\\phi}}{\\phi+\\frac{1}{\\phi}}=1-\\frac{2}{\\phi^{2}+1} \\Longrightarrow v_{f}=1-\\frac{2}{21 !+1}\n$$\n\nThis implies 19 zeros, but you can also use Stirlings approximation to further approximate the factorial.\n\nAlternate 1: Define $u_{k}$ as Spaceship 0's velocity in frame $R_{k}$, given $c=1$. The recurrence relation from the relativistic velocity addition formula is: $u_{k+1}=\\frac{u_{k}(k+3)+(k+1)}{u_{k}(k+1)+(k+3)}$, starting with $u_{1}=\\frac{1}{3}$. The relation can be simplified as: $u_{k+1}=1+\\frac{2\\left(u_{k}-1\\right)}{u_{k}(k+1)+(k+3)}$. Introducing $v_{k}=u_{k}-1$, $v_{k+1}=v_{k} \\frac{2}{v_{k}(k+1)+(2 k+4)}$. Further simplifying with $w_{k}=\\frac{1}{v_{k}}$, we get $w_{k+1}=w_{k}(k+2)+\\frac{k+1}{2}$. By setting $x_{k}=w_{k}+c$, we find $c=-\\frac{1}{2}$ simplifies to $x_{k+1}=x_{k}(k+2)$. This gives $x_{k}=$ $(k+1)(k)(k-1) \\ldots(4)(3) x_{1}$ and using the initial condition $x_{1}=\\frac{1}{u_{1}-1}+\\frac{1}{2}=1$, we obtain $x_{k}=\\frac{(k+1) !}{2}$. Consequently, $u_{k}=\\frac{(k+1) !-1}{(k+1) !+1}$ and substituting $k=20, u_{20} \\approx 1-3.9 \\times 10^{-20}$, yielding 19 significant digits.\n\nAlternate 2: Let $l=k+2$, then $u_{l}=\\frac{l-1}{l+1}$. Let $u_{m}=\\frac{m-1}{m+1}$. Then you can find that the velocity addition of any $l, m$ will be $\\frac{m l-1}{m l+1}$. Using this identity, we can use recurrence to find that $u_{k}=\\frac{(k+1) !-1}{(k+1) !+1}$.""]"	['19']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Numerical	0	Open-ended	Mechanics	Physics	English
+62	The path of an asteroid that comes close to the Earth can be modeled as follows: neglect gravitational effects due to other bodies, and assume the asteroid comes in from far away with some speed $v$ and lever arm distance $r$ to Earth's center. On January 26, 2023, a small asteroid called 2023 BU came to a close distance of $3541 \mathrm{~km}$ to Earth's surface with a speed of $9300 \mathrm{~m} / \mathrm{s}$. Although BU had a very small mass estimated to be about $300,000 \mathrm{~kg}$, if it was much more massive, it could have hit the Earth. How massive would BU have had to have been to make contact with the Earth? Express your answer in scientific notation with 3 significant digits. Use $6357 \mathrm{~km}$ as the radius of the Earth. The parameters that remain constant when the asteroid mass changes are $v$ and $r$, where $v$ is the speed at infinity and $r$ is the impact parameter.	"[""Let $v_{1}=9300 \\mathrm{~m} / \\mathrm{s}, d=3541 \\mathrm{~km}$, and $m=300,000 \\mathrm{~kg}$, and let $M$ and $R$ be the Earth's mass and radius.\n\nFirst we find $v$ and $r$. We use the reference frame of the Earth, where the asteroid has reduced mass $\\mu=\\frac{M m}{M+m}$ and the Earth has mass $M+m$. Then by energy and angular momentum conservation, we have\n\n$$\n\\mu v r=\\mu v_{1}(R+d)\n$$\n\n\n\nand\n\n$$\n\\frac{1}{2} \\mu v^{2}=\\frac{1}{2} \\mu v_{1}^{2}-\\frac{G M m}{R+d}\n$$\n\nWe solve for\n\n$$\nv=\\sqrt{2 G(M+m) \\cdot \\frac{R+d}{r^{2}-(R+d)^{2}}}\n$$\n\nso\n\n$$\nv_{1}=\\sqrt{\\frac{2 G(M+m)}{R+d} \\cdot \\frac{r^{2}}{r^{2}-(R+d)^{2}}},\n$$\n\nand we compute $r=37047 \\mathrm{~km}$ and $v=2490 \\mathrm{~m} / \\mathrm{s}$.\n\nNow we consider when the asteroid is massive enough to touch the Earth. We let $m^{\\prime}$ and $\\mu^{\\prime}$ be the mass of the asteroid and its reduced mass, and using a similar method to above, we arrive at\n\n$$\nv=\\sqrt{2 G\\left(M+m^{\\prime}\\right) \\cdot \\frac{R}{r^{2}-R^{2}}}\n$$\n\nso we can solve for $m^{\\prime}=3.74 \\times 10^{24} \\mathrm{~kg}$.""]"	['$3.74 \\times 10^{24}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	kg	Numerical	1e23	Open-ended	Mechanics	Physics	English
+63	"Pegasi and Betelgeuse are two star systems that can undergo a supernova. Betelgeuse is 548 light-years away from Earth and IK Pegasi is 154 light-years away from Earth. Assume that the two star systems are 500 light-years away from each other.
+Astronomers on Earth observe that the two star systems undergo a supernova explosion 300 years apart. A spaceship, the OPhO Galaxia Explorer which left Earth in an unknown direction before the first supernova observes both explosions occur simultaneously. Assume that this spaceship travels in a straight line at a constant speed $v$. How far are the two star systems according to the OPhO Galaxia Explorer at the moment of the simultaneous supernovae? Answer in light-years.
+
+Note: Like standard relativity problems, we are assuming intelligent observers that know the finite speed of light and correct for it."	"[""For any inertial observer, define the 4-distance between two events as $\\Delta s^{\\mu}=(c \\Delta t, \\Delta \\mathbf{x})$, where $\\Delta t$ and $\\Delta \\mathrm{x}$ are the temporal and spatial intervals measured by the observer. By the properties of 4 -vectors, the following quantity is Lorentz-invariant:\n$$\n\\Delta s^{\\mu} \\Delta s_{\\mu}=c^{2} \\Delta t^{2}-\\|\\Delta \\mathbf{x}\\|^{2}\n$$\n\nIn the spaceship's frame, this is equal to $\\Delta s^{\\mu} \\Delta s_{\\mu}=-\\left\\|\\Delta \\mathbf{x}^{\\prime}\\right\\|^{2}$, since the two supernovas are simultaneous; hence\n\n$$\n\\left\\|\\Delta \\mathbf{x}^{\\prime}\\right\\|=\\sqrt{\\|\\Delta \\mathbf{x}\\|^{2}-c^{2} \\Delta t^{2}}\n$$\n\nSince the observers have already taken into account the delay due to the nonzero speed of light, $c \\Delta t=300 \\mathrm{ly},\\|\\Delta \\mathbf{x}\\|=500 \\mathrm{ly}$, and\n\n$$\n\\left\\|\\Delta \\mathbf{x}^{\\prime}\\right\\|=\\sqrt{\\|\\Delta \\mathbf{x}\\|^{2}-c^{2} \\Delta t^{2}}=400 l y\n$$""]"	['400']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	light-years	Numerical	0	Open-ended	Modern Physics	Physics	English
+64	A ball of mass $1 \mathrm{~kg}$ is thrown vertically upwards and it faces a quadratic drag with a terminal velocity of $20 \mathrm{~m} / \mathrm{s}$. It reaches a maximum height of $30 \mathrm{~m}$ and falls back to the ground. Calculate the energy dissipated until the point of impact (in J).	"[""If we suppose that the magnitude of quadratic drag is $F_{\\mathrm{drag}}=c v^{2}$ for some constant $c$, then when the ball is falling downwards at terminal velocity $v_{t}$ this upwards drag force must cancel gravity to provide zero net force:\n$$\nm g=c v_{t}^{2} \\Longrightarrow c=\\frac{m g}{v_{t}^{2}}\n$$\n\nso we can rewrite our equations of motion in terms of terminal velocity. In the ascending portion of the ball's trajectory, suppose that the ball's speed changes from $v$ to $v+d v$ after traveling an infinitesimal distance from $h$ to $h+d h$. Then:\n\n$d($ kinetic energy $)=-d($ potential energy and energy lost to drag $) \\Longrightarrow m v d v=-\\frac{m g v^{2}}{v_{t}^{2}} d h-m g d h$\n\nwhere we used chain rule and $d W=F \\cdot d x$. This is a separable differential equation for velocity in terms of height (!), which we can solve:\n\n$$\n\\int \\frac{v d v}{1+\\frac{v^{2}}{v_{t}^{2}}}=\\int-g d h+\\text { constant }=\\text { constant }-g h\n$$\n\nWe u-sub the entire denominator $\\left(u=1+v^{2} / v_{t}^{2} \\Longrightarrow d u=2 v / v_{t}^{2} d v\\right)$, which nicely cancels the numerator. Our left-hand integral is\n\n$$\n\\frac{v_{t}^{2}}{2} \\int \\frac{d u}{u}=\\frac{v_{t}^{2}}{2} \\ln \\left|1+\\frac{v^{2}}{v_{t}^{2}}\\right|=\\text { constant }-g h\n$$\n\nNow to find the constant! At our maximum height $h_{0}=30 \\mathrm{~m}$, the speed and left-hand side are zero, so the constant must be $g h_{0}$. Then we can find the initial speed of the projectile by substituting $h=0$, from which:\n\n$$\nv_{\\text {initial }}=v_{t} \\sqrt{e^{\\frac{2 g h_{0}}{v_{t}^{2}}}-1}\n$$\n\nWe can use a similar argument for the downwards trajectory, albeit with drag pointing upwards. We should get that the final speed immediately before impact is\n\n$$\nv_{\\text {final }}=v_{t} \\sqrt{1-e^{\\frac{-2 g h_{0}}{v_{t}^{2}}}}\n$$\n\nfrom which the dissipated energy is the difference in final and initial kinetic energies, or\n\n$$\n\\frac{1}{2} m\\left(v_{\\text {final }}^{2}-v_{\\text {initial }}^{2}\\right) \\approx 515.83 \\text { joules }\n$$""]"	['515.83']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	joules	Numerical	2e-1	Open-ended	Mechanics	Physics	English
+65	Two parallel square plates of side length $1 \mathrm{~m}$ are placed a distance $30 \mathrm{~cm}$ apart whose centers are at $(-15 \mathrm{~cm}, 0,0)$ and $(15 \mathrm{~cm}, 0,0)$ have uniform charge densities $-10^{-6} \mathrm{C} / \mathrm{m}^{2}$ and $10^{-6} \mathrm{C} / \mathrm{m}^{2}$ respectively. Find the magnitude of the component of the electric field perpendicular to axis passing through the centers of the two plates at $(10 \mathrm{~cm}, 1 \mathrm{~mm}, 0)$.	['By symmetry, the electric field due to the portion of the plates between $y=50 \\mathrm{~cm}$ and $y=-49.8 \\mathrm{~cm}$ will cancel out. We only need to consider the electric field from two $2 \\mathrm{~mm}$ wide strips of charge, which are small enough to be approximated as wires with charge per unit length $\\lambda=\\sigma w= \\pm 2 \\times 10^{-9} \\mathrm{C} / \\mathrm{m}^{2}$ at a distance $y=50 \\mathrm{~cm}$ away. The y-component of the electric field from these wires is then\n$$\n\\begin{aligned}\n& E_{y}=\\frac{0.5 \\lambda}{4 \\pi \\epsilon_{0}} \\int_{-0.5}^{0.5}\\left(\\frac{1}{\\left(z^{2}+0.5^{2}+0.05^{2}\\right)^{\\frac{3}{2}}}-\\frac{1}{\\left(z^{2}+0.5^{2}+0.25^{2}\\right)^{\\frac{3}{2}}}\\right) \\mathrm{d} z \\\\\n& E_{y}=\\frac{0.5 \\lambda}{4 \\pi \\epsilon_{0}}\\left(\\frac{z}{\\left(0.5^{2}+0.05^{2}\\right) \\sqrt{z^{2}+0.5^{2}+0.05^{2}}}-\\frac{z}{\\left(0.5^{2}+0.25^{2}\\right) \\sqrt{z^{2}+0.5^{2}+0.25^{2}}}\\right) \\\\\n& E_{y}=\\frac{0.5 \\lambda}{4 \\pi \\epsilon_{0}}\\left(\\frac{1}{\\left(0.5^{2}+0.05^{2}\\right) \\sqrt{0.5^{2}+0.5^{2}+0.05^{2}}}-\\frac{1}{\\left(0.5^{2}+0.25^{2}\\right) \\sqrt{0.5^{2}+0.5^{2}+0.25^{2}}}\\right)=11.9 \\mathrm{~N} / \\mathrm{C}\n\\end{aligned}\n$$']	['$11.9$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	N/C	Numerical	1e-1	Open-ended	Electromagnetism	Physics	English
+66	"A space elevator consists of a heavy counterweight placed near geostationary orbit, a thread that connects it to the ground (assume this is massless), and elevators that run on the threads (also massless). The mass of the counterweight is $10^{7} \mathrm{~kg}$. Mass is continuously delivered to the counterweight at a rate of $0.001 \mathrm{~kg} / \mathrm{s}$. The elevators move upwards at a rate of $20 \mathrm{~m} / \mathrm{s}$. Assume there are many elevators, so their discreteness can be neglected. The elevators are massless. The counterweight orbits the Earth.
+
+
+Find the minimum possible displacement radially of the counterweight. Specify the sign."	['As the orbit is geostationary, we can balance forces to write that\n$$\n\\frac{G M_{E}}{R_{\\mathrm{GS}}^{2}}=\\omega^{2} R_{\\mathrm{GS}} \\Longrightarrow R_{\\mathrm{GS}}=\\left(\\frac{G M_{E} T^{2}}{4 \\pi}\\right)^{1 / 3}\n$$\nThe total mass of masses $=1790 \\mathrm{~kg} \\ll 10^{6} \\mathrm{~kg}$ which means the displacements are small. The total gravity of masses is\n$$\n\\int_{R_{E}}^{R_{E s}} \\frac{G M_{E}}{x^{2}}(\\lambda d x)=\\operatorname{GME} \\lambda\\left(\\frac{1}{R_{E}}-\\frac{1}{R_{\\cos }}\\right)=2650 \\mathrm{~N}\n$$\nwhere $\\lambda=5 \\cdot 10^{-5} \\mathrm{~kg} / \\mathrm{m}$. The total centrifugal of masses is\n$$\n\\int_{R_{E}}^{R_{E}} \\frac{4 \\pi^{2}}{T^{2}} \\times(\\lambda d x)=\\frac{2 \\pi^{2}}{T^{2}} \\lambda\\left(R_{E s}^{2}-R_{E}^{2}\\right)=230 \\mathrm{~V}\n$$\nHence, the total force is $2420 \\mathrm{~N}$. The outwards force is just\n$$\nF_{\\text {out }}=\\omega^{2} a M-\\frac{\\mathrm{d}}{\\mathrm{d} x}\\left(\\frac{G M_{E}}{x^{2}}\\right) a M\n$$\nwhere $a$ is the horizontal displacement which is much less than $R_{E}, R_{\\mathrm{GS}}$. Hence, this can be rewritten as\n$$\nF_{\\text {out }}=\\left(\\frac{4 \\pi^{2}}{T^{2}}-2 \\frac{G M_{E}}{R_{\\mathrm{GS}}^{2}}\\right) a M=2420 \\mathrm{~N}\n$$\nThis implies that $a=15.21 \\mathrm{~km}$.']	['$15.21$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	km	Numerical	1e-2	Open-ended	Mechanics	Physics	English
+67	Consider a spherical shell of thickness $\delta=0.5 \mathrm{~cm}$ and radius $R=5 \mathrm{~cm}$ made of an Ohmic material with resistivity $\rho=10^{-7} \Omega \mathrm{m}$. A spherical laser source with a tuned frequency of $f_{0}$ and intensity $I_{0}=10^{5} \mathrm{~W} / \mathrm{m}^{2}$ is placed at the center of the shell and is turned on. Working in the limit $\delta \ll \frac{c}{f_{0}} \ll R$, approximate the initial average power dissipated by the shell. Neglect inductance.	"[""Note that the proper treatment of this problem requires more advanced electromagnetism. The following might seem sketchy, but this will do. We first consider the case of a thin plane made of an ideal conductor placed at $z=0$. A monochromatic electromagnetic wave with an $E$ field: $\\vec{E}_{i}(t, z)=E_{0} \\cos (k z-\\omega t) \\hat{x}$ is incident on the plane. The corresponding $B$ field is of course $\\vec{B}_{i}(t, z)=\\frac{E_{0}}{c} \\cos (k z-\\omega t) \\hat{y}$. We expect a current density $\\hat{K}(t)=K(t) \\hat{x}$ to form in the direction of the $E$ field, as the electric field accelerates the charges on the sheet $-K(t)$ will oscillate with the $E$ field. We know that perfect conductors completely reflect electromagnetic waves, so we expect no EM field in the $z>0$ region. The total EM field is a combination of the incident wave and the wave generated by the oscillating surface charges, so we are to believe that the latter provides just the right EM wave to cancel out $\\vec{B}_{i}$ and $\\vec{E}_{i}$ in the $z>0$ region. Consider an instance where the EM waves have an orientation shown in the figure below. It's easy to verify that the magnetic field generated by a uniform sheet of current pointing out of the page is given as shown. Since the electric wave generated by $\\vec{K}(t)$ must cancel the $\\vec{E}_{i}$ pointing out of page in $z>0$, we get a poynting vector (for the wave generated by $\\vec{K}$ ) in the $\\hat{z}$ direction in $z>0$. By symmetry, this means that the poynting vector points in the $-\\hat{z}$ direction in $z<0$. Such simple symmetry arguments fully determine the wave generated by an oscillating sheet of charge: $\\left|\\vec{E}_{k}(t, z)\\right|=\\left|\\vec{E}_{i}(t, z)\\right|$, $\\left|\\vec{B}_{k}(t, z)\\right|=\\left|\\vec{B}_{i}(t, z)\\right|$ with the orientations shown below.\n<img_4292>\n\nWe therefore get that the total magnetic field in the $z<0$ region is $\\vec{B}=2 \\vec{B}_{i}(t, z)$. Let us now take an Ampere loop as shown in the figure below:\n\n<img_4285>\n\n\n\nWe have:\n$$\nT_{\\epsilon} \\equiv \\oint_{\\Gamma_{\\epsilon}} \\vec{B}(t,-\\epsilon / 2) \\cdot d \\vec{l}=\\mu_{0} I_{e n c}+\\frac{1}{c^{2}} \\frac{\\partial}{\\partial t} \\int_{S_{\\epsilon}} \\vec{E}(t, z) \\cdot d \\vec{A}\n$$\nwhere $S_{\\epsilon}$ is the surface bounded by $\\Gamma_{\\epsilon}$. Since $\\vec{E}=0$ everywhere and $I_{e n c}=K(t) l$,\n$$\n\\lim _{\\epsilon \\rightarrow 0} T_{\\epsilon}=\\frac{2 E_{0} l}{c} \\cos (\\omega t)=\\mu_{0} K(t) l\n$$\nHence, we have $\\vec{K}(t)=\\frac{2 E_{0}}{\\mu_{0} c} \\cos (\\omega t) \\hat{x}$. The key find in this long introduction is that an oscillating sheet of current $\\vec{K}(t)=K_{0} \\cos (\\omega t) \\hat{x}$ generates an EM wave with an electric field $\\vec{E}_{K}(t, 0)=-\\frac{\\mu_{0} c}{2} \\vec{K}(t)$ near the sheet (at $z=0$ ). We finally return to the original problem.\n\nAll laser beams have poynting vectors pointing radially outwards, so the $E_{i}$ field must point in the tangential direction to the sphere at all normal incidence. Say the $E_{i}$ field points in the $\\hat{\\theta}$ direction, so that the current in the shell flows from the north pole to the south pole. The condition $\\delta \\ll \\frac{c}{f_{0}} \\ll R$ allows us to neglect attenuation due to skin effects - essentially, the electric field, thus the surface current, is approximately uniform in the shell. The condition $\\frac{c}{f_{0}} \\ll R$ allows us to treat the incidence of the laser beams as an EM wave hitting an Ohmic plane of thickness $\\delta$, so we consider that problem first. Denote the total electric field inside the the plane at time $t$ as $\\vec{E}(t)$. The current density is given by Ohm's law: $\\vec{J}(t)=\\vec{E}(t) / \\rho$. The corresponding surface current is of course $\\vec{K}(t)=\\delta \\vec{E} / \\rho(t)$. Carefully note the distinction between $\\vec{E}$ and $\\vec{E}_{i}$. Since the source wave $\\vec{E}_{i}(t)$ is sinusoidal, we expect $\\vec{E}(t)$ to be sinusoidal as well so that $\\vec{E}_{K}(t)=-\\frac{\\mu_{0} c}{2} \\vec{K}(t)$ applies. The total electric field inside the sheet is a sum of $\\vec{E}_{K}$ and $\\vec{E}_{i}$, thus:\n$$\n\\vec{E}(t)=\\vec{E}_{i}(t)-\\frac{\\mu_{0} c \\delta}{2 \\rho} \\vec{E}(t)\n$$\nRearranging for $\\vec{E}(t)$, we get:\n$$\n\\vec{E}(t)=\\frac{\\vec{E}_{i}(t)}{1+\\mu_{0} c \\delta / 2 \\rho}\n$$\nOn the sphere, $\\vec{E}(t)$ points in the $\\hat{\\theta}$ direction and has the same magnitude everywhere on the sphere. The relevant cross-section will be the lateral surface of a truncated cone, as shown below:\n\n<img_4286>\n\n\n\nThe current through this surface is therefore:\n$$\nI(\\theta, t)=\\frac{\\pi \\delta(2 R+\\delta)\\left|\\vec{E}_{i}(t)\\right|}{\\rho+\\mu_{0} c \\delta / 2} \\sin \\theta\n$$\nThe power dissipated by a volume generated by $\\theta \\sim \\theta+d \\theta$ is\n$$\nd P(t)=I^{2}(\\theta, t) \\frac{\\rho(R+\\delta / 2) d \\theta}{\\pi \\delta(2 R+\\delta) \\sin \\theta}\n$$\nWe integrate this expression from $\\theta=0$ to $\\theta=\\pi$ :\n$$\nP(t)=\\frac{\\pi \\delta \\rho(2 R+\\delta)^{2}}{\\left(\\rho+\\mu_{0} c \\delta / 2\\right)^{2}}\\left|\\vec{E}_{i}(t)\\right|^{2}\n$$\nNow, the incident electric field has the form $\\left|\\vec{E}_{i}(t)\\right|^{2}=\\frac{2 I_{0}}{\\epsilon_{0} c} \\cos ^{2}\\left(2 \\pi f_{0} t\\right)$. The average of this function from $t=0$ to $t=\\frac{1}{f_{0}}$ is $1 / 2$, so our final answer is:\n$$\n\\bar{P}=\\frac{\\pi \\delta \\rho(2 R+\\delta)^{2}}{\\left(\\rho+\\mu_{0} c \\delta / 2\\right)^{2}} \\frac{I_{0}}{\\epsilon_{0} c}\n$$\nwhich turns out to be $2.39078 \\times 10^{-15} \\mathrm{~W}$""]"	['$2.39078 \\times 10^{-15}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	W	Numerical	1e-17	Open-ended	Electromagnetism	Physics	English
+68	"A stable star of radius $R$ has a mass density profile $\rho(r)=\alpha(1-r / R)$. Here, ""stable"" means that the star doesn't collapse under its own gravity. If the internal pressure at the core is provided solely by the radiation of photons, calculate the temperature at the core. Assume the star is a perfect black body and treat photons as a classical ideal gas. Use $R=7 \times 10^{5} \mathrm{~km}$ and $\alpha=3 \mathrm{~g} / \mathrm{cm}^{3}$. Round your answer to the nearest kilokelvin. We treat photons as a classical gas here to neglect any relativistic effects."	"[""For a star that doesn't collapse in on itself, there must be some source of pressure $p(r)$ that balances out the pressure due to gravity. Define the function:\n$$\nm(r)=\\int_{0}^{r} 4 \\pi r^{2} \\rho(r) d r\n$$\ninside the star. The gravitational pressure on an infinitesimal shell of radius $r$ and thickness $d r$ is given by:\n$$\nd p_{g}=-\\frac{G m(r) \\rho(r) 4 \\pi r^{2} d r}{r^{2} 4 \\pi r^{2}}=-\\frac{G m(r) \\rho(r)}{r^{2}} d r\n$$\nHence, the pressure source must provide a pressure gradient $\\frac{d p}{d r}$ given by:\n$$\n\\frac{d p}{d r}=\\frac{G m(r) \\rho(r)}{r^{2}}\n$$\nIf $\\rho(r)=\\alpha(1-r / R)$, then $m(r)=\\frac{1}{3} \\pi \\alpha r^{3}(3 r / R-4)$. We need the pressure gradient:\n$$\n\\frac{d p}{d r}=\\frac{1}{3} \\pi G \\alpha^{2} r(3 r / R-4)(r / R-1)\n$$\nWe integrate this equation from $r=0$ to $R$ with the obvious boundary condition $p(R)=0$. We find that $p(0)=\\frac{5}{36} \\pi G \\alpha^{2} R^{2}$. From the Stefan-Boltzmann law, along with elementary kinetic theory, the pressure due to local radiation is given by $\\frac{4 \\sigma}{3 c} T_{c}^{4}$. Plugging in values gives $T_{c}=26718 \\mathrm{kK}$""]"	['$26718$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	kK	Numerical	1e2	Open-ended	Modern Physics	Physics	English
+69	"In this problem, we consider a simple model for a thermoacoustic device. The device uses heavily amplified sound to provide work for a pump that can then extract heat. Sound waves form standing waves in a tube of radius $0.25 \mathrm{~mm}$ that is closed on both sides, and a two-plate stack is inserted in the tube. A temperature gradient forms between the plates of the stack, and the parcel of gas trapped between the plates oscillates sinusoidally between a maximum pressure of $1.03 \mathrm{MPa}$ and a minimum of $0.97 \mathrm{MPa}$. The gas is argon, with density $1.78 \mathrm{~kg} / \mathrm{m}^{3}$ and adiabatic constant $5 / 3$. The speed of sound is $323 \mathrm{~m} / \mathrm{s}$. The heat pump itself operates as follows:
+The parcel of gas starts at minimum pressure. The stack plates adiabatically compress the parcel of gas to its maximum pressure, heating the gas to a temperature higher than that of the hotter stack plate. Then, the gas is allowed to isobarically cool to the temperature of the hotter stack plate. Next, the plates adiabatically expand the gas back to its minimum pressure, cooling it to a temperature lower than that of the colder plate. Finally, the gas is allowed to isobarically heat up to the temperature of the colder stack plate.
+
+Find the power at which the thermoacoustic heat pump emits heat."	['The efficiency of the heat engine is $\\epsilon=1-\\left(\\frac{P_{2}}{P_{1}}\\right)^{\\frac{\\gamma-1}{\\gamma}}=0.0237$. The parcel oscillates between pressures $P_{1}$ and $P_{2}$ sinusoidally with amplitude $P_{0}=\\frac{P_{1}-P_{2}}{2}$. For a sound wave, the pressure amplitude is $\\rho s_{0} \\omega v$, where $s_{0}$ is the position amplitude and $v$ is the speed of sound.\n\nThen the average power with which the sound wave does work on the plates is\n$$\n\\langle P\\rangle=\\frac{1}{2} \\rho \\omega^{2} s_{0}^{2} A v=\\frac{P_{0}^{2}}{2 \\rho v} A\n$$\nwhere $A$ is the area of each plate. From this, the heat power generated by the engine is $\\langle P\\rangle / \\epsilon=$ $6.47 \\mathrm{~W}$.']	['$6.47$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	W	Numerical	3e-2	Open-ended	Thermodynamics	Physics	English
+70	"The following information applies for the next two problems. For your mass spectroscopy practical you are using an apparatus consisting of a solenoid enclosed by a uniformly charged hollow cylinder of charge density $\sigma=50 \mu \mathrm{C} / \mathrm{m}^{2}$ and radius $r_{0}=7 \mathrm{~cm}$. There exists an infinitesimal slit of insular material between the cylinder and solenoid to stop any charge transfer. Also, assume that there is no interaction
+
+
+
+between the solenoid and the cylinder, and that the magnetic field produced by the solenoid can be easily controlled to a value of $B_{0}$.
+
+An electron is released from rest at a distance of $R=10 \mathrm{~cm}$ from the axis. Assume that it is small enough to pass through the cylinder in both directions without exchanging charge. It is observed that the electron reaches a distance $R$ at different points from the axis 7 times before returning to the original position.
+
+
+Calculate $B_{0}$ under the assumption that the path of the electron does not self-intersect with itself."	"['By Gauss\' law, we have\n$$\n\\varepsilon_{0} \\oint \\vec{E} \\cdot \\mathrm{d} \\vec{A}=q_{\\mathrm{enc}} \\Longrightarrow \\varepsilon_{0}(2 \\pi r h) E(r)=\\left(2 \\pi r_{0} h\\right) \\sigma \\Longrightarrow E(r)=\\frac{\\sigma r_{0}}{\\varepsilon_{0} r} \\hat{r}\n$$\nWe find the work done on the electron from a distance $r$ to $R$ is\n$$\nW=-\\int_{R}^{r} q E(r) \\mathrm{d} r=\\int_{r}^{R} \\frac{q \\sigma r_{0}}{\\varepsilon_{0} r} \\mathrm{~d} r=\\frac{q \\sigma r_{0}}{\\varepsilon_{0}} \\ln \\frac{R}{r}\n$$\nTherefore, by conservation of energy:\n$$\n\\frac{1}{2} m v(r)^{2}=\\frac{q \\sigma r_{0}}{\\varepsilon_{0}} \\ln \\frac{R}{r} \\Longrightarrow v(r)=\\sqrt{\\frac{2 q \\sigma r_{0}}{m} \\ln \\frac{R}{r}}\n$$\nSince the electron is moving in a magnetic field, the radius of the electron follows\n$$\n\\frac{m v^{2}}{a}=q v B \\Longrightarrow a=\\frac{m v}{q B}=\\sqrt{\\frac{2 q \\sigma r_{0}}{q B_{0}^{2}} \\ln \\frac{R}{r_{0}}}\n$$\nThis means that $B=\\sqrt{\\frac{2 \\sigma m}{q \\varepsilon_{0} r_{0}} \\ln \\frac{R}{r_{0}}}$. The trajectory of the electron can follow the following paths\n\n<img_4320>\n\nAs the path is specified to be non-intersecting, we analyze the first path. We can say that the radius of each ""circle"" the electron travels through is $s=r \\tan \\frac{\\pi}{8}=\\frac{m v}{q B}$. Hence,\n$$\nB=\\cot \\frac{\\pi}{8} \\sqrt{\\frac{2 \\sigma m}{q \\varepsilon_{0} r_{0}} \\ln \\frac{R}{r_{0}}}\n$$']"	['$\\cot \\frac{\\pi}{8} \\sqrt{\\frac{2 \\sigma m}{q \\varepsilon_{0} r_{0}} \\ln \\frac{R}{r_{0}}}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+71	"In the far future, the Earth received an enormous amount of charge as a result of Mad Scientist ecilA's nefarious experiments. Specifically, the total charge on Earth is $Q=1.0 \times 10^{11} \mathrm{C}$. (compare this with the current $5 \times 10^{5} \mathrm{C}$ ).
+Estimate the maximum height of a ""mountain"" on Earth that has a circular base with diameter $w=1.0$ $\mathrm{km}$, if it has the shape of a spherical sector. You may assume that $h_{\max } \ll w$. The tensile strength of rock is $10 \mathrm{MPa}$."	['You can approximate the mound as a conducting sphere of radius $r$ connecting by a wire to the Earth of radius $R$. We can therefore say that\n$$\n\\sigma_{\\text {Mound }} \\sim \\frac{R}{r} \\sigma_{\\text {Earth }}\n$$\nas voltages are equal and proportional to $Q / R$. The electrostatic pressure can be given as $P=\\frac{\\sigma E}{2}$, where $E / 2$ comes from the fact that the section does not contribute a force on itself. This can be rewritten as\n$$\nP=\\left(\\frac{R}{r} \\frac{Q}{4 \\pi R^{2}}\\right) \\cdot \\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{Q}{R^{2}}=\\frac{Q^{2}}{32 \\pi \\varepsilon_{0} R^{3} r}\n$$\nBy using Pythagorean theorem, we can find that $r^{2}-(r-h)^{2}=w^{2}$ which means that $h=\\frac{w^{2}}{2 r}$. We can hence write the tensile strength as\n$$\nY=\\frac{Q^{2} h}{16 \\pi \\varepsilon_{0} R^{3} w^{2}} \\Longrightarrow h=\\frac{16 \\pi \\varepsilon_{0} R^{3} w^{2} Y}{G^{2}}=115 \\mathrm{~m}\n$$']	['115']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	1e-1	Open-ended	Electromagnetism	Physics	English
+72	"Follin is investigating the electrostatic pendulum. His apparatus consists of an insulating Styrofoam ball with a mass of $14 \mathrm{mg}$ and radius $r=0.5 \mathrm{~cm}$ suspended on a uniform electrically-insulating string of length $1 \mathrm{~m}$ and mass per unit length density of $1.1 \cdot 10^{-5} \mathrm{~kg} / \mathrm{m}$ between two large metal plates separated by a distance $17 \mathrm{~cm}$ with a voltage drop of $10 \mathrm{kV}$ between them, such that when the ball is in equilibrium, its center of mass is exactly equidistant to the two plates. Neglect the possibility of electrical discharge throughout the next two problems.
+Follin then gives the ball a charge $0.15 \mathrm{nC}$. Assuming that the charge is distributed evenly across the surface of the ball, find the subsequent horizontal deflection of the pendulum bob's center of mass from its hanging point at equilibrium."	"[""The force on the Styrofoam ball due to the electric field from the two plates is $F=\\frac{Q V}{d}$. Since the plates are conducting, the ball also experiences an attraction toward the closer plate. This force can be neglected because it is much smaller than $\\frac{Q V}{d}$ (an assumption that will be justified at the end of the solution). The mass of the string, however, needs to be considered. Consider the forces on an infinitesimal segment of the string. The horizontal component of the tension must balance out the electric force on the ball and the vertical component must balance out the weight of everything below it (string and ball). This gives us\n$$\n\\frac{d x}{d h}=\\frac{F}{m g+\\lambda g h}\n$$\nwhere $h$ is the height above the ball, $x$ is the horizontal displacement from equilibrium, $F$ is the electrostatic force, $\\lambda$ is the mass density of the string, and $m$ is the mass of the ball. Separating\n\n\n\nvariables, we have\n$$\nx=F \\int_{0}^{L} \\frac{d h}{m g+\\lambda g h}\n$$\nwhere $L$ is the length of the string. Integrating, we get\n$$\nx=\\frac{F}{\\lambda g} \\ln \\left(\\frac{m+\\lambda L}{m}\\right)=0.0475 \\mathrm{~m}\n$$\nNow, let's justify our assumption that the attraction of the ball towards the plates is negligible. Using the method of image charges, the force due to the closer plate is equivalent to a charge $-Q$ a distance of $d-2 x$ away. Infinitely many image charges exist due to the other plate but they have a smaller effect so they will not be considered. The force due to the image charge is\n$$\nF^{\\prime}=\\frac{Q^{2}}{4 \\pi \\epsilon_{0}(d-2 x)^{2}}=3.59 \\times 10^{-8} \\mathrm{~N}\n$$\nwhich is indeed much less than $F=\\frac{Q V}{d}=8.82 \\times 10^{-6} \\mathrm{~N}$""]"	['$0.0475$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	m	Numerical	3e-3	Open-ended	Electromagnetism	Physics	English
+73	Hoping to get a larger deflection, Follin replaces the insulating Styrofoam ball with a conducting pith ball of mass $250 \mathrm{mg}$ and $2 \mathrm{~cm}$ and daisy chains 4 additional $10 \mathrm{kV}$ High Voltage Power Supplies to increase the voltage drop across the plates to $50 \mathrm{kV}$. Leaving the plate separation and the string unchanged, he repeats the same experiment as before, but forgets to measure the charge on the ball. Nonetheless, once the ball reaches equilibrium, he measures the deflection from the hanging point to be $5.6 \mathrm{~cm}$. Find the charge on the ball.	['When a conducting ball is placed in a uniform electric field, the charges separate so that the sphere becomes a dipole with electric dipole moment\n$$\np=4 \\pi \\epsilon_{0} r^{3} E_{0}\n$$\nLet $Q$ be the charge on the ball. The ball can be approximated as a point dipole with dipole moment $p$ and a point charge of magnitude $Q$ at the center of the ball. Since the plates are conducting and must be equipotential, the charge and dipole will experience an attractive force to the plates. Using the method of image charges, the force caused by the closer plate is equivalent to that of a point charge $-Q$ and a dipole moment $p$ at a distance $d-2 x$ from the center of the ball, where $d=17 \\mathrm{~cm}$ is the separation between the plates and $x=5.6 \\mathrm{~cm}$ is the deflection of the ball from the hanging point. The forces from the farther plate are negligible compared to the force due to the electric field, $Q E_{0}$. The force between a point charge $Q$ and a point dipole moment $p$ a distance $r$ apart is\n$$\nF=\\frac{p Q}{2 \\pi \\epsilon_{0} r^{3}}\n$$\nThe force between two dipoles with dipole moments of magnitude $p$ in the same direction as the vector from one dipole to the other is\n$$\nF=\\frac{3 p^{2}}{2 \\pi \\epsilon_{0} r^{4}}\n$$\nAdding up the charge-image charge, charge-image dipole, dipole-image charge, and dipole-image dipole interactions, we get that the total electrostatic force on the ball is\n$$\nF=Q E_{0}+\\frac{Q^{2}}{4 \\pi \\epsilon_{0}(d-2 x)^{2}}+\\frac{p Q}{\\pi \\epsilon_{0}(d-2 x)^{3}}+\\frac{3 p^{2}}{2 \\pi \\epsilon_{0}(d-2 x)^{4}}\n$$\nFrom the previous problem, we know that the displacement $x$ of the ball from equilibrium as a function of the electrostatic force $F$ is\n$$\nx=\\frac{F}{l g} \\ln \\left(\\frac{m+l}{m}\\right)\n$$\n\nRearranging this as a quadratic in $Q$ and solving, we have\n$$\n\\begin{gathered}\n\\frac{1}{4 \\pi \\epsilon_{0}(d-2 x)^{2}} Q^{2}+\\left(\\frac{p}{\\pi \\epsilon_{0}(d-2 x)^{3}}+E_{0}\\right) Q+\\left(\\frac{3 p^{2}}{2 \\pi \\epsilon_{0}(d-2 x)^{4}}-\\frac{\\lg x}{\\ln \\left(\\frac{m+l}{m}\\right)}\\right)=0 \\\\\nQ=\\frac{-\\left(\\frac{p}{\\pi \\epsilon_{0}(d-2 x)^{3}}+E_{0}\\right)+\\sqrt{\\left(\\frac{p}{\\pi \\epsilon_{0}(d-2 x)^{3}}+E_{0}\\right)^{2}-4\\left(\\frac{1}{4 \\pi \\epsilon_{0}(d-2 x)^{2}}\\right)\\left(\\frac{3 p^{2}}{2 \\pi \\epsilon_{0}(d-2 x)^{4}}-\\frac{\\lg x}{\\ln \\left(\\frac{m+l}{m}\\right)}\\right)}}{2 \\frac{1}{4 \\pi \\epsilon_{0}(d-2 x)^{2}}} .\n\\end{gathered}\n$$\nPlugging gives $Q=4.48 \\times 10^{-10} \\mathrm{C}$.']	['$4.48 \\times 10^{-10}$']	"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
+- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
+- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
+- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
+- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
+- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
+- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
+- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
+- Universal Gravitational constant,
+
+$$
+G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
+$$
+
+- Solar Mass
+
+$$
+M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
+$$
+
+- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
+- 1 unified atomic mass unit,
+
+$$
+1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
+$$
+
+- Planck's constant,
+
+$$
+h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
+$$
+
+- Permittivity of free space,
+
+$$
+\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
+$$
+
+- Coulomb's law constant,
+
+$$
+k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
+$$
+
+- Permeability of free space,
+
+$$
+\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
+$$
+
+- Magnetic constant,
+
+$$
+\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
+$$
+
+- 1 atmospheric pressure,
+
+$$
+1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
+$$
+
+- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
+- Stefan-Boltzmann constant,
+
+$$
+\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
+$$"	[]	Text-only	Competition	False	C	Numerical	1e-11	Open-ended	Electromagnetism	Physics	English
+74	(a) Where should a pin be placed on the optical axis such that its image is formed at the same place?	['Image of object will coincide with it if ray of light after refraction from the concave surface fall normally on concave mirror so formed by silvering the convex surface. Or image after refraction from concave surface should form at centre of curvature of concave mirror or at a distance of $20 \\mathrm{~cm}$ on same side of the combination. Let $x$ be the distance of pin from the given optical system.\n\nUsing,\n\n$$\n\\frac{\\mu_{2}}{\\nu}-\\frac{\\mu_{1}}{u}=\\frac{\\mu_{2}-\\mu_{1}}{R}\n\\tag{8}\n$$\n\nWith proper signs,\n\n$$\n\\frac{1.5}{-20}-\\frac{1}{-x}=\\frac{1.5-1}{-60}\n\\tag{9}\n$$\n\nSolve to get $x=15 \\mathrm{~cm}$']	['15']	"3. Stick a pin there 
+
+The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature $20 \mathrm{~cm}$. The concave surface has a radius of curvature $60 \mathrm{~cm}$. The convex side is silvered and placed on a horizontal surface."	[]	Text-only	Competition	False	cm	Numerical	5e-1	Open-ended	Optics	Physics	English
+75	"(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.
+
+Plugging in this guess, what must $\lambda$ be?"	['We can plug in $\\eta=\\alpha e^{\\lambda t}$ :\n\n$$\n0=\\lambda^{2} \\alpha e^{\\lambda t}-2 i \\lambda \\Omega \\alpha e^{\\lambda t}-\\Omega^{2} \\alpha e^{\\lambda t}\n\\tag{23}\n$$\n\nThen we can cancel common factors and find that\n\n$$\n0=\\lambda^{2}-2 i \\lambda \\Omega-\\Omega^{2}=(\\lambda-i \\Omega)^{2}\n\\tag{24}\n$$\n\nWe see that $\\lambda=i \\Omega$.']	['$\\lambda=i \\Omega$']	"4. A complex dance 
+
+In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.
+
+Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:
+
+$$
+e^{i \theta}=\cos \theta+i \sin \theta
+\tag{1}
+$$
+Context question:
+(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form
+
+$$
+\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
+\tag{2}
+$$
+
+For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,
+
+$$
+\begin{aligned}
+& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
+& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
+\end{aligned}
+\tag{3}
+$$
+
+where dots represent time derivatives.
+
+Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:
+
+$$
+0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
+\tag{4}
+$$
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+76	"(c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.
+
+This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution."	['Using our answer to part (b) and $\\alpha=A e^{i \\phi}$, we have\n\n$$\n\\eta(t)=A e^{i(\\Omega t+\\phi)}\n\\tag{25}\n$$\n\nUsing the Euler identity, we have\n\n$$\nx(t)+i y(t)=A \\cos (\\Omega t+\\phi)+i A \\sin (\\Omega t+\\phi)\n\\tag{26}\n$$\n\nThe real and imaginary parts become\n\n$$\n\\begin{aligned}\n& x(t)=A \\cos (\\Omega t+\\phi) \\\\\n& y(t)=A \\sin (\\Omega t+\\phi)\n\\end{aligned}\n\\tag{27}\n$$']	['$x(t)=A \\cos (\\Omega t+\\phi)$ , $y(t)=A \\sin (\\Omega t+\\phi)$']	"4. A complex dance 
+
+In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.
+
+Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:
+
+$$
+e^{i \theta}=\cos \theta+i \sin \theta
+\tag{1}
+$$
+Context question:
+(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form
+
+$$
+\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
+\tag{2}
+$$
+
+For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,
+
+$$
+\begin{aligned}
+& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
+& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
+\end{aligned}
+\tag{3}
+$$
+
+where dots represent time derivatives.
+
+Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:
+
+$$
+0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
+\tag{4}
+$$
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.
+
+Plugging in this guess, what must $\lambda$ be?
+Context answer:
+\boxed{$\lambda=i \Omega$}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Mechanics	Physics	English
+77	"(d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by
+
+$$
+\frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}}
+\tag{5}
+$$
+
+A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation."""	['Consider the given differential equation:\n\n$$\n\\frac{\\partial \\psi}{\\partial t}=a \\frac{\\partial^{2} \\psi}{\\partial x^{2}}\n\\tag{28}\n$$\n\nWe can plug in $\\psi(x, t)=A e^{i k x-i \\omega t}$ to find\n\n$$\n-i \\omega A e^{i k x-i \\omega t}=-k^{2} a A e^{i k x-i \\omega t}\n\\tag{29}\n$$\n\nso that\n\n$$\n\\omega=-i k^{2} a\n\\tag{30}\n$$']	['$\\omega=-i k^{2} a$']	"4. A complex dance 
+
+In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.
+
+Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:
+
+$$
+e^{i \theta}=\cos \theta+i \sin \theta
+\tag{1}
+$$
+Context question:
+(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form
+
+$$
+\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
+\tag{2}
+$$
+
+For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,
+
+$$
+\begin{aligned}
+& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
+& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
+\end{aligned}
+\tag{3}
+$$
+
+where dots represent time derivatives.
+
+Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:
+
+$$
+0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
+\tag{4}
+$$
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.
+
+Plugging in this guess, what must $\lambda$ be?
+Context answer:
+\boxed{$\lambda=i \Omega$}
+
+
+Context question:
+(c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.
+
+This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.
+Context answer:
+\boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+78	"(e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by
+
+$$
+i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}}
+\tag{6}
+$$
+
+Using your answer to part (d), what is the dispersion relation of the Schrödinger equation?"	['We see that the free Schrödinger equation takes the form of Equation 28, but with\n\n$$\na=\\frac{i \\hbar}{2 m}\n\\tag{31}\n$$\n\nThen, using our answer from part (d), we have\n\n$$\n\\omega=\\frac{\\hbar k^{2}}{2 m}\n\\tag{32}\n$$']	['$\\omega=\\frac{\\hbar k^{2}}{2 m}$']	"4. A complex dance 
+
+In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.
+
+Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:
+
+$$
+e^{i \theta}=\cos \theta+i \sin \theta
+\tag{1}
+$$
+Context question:
+(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form
+
+$$
+\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
+\tag{2}
+$$
+
+For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,
+
+$$
+\begin{aligned}
+& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
+& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
+\end{aligned}
+\tag{3}
+$$
+
+where dots represent time derivatives.
+
+Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:
+
+$$
+0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
+\tag{4}
+$$
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.
+
+Plugging in this guess, what must $\lambda$ be?
+Context answer:
+\boxed{$\lambda=i \Omega$}
+
+
+Context question:
+(c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.
+
+This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.
+Context answer:
+\boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$}
+
+
+Context question:
+(d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by
+
+$$
+\frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}}
+\tag{5}
+$$
+
+A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation.""
+Context answer:
+\boxed{$\omega=-i k^{2} a$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+79	"(g) The theory of relativity instead posits that the energy of a particle is given by $E=\sqrt{p^{2} c^{2}+m^{2} c^{4}}$. In accordance with this, we can try to guess a relativistic version of the Schrödinger equation:
+
+$$
+\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}-\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{m^{2} c^{2}}{\hbar^{2}} \phi=0
+\tag{7}
+$$
+
+This is called the Klein-Gordon equation. Using the same guess as before, find $\omega$ in terms of $\mathrm{k}$.
+
+Hint: If you are careful, you should find that there is an infinite continuum of energy states extending down to negative infinity. This apparently mathematical issue hints at the existence of antimatter, and ultimately demonstrates to us that we must formulate quantum field theory to properly describe relativistic quantum physics."	['We consider the Klein-Gordon equation:\n\n$$\n\\frac{1}{c^{2}} \\frac{\\partial^{2} \\phi}{\\partial t^{2}}-\\frac{\\partial^{2} \\phi}{\\partial x^{2}}+\\frac{m^{2} c^{2}}{\\hbar^{2}} \\phi=0\n\\tag{35}\n$$\n\nWe can, as before, plug in a guess $\\phi(x, t)=A e^{i k x-i \\omega t}$. This yields\n\n$$\n-\\frac{\\omega^{2}}{c^{2}} A e^{i k x-i \\omega t}+k^{2} A e^{i k x-i \\omega t}+\\frac{m^{2} c^{2}}{\\hbar^{2}} A e^{i k x-i \\omega t}=0\n\\tag{36}\n$$\n\nCancelling out common terms, we see that\n\n$$\n\\omega^{2}=k^{2} c^{2}+\\frac{m^{2} c^{4}}{\\hbar^{2}}\n\\tag{37}\n$$\n\nor\n\n$$\n\\omega= \\pm \\sqrt{k^{2} c^{2}+\\frac{m^{2} c^{4}}{\\hbar^{2}}}\n\\tag{38}\n$$\n\nNote that both positive and negative $\\omega$ solve the Klein-Gordon equation.']	['$\\sqrt{k^{2} c^{2}+\\frac{m^{2} c^{4}}{\\hbar^{2}}}$, $-\\sqrt{k^{2} c^{2}+\\frac{m^{2} c^{4}}{\\hbar^{2}}}$']	"4. A complex dance 
+
+In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.
+
+Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity:
+
+$$
+e^{i \theta}=\cos \theta+i \sin \theta
+\tag{1}
+$$
+Context question:
+(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form
+
+$$
+\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
+\tag{2}
+$$
+
+For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,
+
+$$
+\begin{aligned}
+& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
+& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
+\end{aligned}
+\tag{3}
+$$
+
+where dots represent time derivatives.
+
+Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:
+
+$$
+0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
+\tag{4}
+$$
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.
+
+Plugging in this guess, what must $\lambda$ be?
+Context answer:
+\boxed{$\lambda=i \Omega$}
+
+
+Context question:
+(c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.
+
+This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.
+Context answer:
+\boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$}
+
+
+Context question:
+(d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by
+
+$$
+\frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}}
+\tag{5}
+$$
+
+A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation.""
+Context answer:
+\boxed{$\omega=-i k^{2} a$}
+
+
+Context question:
+(e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by
+
+$$
+i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}}
+\tag{6}
+$$
+
+Using your answer to part (d), what is the dispersion relation of the Schrödinger equation?
+Context answer:
+\boxed{$\omega=\frac{\hbar k^{2}}{2 m}$}
+
+
+Context question:
+(f) If the energy of a wave is $E=\hbar \omega$ and the momentum is $p=\hbar k$, show that the dispersion relation found in part (e) resembles the classical expectation for the kinetic energy of a particle, $\mathrm{E}=\mathrm{mv}^{2} / \mathbf{2}$.
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Modern Physics	Physics	English
+80	"(c) The uniqueness theorem allows us to use ""image"" charges in certain settings to describe the system. Consider one such example: There is a point-like charge $q$ at a distance $L$ from a metallic sphere of radius $R$ attached to the ground. As you argued in part (a), sphere will be polarized to make sure the electric potential is constant throughout its body. Since it is attached to the ground, the constant potential will be zero. Place an image charge inside the sphere to counter the non-uniform potential of the outer charge $q$ on the surface. Where should this charge be placed, and what is its value?"	['Take the origin of our coordinate system at the center of the sphere and align the axes so that the coordinates of our charge is $(x, y)=(L, 0)$. Now suppose we place an image charge $q_{0}$ at $(x, 0)$ where $0<r<R$. Then the potential at the point $(R \\cos \\theta, R \\sin \\theta)$ on the surface becomes\n\n$$\nV(\\theta)=\\frac{q_{0}}{x_{\\theta}}+\\frac{q}{L_{\\theta}}\n$$\n\nwhere\n\n$$\n\\begin{aligned}\nx_{\\theta} & =\\sqrt{x^{2}+R^{2}-2 x R \\cos \\theta} \\\\\nL_{\\theta} & =\\sqrt{L^{2}+R^{2}-2 L R \\cos \\theta}\n\\end{aligned}\n$$\n\nWe want this image charge to satisfy $V(\\theta)=0 \\forall \\theta \\in[0,2 \\pi]$. Then we have,\n\n$$\n\\begin{aligned}\n\\frac{q}{q_{0}} & =-\\frac{\\sqrt{L^{2}+R^{2}-2 L R \\cos \\theta}}{\\sqrt{x^{2}+R^{2}-2 x R \\cos \\theta}} \\\\\n& =-\\sqrt{\\frac{L}{x} \\frac{\\sqrt{\\frac{L^{2}+R^{2}}{L}-2 R \\cos \\theta}}{\\sqrt{\\frac{x^{2}+R^{2}}{x}-2 R \\cos \\theta}}}\n\\end{aligned}\n$$\n\nConsequently,\n\n$$\n\\begin{aligned}\n& \\frac{L^{2}+R^{2}}{L}=\\frac{x^{2}+R^{2}}{x} \\text { and } \\\\\n& \\frac{q}{q_{0}}=-\\sqrt{\\frac{L}{x}}\n\\end{aligned}\n$$\n\nSolution to the first equation with $x<R$ condition gives $x=\\frac{R^{2}}{L}$, and from the second equation, $q_{0}$ can be found as $q_{0}=-q \\frac{R}{L}$. By azimuthal symmetry, we conclude that if the\n\n\n\npotential is zero on surface crossing the $x-y$ plane with the image charge $q_{0}=-q \\frac{R}{L}$ at $x=\\frac{R^{2}}{L}$, then it is zero everywhere throughout the surface.']	['$x=\\frac{R^{2}}{L}$ , $q_{0}=-q \\frac{R}{L}$']	"5. Polarization and Oscillation 
+
+In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations.
+
+Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system.
+Context question:
+(a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.
+Context answer:
+开放性回答
+
+
+Context question:
+(b) Laplace's equation is a second order differential equation
+
+$$
+\nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0
+\tag{8}
+$$
+
+Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary.
+
+Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$.
+
+Hint: Consider $\phi=\phi_{1}-\phi_{2}$.
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Electromagnetism	Physics	English
+81	(e) Find the force of attraction between the charge and the sphere.	['From part (d) and (c), this is simply\n\n$$\nF=\\frac{1}{4 \\pi \\epsilon_{0}} \\frac{q^{2} \\frac{R}{L}}{\\left(L-\\frac{R^{2}}{L}\\right)^{2}}\n$$']	['$F=\\frac{1}{4 \\pi \\epsilon_{0}} \\frac{q^{2} \\frac{R}{L}}{\\left(L-\\frac{R^{2}}{L}\\right)^{2}}$']	"5. Polarization and Oscillation 
+
+In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations.
+
+Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system.
+Context question:
+(a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.
+Context answer:
+开放性回答
+
+
+Context question:
+(b) Laplace's equation is a second order differential equation
+
+$$
+\nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0
+\tag{8}
+$$
+
+Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary.
+
+Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$.
+
+Hint: Consider $\phi=\phi_{1}-\phi_{2}$.
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+(c) The uniqueness theorem allows us to use ""image"" charges in certain settings to describe the system. Consider one such example: There is a point-like charge $q$ at a distance $L$ from a metallic sphere of radius $R$ attached to the ground. As you argued in part (a), sphere will be polarized to make sure the electric potential is constant throughout its body. Since it is attached to the ground, the constant potential will be zero. Place an image charge inside the sphere to counter the non-uniform potential of the outer charge $q$ on the surface. Where should this charge be placed, and what is its value?
+Context answer:
+\boxed{$x=\frac{R^{2}}{L}$ , $q_{0}=-q \frac{R}{L}$}
+
+
+Context question:
+(d) Argue from the uniqueness theorem that the electric field created by this image charge outside the sphere will be the same as the field created by the complicated polarization of the sphere.
+Context answer:
+开放性回答
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+82	(a) What value of $\theta$ maximizes the range?	['From the equation of motion $\\frac{d \\mathbf{v}}{d t}=\\mathbf{g}$ we find that $\\mathbf{v}=\\mathbf{v}_{0}+\\mathbf{g} t$. Then the position of the mass after time $t$ is given by $\\mathbf{r}=\\int_{0}^{t} \\mathbf{v} d t=\\int_{0}^{t}\\left(\\mathbf{v}_{0}+\\mathbf{g} t\\right) d t=\\mathbf{v}_{0} t+\\frac{1}{2} \\mathbf{g} t^{2}$. Denoting the final time as $T$ and the position as $\\mathbf{R}$, we find that\n\n$$\n\\mathbf{R e}_{\\mathbf{y}}=0=v_{0} \\sin \\theta T-\\frac{1}{2} g T^{2}\n\\tag{1}\n$$\n$$\n\\mathbf{R e}_{\\mathbf{x}}=L=v_{0} \\cos \\theta T\n\\tag{2}\n$$\n\nwhere $L$ is the range of the mass. From equations (1) and (2), we get\n\n$$\nL=\\frac{2 v_{0}^{2}}{g} \\sin \\theta \\cos \\theta=\\frac{v_{0}^{2}}{g} \\sin (2 \\theta)\n$$\n\nMaximum $L$ is achieved when $\\sin (2 \\theta)=1$. Therefore, $\\theta=\\frac{\\pi}{4}$']	['$\\frac{\\pi}{4}$']	"2. Trajectory of a point mass 
+
+A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,"	[]	Text-only	Competition	False		Numerical	1e-8	Open-ended	Mechanics	Physics	English
+83	(b) What value of $\theta$ maximizes the surface area under the trajectory curve?	['The surface area covered by the trajectory can be found by\n\n$$\n\\mathbf{A}=\\frac{1}{2} \\int \\mathbf{r} \\times \\mathbf{d} \\mathbf{r}=\\frac{1}{2} \\int(\\mathbf{r} \\times \\mathbf{v}) d t\n$$\n\nSince the motion occurs in a plane, the surface area is always orthogonal to the plane. Therefore, the area covered by the trajectory can be found by calculating the magnitude of the vector $\\mathbf{A}$. Now, inserting the equations for the position vector and the velocity as found in (1), we get\n\n$$\n\\begin{aligned}\n\\mathbf{A}=\\frac{1}{2} \\int(\\mathbf{r} \\times \\mathbf{v}) d t & =\\frac{1}{4} \\int\\left(\\mathbf{v}_{\\mathbf{0}} \\times \\mathbf{g}\\right) t^{2} d t \\\\\n& =\\frac{1}{4}\\left(\\mathbf{v}_{\\mathbf{0}} \\times \\mathbf{g}\\right) \\int t^{2} d t \\\\\n& \\left.=\\frac{1}{12} \\mathbf{v}_{\\mathbf{0}} \\times \\mathbf{g}\\right) T^{3}\n\\end{aligned}\n$$\n\n\n\nAs found in (1), $T$ is proportional to $\\sin \\theta$, and $\\mathbf{v}_{\\mathbf{0}} \\times \\mathbf{g}$ is proportional to $\\cos \\theta$. Therefore, the angle that maximizes the area maximizes the function\n\n$$\nf(\\theta)=\\cos \\theta \\sin ^{3} \\theta\n$$\n\nTaking derivative of $f(\\theta)$, we get the equation\n\n$$\n\\frac{d f}{d \\theta}=\\sin ^{2} \\theta\\left(3 \\cos ^{2} \\theta-\\sin ^{2} \\theta\\right)\n$$\n\nwhich gives zero at $\\theta=0$ or $\\theta=\\arctan \\sqrt{3}=\\pi / 3$. Since $\\theta=0$ corresponds to the minimal area (no trajectory, $\\mathrm{A}=0$ ), the answer is $\\theta=\\pi / 3$']	['$\\pi / 3$']	"2. Trajectory of a point mass 
+
+A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,
+Context question:
+(a) What value of $\theta$ maximizes the range?
+Context answer:
+\boxed{$\frac{\pi}{4}$}
+"	[]	Text-only	Competition	False	rads	Numerical	1e-8	Open-ended	Mechanics	Physics	English
+84	"(c) What is the answer for (a), if the point mass is thrown from an apartment of height $h$ ?
+
+
+Now assume that we have a frictional force that is proportional to the velocity vector, such that the equation of motion is as follows
+
+$$
+\frac{d \vec{v}}{d t}=\vec{g}-\beta \vec{v}
+$$"	['If the initial position of the mass is $\\mathbf{r}_{\\mathbf{0}}=h \\mathbf{e}_{\\mathbf{y}}$, equation (1) and (2) becomes\n\n$$\n\\mathbf{R e}_{\\mathbf{y}}=-h=v_{0} \\sin \\theta T-\\frac{1}{2} g T^{2}\n\\tag{3}\n$$\n$$\n\\mathbf{R e}_{\\mathbf{x}}=L=v_{0} \\cos \\theta T\n\\tag{4}\n$$\n\nMaximum $L$ occurs when $\\frac{d L}{d \\theta}=0$. Therefore, from (4), we obtain\n\n$$\n\\frac{d L}{d \\theta}=-v_{0} \\sin \\theta+v_{0} \\cos \\theta \\frac{d T}{d \\theta}=0\n\\tag{5}\n$$\n\nTaking derivative of both sides of (3) to find $\\frac{d T}{d \\theta}$ and inserting it into (5), we get\n\n$$\n\\begin{aligned}\n0 & =v_{0} \\cos \\theta T+\\frac{v_{0} \\sin \\theta-g T}{\\cos \\theta} \\sin \\theta T \\\\\n& \\Longrightarrow v_{0}=g T \\sin \\theta\n\\end{aligned}\n$$\n\nInserting this equation to (3), we obtain\n\n$$\n-h=\\frac{v_{0}^{2}}{g}-\\frac{1}{2} \\frac{v_{0}^{2}}{g \\sin \\theta}\n$$\n\nTherefore, the answer is $\\theta=\\arcsin \\left(\\frac{1}{\\sqrt{2}} \\frac{v_{0}}{\\sqrt{v_{0}^{2}+g h}}\\right)$']	['$\\arcsin \\left(\\frac{1}{\\sqrt{2}} \\frac{v_{0}}{\\sqrt{v_{0}^{2}+g h}}\\right)$']	"2. Trajectory of a point mass 
+
+A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,
+Context question:
+(a) What value of $\theta$ maximizes the range?
+Context answer:
+\boxed{$\frac{\pi}{4}$}
+
+
+Context question:
+(b) What value of $\theta$ maximizes the surface area under the trajectory curve?
+Context answer:
+\boxed{$\pi / 3$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+85	(d) Supposing that $\beta<<\frac{g}{v_{0}}$, find the duration of the motion $T$.	['Under the presence of friction, equation of motion for $y$ component becomes\n\n$$\n\\frac{d v_{y}}{d t}=-g-\\beta v_{y}\n\\tag{6}\n$$\n\nSolving (6), we get\n\n$$\nv_{y}=v_{0 y}-\\left(v_{0 y}+\\frac{g}{\\beta}\\right)\\left(1-e^{-\\beta t}\\right)\n\\tag{7}\n$$\n\n\n\n$y$ component of the final position vector can be found by integrating $v_{y}$,\n\n$$\ny=0=-\\frac{g T}{\\beta}+\\frac{\\left(v_{0 y}+\\frac{g}{\\beta}\\right)}{\\beta}\\left(1-e^{-\\beta T}\\right)\n\\tag{8}\n$$\n\nSince $\\beta$ is small compared to the dimensions of the system, we can use Taylor expansion for (8) to find an analytic equation for $T$.\n\n$$\n\\begin{aligned}\n& g T=\\left(v_{0 y}+\\frac{g}{\\beta}\\right)\\left(1-e^{-\\beta T}\\right) \\approx\\left(v_{0 y}+\\frac{g}{\\beta}\\right)\\left(\\beta T-\\frac{1}{2} \\beta^{2} T^{2}\\right) \\\\\n& \\Longrightarrow T=\\frac{2 v_{0 y}}{\\beta v_{0 y}+g}\n\\end{aligned}\n$$\n\nTherefore, the answer is\n\n$$\nT=\\frac{2 v_{0} \\sin \\theta}{g} \\frac{1}{1+\\frac{\\beta v_{0} \\sin \\theta}{g}}\n$$\n\nwhich is less than the value found for frictionless trajectory, as expected.']	['$T=\\frac{2 v_{0} \\sin \\theta}{g} \\frac{1}{1+\\frac{\\beta v_{0} \\sin \\theta}{g}}$']	"2. Trajectory of a point mass 
+
+A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,
+Context question:
+(a) What value of $\theta$ maximizes the range?
+Context answer:
+\boxed{$\frac{\pi}{4}$}
+
+
+Context question:
+(b) What value of $\theta$ maximizes the surface area under the trajectory curve?
+Context answer:
+\boxed{$\pi / 3$}
+
+
+Context question:
+(c) What is the answer for (a), if the point mass is thrown from an apartment of height $h$ ?
+
+
+Now assume that we have a frictional force that is proportional to the velocity vector, such that the equation of motion is as follows
+
+$$
+\frac{d \vec{v}}{d t}=\vec{g}-\beta \vec{v}
+$$
+Context answer:
+\boxed{$\arcsin \left(\frac{1}{\sqrt{2}} \frac{v_{0}}{\sqrt{v_{0}^{2}+g h}}\right)$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+86	(a) Compute the electric potential inside and outside the sphere.	"[""From Gauss's theorem:\n\n$$\n\\frac{Q}{\\varepsilon_{0}}=4 \\pi R^{2} E\n$$\n\nSo:\n\n$$\nE=\\frac{Q}{4 \\pi \\varepsilon_{0}}\n$$\n\nBut:\n\n$$\n\\mathbf{E}=-\\nabla \\Phi\n$$\n\nIn this case $E=-\\frac{\\partial \\Phi}{\\partial r}$. To obtain a zero potential at infinity we get:\n\n$$\n\\Phi_{+}=\\frac{Q}{4 \\pi \\varepsilon_{0} r}\n$$\n\nThe electric field inside the sphere is 0 (no charges), so the potential is constant. Because it is continuous on the surface of the sphere (no jumps):\n\n$$\n\\Phi_{-}=\\frac{Q}{4 \\pi \\varepsilon_{0} R}\n$$""]"	['$\\Phi_{-}=\\frac{Q}{4 \\pi \\varepsilon_{0} R}$,$\\Phi_{+}=\\frac{Q}{4 \\pi \\varepsilon_{0} r}$']	"Computing electric fields
+
+Electrostatics relies on multiple methods for computing electric fields and potentials. In this problem, we will explore two of them, Gauss's Law and Legendre polynomials.
+
+Uniform charge distributions
+
+Let us consider a hollow conducting sphere of radius $R$ charged with the electric charge $Q$, uniformly distributed on its surface. In order to calculate its potential, we can use Gauss's Law, which states that the flux of the electric field $d F=\mathbf{E} \cdot \mathbf{d} \mathbf{A}$ across a closed surface is proportional to the charge enclosed by that surface: $F=Q / \varepsilon_{0}$. We have denoted $\mathbf{d A}=d A \mathbf{n}$ the elementary oriented (towards the exterior) surface element."	[]	Text-only	Competition	True		Expression		Open-ended	Electromagnetism	Physics	English
+87	"(a) Knowing that Legendre polynomials are orthogonal $\left(\left\langle P_{m}(x), P_{n}(x)\right\rangle=0\right.$ if $m \neq n)$ and $\operatorname{deg} P_{n}(x)=n$, obtain $P_{2}(x)$ and $P_{3}(x)$. For reaching the usual and most convenient form of these polynomials, divide your results by the norm: $\left\|P_{n}(x)\right\|=\frac{2}{2 n+1}$.
+
+Let us now consider a sphere of radius $R$ centered at the origin. Suppose a point charge $q$ is put at the origin and that this is the only charge inside or outside the sphere. Furthermore, the potential is $\Phi=V_{0} \cos \theta$ on the surface of the sphere.
+
+We know that we can write the potential induced by the charge on the sphere (without taking into account $q$ ) in the following way:
+
+$$
+\begin{array}{ll}
+\Phi_{-}=\sum_{n=0}^{\infty} A_{n} r^{n} P_{n}(\cos \theta), & r<R \\
+\Phi_{+}=\sum_{n=0}^{\infty} \frac{B_{n}}{r^{n+1}} P_{n}(\cos \theta), & r>R
+\end{array}
+$$"	['$$\nP_{2}(x)=C_{2}\\left(x^{2}+\\lambda_{1} x+\\lambda_{0}\\right)\n$$\n\nBecause $\\left\\langle P_{2}(x), P_{1}(x)\\right\\rangle=\\int_{-1}^{1} P_{2}(x) P_{1}(x) d x=0$ we get $\\lambda_{1}=0$. Because $\\int_{-1}^{1} x^{2} d x=2 / 3$ and $\\int_{-1}^{1} d x=$ 2 , we get $\\lambda_{0}=-1 / 3$. The condition for the norm implies $P_{2}(x)=\\frac{1}{2}\\left(3 x^{2}-1\\right)$. Similarly:\n\n$$\nP_{3}(x)=C_{3}\\left(x^{3}+\\lambda_{2} x^{2}+\\lambda_{1} x+\\lambda_{0}\\right)\n$$\n\nThrough an identical reasoning:\n\n$$\nP_{3}=\\frac{1}{2}\\left(5 x^{3}-3 x\\right)\n$$']	['$P_{2}(x)=C_{2}\\left(x^{2}+\\lambda_{1} x+\\lambda_{0}\\right)$ , $P_{3}=\\frac{1}{2}\\left(5 x^{3}-3 x\\right)$']	"Computing electric fields
+
+Electrostatics relies on multiple methods for computing electric fields and potentials. In this problem, we will explore two of them, Gauss's Law and Legendre polynomials.
+
+Uniform charge distributions
+
+Let us consider a hollow conducting sphere of radius $R$ charged with the electric charge $Q$, uniformly distributed on its surface. In order to calculate its potential, we can use Gauss's Law, which states that the flux of the electric field $d F=\mathbf{E} \cdot \mathbf{d} \mathbf{A}$ across a closed surface is proportional to the charge enclosed by that surface: $F=Q / \varepsilon_{0}$. We have denoted $\mathbf{d A}=d A \mathbf{n}$ the elementary oriented (towards the exterior) surface element.
+Context question:
+(a) Compute the electric potential inside and outside the sphere.
+Context answer:
+\boxed{$\Phi_{-}=\frac{Q}{4 \pi \varepsilon_{0} R}$,$\Phi_{+}=\frac{Q}{4 \pi \varepsilon_{0} r}$}
+
+
+Extra Supplementary Reading Materials:
+
+Legendre polynomials and non-uniform charge distributions
+
+Legendre polynomials are a type of orthogonal polynomials essential in mathematical physics. One of their applications is in computing electric potentials for more complicated charge configurations. We will denote the $n$-th Legendre polynomial (having degree $n$ ) as $P_{n}$. Legendre polynomials are defined on $[-1,1]$ and we can express their scalar product as
+
+$$
+\left\langle P_{m}(x), P_{n}(x)\right\rangle=\int_{-1}^{1} P_{m}(x) P_{n}(x) d x
+\tag{1}
+$$
+
+The first two Legendre polynomials are $P_{0}(x)=1$ and $P_{1}(x)=x$."	[]	Text-only	Competition	True		Expression		Open-ended	Electromagnetism	Physics	English
+88	(b) Compute the electric potential both inside and outside the sphere.	"[""The potential is given by either Poisson's or Laplace's equation:\n\n$$\n\\begin{array}{ll}\n\\nabla^{2} \\Phi_{-}=-\\frac{q}{\\varepsilon_{0}} \\delta(r), & r<R \\\\\n\\nabla^{2} \\Phi_{+}=0, & r>R\n\\end{array}\n$$\n\n\n\nThe general solutions finite in the respective regions, taking account of the symmetry, are\n\n$$\n\\begin{aligned}\n& \\Phi_{-}=\\frac{q}{4 \\pi \\varepsilon_{0} r}+\\sum_{n=0}^{\\infty} A_{n} r^{n} P_{n}(\\cos \\theta), \\quad r<R \\\\\n& \\Phi_{+}=\\sum_{n=0}^{\\infty} \\frac{B_{n}}{r^{n+1}} P_{n}(\\cos \\theta), \\quad r>R\n\\end{aligned}\n$$\n\nThen from the condition $\\Phi_{-}=\\Phi_{+}=V_{0} \\cos \\theta$ at $r=R$ we obtain $A_{0}=-\\frac{q}{4 \\pi \\varepsilon_{0} R}, A_{1}=\\frac{V_{0}}{R}, B_{1}=$ $V_{0} R^{2}, B_{0}=0, A_{n}=B_{n}=0$ for $n \\neq 0,1$, and hence\n\n$$\n\\begin{aligned}\n& \\Phi_{-}=\\frac{q}{4 \\pi \\varepsilon_{0} r}-\\frac{q}{4 \\pi \\varepsilon_{0} R}+\\frac{V_{0} \\cos \\theta}{R} r, \\quad r<R ; \\\\\n& \\Phi_{+}=\\frac{V_{0} R^{2}}{r^{2}} \\cos \\theta, \\quad r>R .\n\\end{aligned}\n$$""]"	['$\\Phi_{-}=\\frac{q}{4 \\pi \\varepsilon_{0} r}-\\frac{q}{4 \\pi \\varepsilon_{0} R}+\\frac{V_{0} \\cos \\theta}{R} r$,$\\Phi_{+}=\\frac{V_{0} R^{2}}{r^{2}} \\cos \\theta$']	"Computing electric fields
+
+Electrostatics relies on multiple methods for computing electric fields and potentials. In this problem, we will explore two of them, Gauss's Law and Legendre polynomials.
+
+Uniform charge distributions
+
+Let us consider a hollow conducting sphere of radius $R$ charged with the electric charge $Q$, uniformly distributed on its surface. In order to calculate its potential, we can use Gauss's Law, which states that the flux of the electric field $d F=\mathbf{E} \cdot \mathbf{d} \mathbf{A}$ across a closed surface is proportional to the charge enclosed by that surface: $F=Q / \varepsilon_{0}$. We have denoted $\mathbf{d A}=d A \mathbf{n}$ the elementary oriented (towards the exterior) surface element.
+Context question:
+(a) Compute the electric potential inside and outside the sphere.
+Context answer:
+\boxed{$\Phi_{-}=\frac{Q}{4 \pi \varepsilon_{0} R}$,$\Phi_{+}=\frac{Q}{4 \pi \varepsilon_{0} r}$}
+
+
+Extra Supplementary Reading Materials:
+
+Legendre polynomials and non-uniform charge distributions
+
+Legendre polynomials are a type of orthogonal polynomials essential in mathematical physics. One of their applications is in computing electric potentials for more complicated charge configurations. We will denote the $n$-th Legendre polynomial (having degree $n$ ) as $P_{n}$. Legendre polynomials are defined on $[-1,1]$ and we can express their scalar product as
+
+$$
+\left\langle P_{m}(x), P_{n}(x)\right\rangle=\int_{-1}^{1} P_{m}(x) P_{n}(x) d x
+\tag{1}
+$$
+
+The first two Legendre polynomials are $P_{0}(x)=1$ and $P_{1}(x)=x$.
+Context question:
+(a) Knowing that Legendre polynomials are orthogonal $\left(\left\langle P_{m}(x), P_{n}(x)\right\rangle=0\right.$ if $m \neq n)$ and $\operatorname{deg} P_{n}(x)=n$, obtain $P_{2}(x)$ and $P_{3}(x)$. For reaching the usual and most convenient form of these polynomials, divide your results by the norm: $\left\|P_{n}(x)\right\|=\frac{2}{2 n+1}$.
+
+Let us now consider a sphere of radius $R$ centered at the origin. Suppose a point charge $q$ is put at the origin and that this is the only charge inside or outside the sphere. Furthermore, the potential is $\Phi=V_{0} \cos \theta$ on the surface of the sphere.
+
+We know that we can write the potential induced by the charge on the sphere (without taking into account $q$ ) in the following way:
+
+$$
+\begin{array}{ll}
+\Phi_{-}=\sum_{n=0}^{\infty} A_{n} r^{n} P_{n}(\cos \theta), & r<R \\
+\Phi_{+}=\sum_{n=0}^{\infty} \frac{B_{n}}{r^{n+1}} P_{n}(\cos \theta), & r>R
+\end{array}
+$$
+Context answer:
+\boxed{$P_{2}(x)=C_{2}\left(x^{2}+\lambda_{1} x+\lambda_{0}\right)$ , $P_{3}=\frac{1}{2}\left(5 x^{3}-3 x\right)$}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Electromagnetism	Physics	English
+89	"(b) Any real inductor has undesired, or parasitic, resistance. We can model the real inductor as an ideal inductor $L$ in series with a parasitic resistance $R$.
+
+Due to the thermal noise $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ of its parasitic resistance, this (real) inductor will support a nonzero per-frequency mean-squared current, $\frac{d\left\langle I^{2}\right\rangle}{d f}$, even when both sides of the inductor are grounded. Compute $\frac{d\left\langle I^{2}\right\rangle}{d f}$ as a function of $f, L, T$ and $R$."	['The total impedance of the circuit is\n\n$$\nZ=R+i \\omega L=R+2 \\pi i f L \\Rightarrow|Z|=\\sqrt{R^{2}+4 \\pi^{2} f^{2} L^{2}}\n\\tag{9}\n$$\n\nTherefore, the circuit equation $V=Z I \\Rightarrow\\left\\langle V^{2}\\right\\rangle=\\left|Z^{2}\\right|\\left\\langle I^{2}\\right\\rangle$ implies\n\n$$\n\\frac{d\\left\\langle I^{2}\\right\\rangle}{d f}=\\frac{1}{\\left|Z^{2}\\right|} \\frac{d\\left\\langle V^{2}\\right\\rangle}{d f}=\\frac{4 k T R}{R^{2}+4 \\pi^{2} f^{2} L^{2}}\n\\tag{10}\n$$']	['$\\frac{4 k T R}{R^{2}+4 \\pi^{2} f^{2} L^{2}}$']	"2. Johnson-Nyquist noise
+
+In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :
+
+$$
+\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
+\tag{2}
+$$
+
+Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.
+
+Electromagnetic modes in a resistor
+
+We first establish the properties of thermally excited electromagnetic modes
+
+$$
+V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
+\tag{3}
+$$
+
+in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
+Context question:
+(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.
+
+Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is
+
+$$
+\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
+\tag{4}
+$$
+
+In the low-energy limit $\hbar \omega_{n} \ll k T$, show that
+
+$$
+\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
+\tag{5}
+$$
+
+You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is
+
+$$
+P[f, f+d f] \approx k T d f .
+\tag{6}
+$$
+
+Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)
+Context answer:
+\boxed{证明题}
+
+
+Extra Supplementary Reading Materials:
+
+Nyquist equivalent noisy voltage source 
+
+The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$.
+Context question:
+(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is
+
+$$
+P_{\max }=\frac{V^{2}}{4 R} .
+\tag{7}
+$$
+
+Give the optimal value of $r$ in terms of $R$ and $V$.
+Context answer:
+证明题
+
+
+Context question:
+(b) If the average power per frequency interval delivered to the resistor $r$ is $\frac{d\left\langle P_{\max }\right\rangle}{d f}=$ $\frac{d E}{d f}=k T$, show that $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$.
+Context answer:
+\boxed{证明题}
+
+
+Extra Supplementary Reading Materials:
+
+Other circuit elements
+
+We derived the Johnson-Nyquist noise due to a resistor, $R$. It turns out the equation $\frac{d\left\langle V^{2}\right\rangle}{d f}=$ $4 k T R$ is not generalizable to inductors or capacitors.
+Context question:
+(a) Explain why no Johnson-Nyquist noise is produced by ideal inductors or capacitors. There are multiple explanations; any explanation will be accepted. (Hint: the impedance of an ideal inductor or capacitor is purely imaginary.)
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+90	(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.	['$$\n\\frac{d^{2}}{d t^{2}}\\left(\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right)=-\\frac{G\\left(M_{1}+M_{2}\\right)}{\\left|\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right|^{3}}\\left(\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right)\n$$\n\nUsing polar coordinate, the time derivative of a position vector $\\vec{r}$ with constant magnitude,\n\n$$\n\\begin{aligned}\n& \\frac{d}{d t} \\vec{r}=\\frac{d}{d t} r \\hat{r}=r \\frac{d}{d t} \\hat{r}=\\vec{\\omega} \\times \\vec{r} \\\\\n& \\frac{d 2}{d t^{2}} \\vec{r}=\\vec{\\omega} \\times(\\vec{\\omega} \\times \\vec{r})=-\\omega^{2} \\vec{r}\n\\end{aligned}\n$$\n\nSince $\\left|\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right|$ is constant,\n\n$$\n\\begin{gathered}\n\\omega=\\sqrt{\\frac{G\\left(M_{1}+M_{2}\\right)}{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}} \\\\\nT=2 \\pi / \\omega=2 \\pi \\sqrt{\\frac{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}{G\\left(M_{1}+M_{2}\\right)}}\n\\end{gathered}\n$$']	['$\\omega=\\sqrt{\\frac{G\\left(M_{1}+M_{2}\\right)}{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}}$ , $T=2 \\pi \\sqrt{\\frac{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}{G\\left(M_{1}+M_{2}\\right)}}$']	"3. The circular restricted three-body problem
+
+In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.
+
+Two-body problem
+
+Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
+Context question:
+(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
+Context answer:
+$$
+\begin{aligned}
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
+\end{aligned}
+$$
+"	[]	Text-only	Competition	True		Expression		Open-ended	Mechanics	Physics	English
+91	(a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.	['$$\n\\begin{gathered}\nM_{1}(-\\alpha R)+M_{2}((1-\\alpha) R)=0 \\\\\n\\alpha=\\frac{M_{2}}{M_{1}+M_{2}}\n\\end{gathered}\n$$']	['$\\alpha=\\frac{M_{2}}{M_{1}+M_{2}}$']	"3. The circular restricted three-body problem
+
+In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.
+
+Two-body problem
+
+Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
+Context question:
+(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
+Context answer:
+$$
+\begin{aligned}
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
+\end{aligned}
+$$
+
+
+Context question:
+(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.
+Context answer:
+\boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$}
+
+
+Extra Supplementary Reading Materials:
+
+Circular restricted three-body problem
+
+Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$.
+
+The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$."	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+92	(b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given.	['$$\n\\begin{gathered}\n\\rho_{1}(t)=\\sqrt{(x(t)+\\alpha R)^{2}+(y(t))^{2}} \\\\\n\\rho_{2}(t)=\\sqrt{(x(t)-(1-\\alpha) R)^{2}+(y(t))^{2}}\n\\end{gathered}\n$$']	['$\\rho_{1}(t)=\\sqrt{(x(t)+\\alpha R)^{2}+(y(t))^{2}}$ , $\\rho_{2}(t)=\\sqrt{(x(t)-(1-\\alpha) R)^{2}+(y(t))^{2}}$']	"3. The circular restricted three-body problem
+
+In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.
+
+Two-body problem
+
+Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
+Context question:
+(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
+Context answer:
+$$
+\begin{aligned}
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
+\end{aligned}
+$$
+
+
+Context question:
+(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.
+Context answer:
+\boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$}
+
+
+Extra Supplementary Reading Materials:
+
+Circular restricted three-body problem
+
+Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$.
+
+The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$.
+Context question:
+(a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.
+Context answer:
+\boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Mechanics	Physics	English
+93	(c) By considering the centrifugal acceleration $\omega^{2} \vec{r}$ and Coriolis acceleration $-2 \omega \times$ $\vec{v}$, find the acceleration $\frac{d^{2}}{d t^{2}} \vec{r}$ of the third mass in terms of the coordinates and constants given, including $\rho_{1}$ and $\rho_{2}$.	['$$\n\\begin{gathered}\nm \\frac{d^{2}}{d t^{2}} \\vec{r}=-G m M_{1} \\frac{\\vec{r}-\\overrightarrow{r_{1}}}{\\rho_{1}^{3}}-G m M_{2} \\frac{\\vec{r}-\\overrightarrow{r_{2}}}{\\rho_{2}^{3}}+m \\omega^{2} \\vec{r}-2 m \\omega \\times \\vec{r} \\\\\n\\frac{d^{2}}{d t^{2}} \\vec{r}=-G M_{1} \\frac{\\vec{r}-\\overrightarrow{r_{1}}}{\\rho_{1}^{3}}-G M_{2} \\frac{\\vec{r}-\\overrightarrow{r_{2}}}{\\rho_{2}^{3}}+\\omega^{2} \\vec{r}-2 \\omega \\times \\vec{r}\n\\end{gathered}\n$$']	['$\\frac{d^{2}}{d t^{2}} \\vec{r}=-G M_{1} \\frac{\\vec{r}-\\vec{r_{1}}}{\\rho_{1}^{3}}-G M_{2} \\frac{\\vec{r}-\\vec{r_{2}}}{\\rho_{2}^{3}}+\\omega^{2} \\vec{r}-2 \\omega \\times \\vec{r}$']	"3. The circular restricted three-body problem
+
+In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.
+
+Two-body problem
+
+Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
+Context question:
+(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
+Context answer:
+$$
+\begin{aligned}
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
+\end{aligned}
+$$
+
+
+Context question:
+(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.
+Context answer:
+\boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$}
+
+
+Extra Supplementary Reading Materials:
+
+Circular restricted three-body problem
+
+Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$.
+
+The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$.
+Context question:
+(a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.
+Context answer:
+\boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$}
+
+
+Context question:
+(b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given.
+Context answer:
+\boxed{$\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+94	(d) Express $\frac{d^{2} x}{d t^{2}}$ and $\frac{d^{2} y}{d t^{2}}$ in terms of $U$, where $U=-\frac{G M_{1}}{\rho_{1}}-\frac{G M_{2}}{\rho_{2}}-\frac{\omega^{2}}{2}\left(x^{2}+y^{2}\right)$.	['$$\n\\begin{gathered}\n\\ddot{x}=-\\frac{G M_{1}(x+\\alpha R)}{\\rho_{1}^{3}}-\\frac{G M_{2}(x-(1-\\alpha) R)}{\\rho_{2}^{3}}+\\omega^{2} x+2 \\omega \\dot{y} \\\\\n\\ddot{y}=-\\frac{G M_{1} y}{\\rho_{1}^{3}}-\\frac{G M_{2} y}{\\rho_{2}^{3}}+\\omega^{2} y-2 \\omega \\dot{x}\n\\end{gathered}\n$$\n\nThen,\n\n$$\n\\begin{aligned}\n\\ddot{x} & =2 \\omega \\dot{y}-\\frac{\\partial U}{\\partial x} \\\\\n\\ddot{y} & =-2 \\omega \\dot{x}-\\frac{\\partial U}{\\partial y}\n\\end{aligned}\n$$']	['$\\ddot{x} =2 \\omega \\dot{y}-\\frac{\\partial U}{\\partial x}$ , $\\ddot{y} =-2 \\omega \\dot{x}-\\frac{\\partial U}{\\partial y}$']	"3. The circular restricted three-body problem
+
+In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.
+
+Two-body problem
+
+Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
+Context question:
+(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
+Context answer:
+$$
+\begin{aligned}
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
+\end{aligned}
+$$
+
+
+Context question:
+(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.
+Context answer:
+\boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$}
+
+
+Extra Supplementary Reading Materials:
+
+Circular restricted three-body problem
+
+Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$.
+
+The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$.
+Context question:
+(a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.
+Context answer:
+\boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$}
+
+
+Context question:
+(b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given.
+Context answer:
+\boxed{$\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$}
+
+
+Context question:
+(c) By considering the centrifugal acceleration $\omega^{2} \vec{r}$ and Coriolis acceleration $-2 \omega \times$ $\vec{v}$, find the acceleration $\frac{d^{2}}{d t^{2}} \vec{r}$ of the third mass in terms of the coordinates and constants given, including $\rho_{1}$ and $\rho_{2}$.
+Context answer:
+\boxed{$\frac{d^{2}}{d t^{2}} \vec{r}=-G M_{1} \frac{\vec{r}-\vec{r_{1}}}{\rho_{1}^{3}}-G M_{2} \frac{\vec{r}-\vec{r_{2}}}{\rho_{2}^{3}}+\omega^{2} \vec{r}-2 \omega \times \vec{r}$}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Mechanics	Physics	English
+95	(e) Hence, write down an expression of the motion of $m$ which is a constant.	['From d),\n\n$$\n\\begin{gathered}\n\\ddot{x} \\dot{x}=2 \\omega \\dot{x} \\dot{y}-\\dot{x} \\frac{\\partial U}{\\partial x} \\\\\n\\ddot{y} \\dot{y}=-2 \\omega \\dot{x} \\dot{y}-\\dot{y} \\frac{\\partial U}{\\partial y} \\\\\n\\ddot{x} \\dot{x}+\\ddot{y} \\dot{y}=-\\dot{x} \\frac{\\partial U}{\\partial x}-\\dot{y} \\frac{\\partial U}{\\partial y} \\\\\n\\frac{d}{d t}\\left[\\frac{1}{2}\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)\\right]=-\\left(\\dot{x} \\frac{\\partial U}{\\partial x}+\\dot{y} \\frac{\\partial U}{\\partial y}\\right)=-\\frac{d U}{d t} \\\\\n\\frac{d}{d t}\\left[\\frac{1}{2}\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)+U\\right]=0 \\\\\n\\text { Constant }=-2 U-v^{2}\n\\end{gathered}\n$$']	['$-2 U-v^{2}$']	"3. The circular restricted three-body problem
+
+In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion.
+
+Two-body problem
+
+Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system.
+Context question:
+(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.
+Context answer:
+$$
+\begin{aligned}
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\
+& \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right)
+\end{aligned}
+$$
+
+
+Context question:
+(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.
+Context answer:
+\boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$}
+
+
+Extra Supplementary Reading Materials:
+
+Circular restricted three-body problem
+
+Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$.
+
+The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$.
+Context question:
+(a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.
+Context answer:
+\boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$}
+
+
+Context question:
+(b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given.
+Context answer:
+\boxed{$\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$}
+
+
+Context question:
+(c) By considering the centrifugal acceleration $\omega^{2} \vec{r}$ and Coriolis acceleration $-2 \omega \times$ $\vec{v}$, find the acceleration $\frac{d^{2}}{d t^{2}} \vec{r}$ of the third mass in terms of the coordinates and constants given, including $\rho_{1}$ and $\rho_{2}$.
+Context answer:
+\boxed{$\frac{d^{2}}{d t^{2}} \vec{r}=-G M_{1} \frac{\vec{r}-\vec{r_{1}}}{\rho_{1}^{3}}-G M_{2} \frac{\vec{r}-\vec{r_{2}}}{\rho_{2}^{3}}+\omega^{2} \vec{r}-2 \omega \times \vec{r}$}
+
+
+Context question:
+(d) Express $\frac{d^{2} x}{d t^{2}}$ and $\frac{d^{2} y}{d t^{2}}$ in terms of $U$, where $U=-\frac{G M_{1}}{\rho_{1}}-\frac{G M_{2}}{\rho_{2}}-\frac{\omega^{2}}{2}\left(x^{2}+y^{2}\right)$.
+Context answer:
+\boxed{$\ddot{x} =2 \omega \dot{y}-\frac{\partial U}{\partial x}$ , $\ddot{y} =-2 \omega \dot{x}-\frac{\partial U}{\partial y}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+96	a - Calculate the flattening factor, given that the gravitational constant is $6.67 \times 10^{-11}$ N.m $\cdot \mathrm{kg}^{-2}$.	"['a - 1st method\n\nFor equilibrium we have $\\mathrm{F}_{\\mathrm{c}}=\\mathrm{F}_{\\mathrm{g}}+\\mathrm{N}$ where $\\mathrm{N}$ is normal to the surface.\n\nResolving into horizontal and vertical components, we find:\n\n<img_4429>\n\n$$\n\\begin{aligned}\n& F_{g} \\cdot \\cos (\\phi)=F_{c}+N \\cdot \\sin (\\alpha) \\\\\n& \\quad F_{g} \\cdot \\sin (\\phi)=N \\cdot \\cos (\\alpha)\n\\end{aligned} \\rightarrow F_{g} \\cdot \\cos (\\phi)=F_{c}+F_{g} \\cdot \\sin (\\phi) \\cdot \\operatorname{tg}(\\alpha)\n$$\n\nFrom:\n\n$$\nF_{g}=\\frac{G \\cdot M}{r^{2}}, \\quad F_{c}=\\omega^{2} \\cdot r, x=r \\cdot \\cos (\\phi), y=r \\cdot \\sin (\\phi) \\text { en } \\operatorname{tg}(\\alpha)=\\frac{d y}{d x}\n$$\n\nwe find:\n\n$$\ny \\cdot d y+\\left(1-\\frac{\\omega^{2} \\cdot r^{3}}{G \\cdot M}\\right) \\cdot x \\cdot d x=0\n$$\n\nwhere:\n\n$$\n\\frac{\\omega^{2} \\cdot r^{3}}{G \\cdot M} \\approx 7 \\cdot 10^{-4}\n$$\n\nThis means that, although $r$ depends on $x$ and $y$, the change in the factor in front of $x d x$ is so slight that we can take it to be constant. The solution of Eq. (1) is then an ellipse:\n\n\n\n$$\n\\frac{x^{2}}{r_{e}^{2}}+\\frac{y^{2}}{r_{p}^{2}}=1 \\rightarrow \\frac{r_{p}}{r_{e}}=\\sqrt{1-\\frac{\\omega^{2} \\cdot r^{3}}{G \\cdot M}} \\approx 1-\\frac{\\omega^{2} \\cdot r^{3}}{2 \\cdot G \\cdot M}\n$$\n\nand from this it follows that:\n\n$$\n\\epsilon=\\frac{r_{e}-r_{p}}{r_{e}}=\\frac{\\omega^{2} \\cdot r^{3}}{2 \\cdot G \\cdot M} \\approx 3,7 \\cdot 10^{-4}\n$$'
+ '2nd method \n\nFor a point mass of $1 \\mathrm{~kg}$ on the surface,\n\n$$\nU_{p o t}=-\\frac{G \\cdot M}{r} \\quad U_{k i n}=\\frac{1}{2} \\cdot \\omega^{2} \\cdot r^{2} \\cdot \\cos ^{2}(\\phi)\n$$\n\nThe form of the surface is such that $U_{\\text {pot }}-U_{\\text {kin }}=$ constant. For the equator $(\\Phi=0$, $\\left.r=r_{e}\\right)$ and for the pole $\\left(\\Phi=\\pi / 2, r=r_{p}\\right)$ we have:\n\n$$\n\\frac{G \\cdot M}{r_{p}}=\\frac{G \\cdot M}{r_{e}}+\\frac{1}{2} \\cdot \\omega^{2} \\cdot r_{e}^{2} \\rightarrow \\frac{r_{e}}{r_{p}}=1+\\frac{\\omega^{2} \\cdot r_{e}^{3}}{2 \\cdot G \\cdot M}\n$$\n\nThus:\n\n$$\n\\epsilon=\\frac{r_{e}-r_{p}}{r_{e}}=\\frac{1+\\frac{\\omega^{2} \\cdot r_{e}^{3}}{2 \\cdot G \\cdot M}-1}{1+\\frac{\\omega^{2} \\cdot r_{e}^{3}}{2 \\cdot G \\cdot M}} \\approx \\frac{\\omega^{2} \\cdot r_{e}^{3}}{2 \\cdot G \\cdot M} \\approx 3,7 \\cdot 10^{-4}\n$$']"	['$3.7 \\times 10^{-4}$']	"The rotating neutron star. 
+
+A 'millisecond pulsar' is a source of radiation in the universe that emits very short pulses with a period of one to several milliseconds. This radiation is in the radio range of wavelengths; and a suitable radio receiver can be used to detect the separate pulses and thereby to measure the period with great accuracy.
+
+These radio pulses originate from the surface of a particular sort of star, the so-called neutron star. These stars are very compact: they have a mass of the same order of magnitude as that of the sun, but their radius is only a few tens of kilometers. They spin very quickly. Because of the fast rotation, a neutron star is slightly flattened (oblate). Assume the axial cross-section of the surface to be an ellipse with almost equal axes. Let $r_{p}$ be the polar and $r_{e}$ the equatorial radii; and let us define the flattening factor by:
+
+$$
+\epsilon=\frac{\left(r_{e}-r_{p}\right)}{r_{p}}
+$$
+
+
+
+| a mass of | $2.0 \cdot 10^{30} \mathrm{~kg}$, |
+| :--- | :--- |
+| an average radius of | $1.0 \cdot 10^{4} \mathrm{~m}$, |
+| and a rotation period of | $2.0 \cdot 10^{-2} \mathrm{~s}$. |"	[]	Text-only	Competition	False		Numerical	1e-5	Open-ended	Mechanics	Physics	English
+97	2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.	['The bubble is surrounded by air.\n\n<img_4487>\n\n$P_{a}, T_{a}, \\rho_{a}$\n\nCutting the sphere in half and using the projected area to balance the forces give\n\n$$\n\\begin{aligned}\nP_{i} \\pi R_{0}^{2} & =P_{a} \\pi R_{0}^{2}+2\\left(2 \\pi R_{0} \\gamma\\right) \\\\\nP_{i} & =P_{a}+\\frac{4 \\gamma}{R_{0}}\n\\end{aligned}\n$$\n\nThe pressure and density are related by the ideal gas law:\n\n$P V=n R T$ or $P=\\frac{\\rho R T}{M}$, where $M=$ the molar mass of air.\n\nApply the ideal gas law to the air inside and outside the bubble, we get\n\n$$\n\\begin{aligned}\n& \\rho_{i} T_{i}=P_{i} \\frac{M}{R} \\\\\n& \\rho_{a} T_{a}=P_{a} \\frac{M}{R}, \\\\\n& \\frac{\\rho_{i} T_{i}}{\\rho_{a} T_{a}}=\\frac{P_{i}}{P_{a}}=\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right]\n\\end{aligned}\n$$']	['$1+\\frac{4 \\gamma}{R_{0} P_{a}}$']	"An Electrified Soap Bubble
+
+A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.
+
+The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+98	2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.	['Using $\\gamma=0.025 \\mathrm{Nm}^{-1}, R_{0}=1.0 \\mathrm{~cm}$ and $P_{a}=1.013 \\times 10^{5} \\mathrm{Nm}^{-2}$, the numerical value of the ratio is\n\n$$\n\\frac{\\rho_{i} T_{i}}{\\rho_{a} T_{a}}=1+\\frac{4 \\gamma}{R_{0} P_{a}}=1+0.0001\n$$\n\n(The effect of the surface tension is very small.)']	['1.0001']	"An Electrified Soap Bubble
+
+A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.
+
+The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
+Context question:
+2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
+Context answer:
+\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}
+"	[]	Text-only	Competition	False		Numerical	5e-5	Open-ended	Thermodynamics	Physics	English
+99	2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.	['Let $W=$ total weight of the bubble, $F=$ buoyant force due to air around the bubble\n\n$$\n\\begin{aligned}\nW & =(\\text { mass of film }+ \\text { mass of air }) g \\\\\n& =\\left(4 \\pi R_{0}^{2} \\rho_{s} t+\\frac{4}{3} \\pi R_{0}^{3} \\rho_{i}\\right) g \\\\\n& =4 \\pi R_{0}^{2} \\rho_{s} t g+\\frac{4}{3} \\pi R_{0}^{3} \\frac{\\rho_{a} T_{a}}{T_{i}}\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right] g\n\\end{aligned}\n$$\n\nThe buoyant force due to air around the bubble is\n\n$$\nB=\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a} g\n$$\n\nIf the bubble floats in still air,\n\n$$\n\\begin{aligned}\nB & \\geq W \\\\\n\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a} g & \\geq 4 \\pi R_{0}^{2} \\rho_{s} \\operatorname{tg}+\\frac{4}{3} \\pi R_{0}^{3} \\frac{\\rho_{a} T_{a}}{T_{i}}\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right] g\n\\end{aligned}\n$$\n\nRearranging to give\n\n$$\n\\begin{aligned}\nT_{i} & \\geq \\frac{R_{0} \\rho_{a} T_{a}}{R_{0} \\rho_{a}-3 \\rho_{s} t}\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right] \\\\\n& \\geq 307.1 \\mathrm{~K}\n\\end{aligned}\n$$\n\nThe air inside must be about $7.1^{\\circ} \\mathrm{C}$ warmer.']	['307.1']	"An Electrified Soap Bubble
+
+A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.
+
+The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
+Context question:
+2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
+Context answer:
+\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}
+
+
+Context question:
+2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.
+Context answer:
+\boxed{1.0001}
+"	[]	Text-only	Competition	False	K	Numerical	1e-1	Open-ended	Thermodynamics	Physics	English
+100	2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$.	"[""Ignore the radius change $\\rightarrow$ Radius remains $R_{0}=1.0 \\mathrm{~cm}$\n\n(The radius actually decreases by $0.8 \\%$ when the temperature decreases from $307.1 \\mathrm{~K}$ to $300 \\mathrm{~K}$. The film itself also becomes slightly thicker.)\n\nThe drag force from Stokes' Law is $F=6 \\pi \\eta R_{0} u$\n\nIf the bubble floats in the updraught,\n\n$F \\geq W-B$\n\n$6 \\pi \\eta R_{0} u \\geq\\left(4 \\pi R_{0}^{2} \\rho_{s} t+\\frac{4}{3} \\pi R_{0}^{3} \\rho_{i}\\right) g-\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a} g$\n\nWhen the bubble is in thermal equilibrium $T_{i}=T_{a}$.\n\n$6 \\pi \\eta R_{0} u \\geq\\left(4 \\pi R_{0}^{2} \\rho_{s} t+\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a}\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right]\\right) g-\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a} g$\n\nRearranging to give\n\n$u \\geq \\frac{4 R_{0} \\rho_{s} t g}{6 \\eta}+\\frac{\\frac{4}{3} R_{0}^{2} \\rho_{a} g\\left(\\frac{4 \\gamma}{R_{0} P_{a}}\\right)}{6 \\eta}$""]"	['$\\frac{4 R_{0} \\rho_{s} t g}{6 \\eta}+\\frac{\\frac{4}{3} R_{0}^{2} \\rho_{a} g\\left(\\frac{4 \\gamma}{R_{0} P_{a}}\\right)}{6 \\eta}$']	"An Electrified Soap Bubble
+
+A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.
+
+The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
+Context question:
+2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
+Context answer:
+\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}
+
+
+Context question:
+2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.
+Context answer:
+\boxed{1.0001}
+
+
+Context question:
+2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.
+Context answer:
+\boxed{307.1}
+
+
+Extra Supplementary Reading Materials:
+
+After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+101	2.5 Calculate the numerical value for $u$ using $\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}$.	['The numerical value is $u \\geq 0.36 \\mathrm{~m} / \\mathrm{s}$.']	['$0.36$']	"An Electrified Soap Bubble
+
+A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.
+
+The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
+Context question:
+2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
+Context answer:
+\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}
+
+
+Context question:
+2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.
+Context answer:
+\boxed{1.0001}
+
+
+Context question:
+2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.
+Context answer:
+\boxed{307.1}
+
+
+Extra Supplementary Reading Materials:
+
+After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground.
+Context question:
+2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$.
+Context answer:
+\boxed{$\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$}
+"	[]	Text-only	Competition	False	$ \mathrm{~m} / \mathrm{s}$	Numerical	1e-2	Open-ended	Thermodynamics	Physics	English
+102	2.6 If this spherical bubble is now electrified uniformly with a total charge $q$, find an equation describing the new radius $R_{1}$ in terms of $R_{0}, P_{a}, q$ and the permittivity of free space $\varepsilon_{0}$.	"[""When the bubble is electrified, the electrical repulsion will cause the bubble to expand in size and thereby raise the buoyant force.\n\nThe force/area is (e-field on the surface $\\times$ charge/area)\n\n\nConsider a very thin pill box on the soap surface.\n\n<img_4490>\n\n$E=$ electric field on the film surface that results from all other parts of the soap film, excluding the surface inside the pill box itself.\n\n$E_{q}=$ total field just outside the pill box $=\\frac{q}{4 \\pi \\varepsilon_{0} R_{1}^{2}}=\\frac{\\sigma}{\\varepsilon_{0}}$\n\n$=E+$ electric field from surface charge $\\sigma$\n\n$=E+E_{\\sigma}$\n\nUsing Gauss's Law on the pill box, we have $E_{\\sigma}=\\frac{\\sigma}{2 \\varepsilon_{0}}$ perpendicular to the film as a result of symmetry.\n\nTherefore, $E=E_{q}-E_{\\sigma}=\\frac{\\sigma}{\\varepsilon_{0}}-\\frac{\\sigma}{2 \\varepsilon_{0}}=\\frac{\\sigma}{2 \\varepsilon_{0}}=\\frac{1}{2 \\varepsilon_{0}} \\frac{q}{4 \\pi R_{1}^{2}}$\n\n\nThe repulsive force per unit area of the surface of bubble is\n\n$$\n\\left(\\frac{q}{4 \\pi R_{1}^{2}}\\right) E=\\frac{\\left(q / 4 \\pi R_{1}^{2}\\right)^{2}}{2 \\varepsilon_{0}}\n$$\n\nLet $P_{i}^{\\prime}$ and $\\rho_{i}^{\\prime}$ be the new pressure and density when the bubble is electrified.\n\nThis electric repulsive force will augment the gaseous pressure $P_{i}^{\\prime}$.\n\n$P_{i}^{\\prime}$ is related to the original $P_{i}$ through the gas law.\n\n$P_{i}^{\\prime} \\frac{4}{3} \\pi R_{1}^{3}=P_{i} \\frac{4}{3} \\pi R_{0}^{3}$\n\n$P_{i}^{\\prime}=\\left(\\frac{R_{0}}{R_{1}}\\right)^{3} P_{i}=\\left(\\frac{R_{0}}{R_{1}}\\right)^{3} P_{a}$\n\nIn the last equation, the surface tension term has been ignored.\n\nFrom balancing the forces on the half-sphere projected area, we have (again ignoring the surface tension term)\n\n$$\n\\begin{aligned}\n& P_{i}^{\\prime}+\\frac{\\left(q / 4 \\pi R_{1}^{2}\\right)^{2}}{2 \\varepsilon_{0}}=P_{a} \\\\\n& P_{a}\\left(\\frac{R_{0}}{R_{1}}\\right)^{3}+\\frac{\\left(q / 4 \\pi R_{1}^{2}\\right)^{2}}{2 \\varepsilon_{0}}=P_{a}\n\\end{aligned}\n$$\n\n\n\nRearranging to get\n\n$$\n\\left(\\frac{R_{1}}{R_{0}}\\right)^{4}-\\left(\\frac{R_{1}}{R_{0}}\\right)-\\frac{q^{2}}{32 \\pi^{2} \\varepsilon_{0} R_{0}^{4} P_{a}}=0\n\\tag{17}\n$$\n\nNote that (17) yields $\\frac{R_{1}}{R_{0}}=1$ when $q=0$, as expected.""]"	['$\\left(\\frac{R_{1}}{R_{0}}\\right)^{4}-\\left(\\frac{R_{1}}{R_{0}}\\right)-\\frac{q^{2}}{32 \\pi^{2} \\varepsilon_{0} R_{0}^{4} P_{a}}=0$']	"An Electrified Soap Bubble
+
+A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.
+
+The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
+Context question:
+2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
+Context answer:
+\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}
+
+
+Context question:
+2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.
+Context answer:
+\boxed{1.0001}
+
+
+Context question:
+2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.
+Context answer:
+\boxed{307.1}
+
+
+Extra Supplementary Reading Materials:
+
+After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground.
+Context question:
+2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$.
+Context answer:
+\boxed{$\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$}
+
+
+Context question:
+2.5 Calculate the numerical value for $u$ using $\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}$.
+Context answer:
+\boxed{$0.36$}
+
+
+Extra Supplementary Reading Materials:
+
+The above calculations suggest that the terms involving the surface tension $\gamma$ add very little to the accuracy of the result. In all of the questions below, you can neglect the surface tension terms."	[]	Text-only	Competition	False		Equation		Open-ended	Thermodynamics	Physics	English
+103	"2.7 Assume that the total charge is not too large (i.e. $\frac{q^{2}}{\varepsilon_{0} R_{0}^{4}}<<P_{a}$ ) and the bubble only experiences a small increase in its radius, find $\Delta R$ where $R_{1}=R_{0}+\Delta R$.
+
+Given that $(1+x)^{n} \approx 1+n x$ where $x \ll 1$."	['Approximate solution for $R_{1}$ when $\\frac{q^{2}}{32 \\pi^{2} \\varepsilon_{0} R_{0}^{4} P_{a}}<<1$\n\nWrite $R_{1}=R_{0}+\\Delta R, \\Delta R<<R_{0}$\n\nTherefore, $\\frac{R_{1}}{R_{0}}=1+\\frac{\\Delta R}{R_{0}},\\left(\\frac{R_{1}}{R_{0}}\\right)^{4} \\approx 1+4 \\frac{\\Delta R}{R_{0}}$\n\nEq. (17) gives:\n\n$$\n\\begin{aligned}\n& \\Delta R \\approx \\frac{q^{2}}{96 \\pi^{2} \\varepsilon_{0} R_{0}^{3} P_{a}} \\\\\n& R_{1} \\approx R_{0}+\\frac{q^{2}}{96 \\pi^{2} \\varepsilon_{0} R_{0}^{3} P_{a}} \\approx R_{0}\\left(1+\\frac{q^{2}}{96 \\pi^{2} \\varepsilon_{0} R_{0}^{4} P_{a}}\\right)\n\\end{aligned}\n$$']	['$\\frac{q^{2}}{96 \\pi^{2} \\varepsilon_{0} R_{0}^{3} P_{a}}$']	"An Electrified Soap Bubble
+
+A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$.
+
+The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.
+Context question:
+2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.
+Context answer:
+\boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$}
+
+
+Context question:
+2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.
+Context answer:
+\boxed{1.0001}
+
+
+Context question:
+2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.
+Context answer:
+\boxed{307.1}
+
+
+Extra Supplementary Reading Materials:
+
+After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground.
+Context question:
+2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$.
+Context answer:
+\boxed{$\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$}
+
+
+Context question:
+2.5 Calculate the numerical value for $u$ using $\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}$.
+Context answer:
+\boxed{$0.36$}
+
+
+Extra Supplementary Reading Materials:
+
+The above calculations suggest that the terms involving the surface tension $\gamma$ add very little to the accuracy of the result. In all of the questions below, you can neglect the surface tension terms.
+Context question:
+2.6 If this spherical bubble is now electrified uniformly with a total charge $q$, find an equation describing the new radius $R_{1}$ in terms of $R_{0}, P_{a}, q$ and the permittivity of free space $\varepsilon_{0}$.
+Context answer:
+\boxed{$\left(\frac{R_{1}}{R_{0}}\right)^{4}-\left(\frac{R_{1}}{R_{0}}\right)-\frac{q^{2}}{32 \pi^{2} \varepsilon_{0} R_{0}^{4} P_{a}}=0$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+104	A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.	"[""The width of the potential well $(L)$ should be equal to the half of the wavelength of the de Broglie standing wave $\\lambda_{\\mathrm{dB}}=h / p$, here $h$ is the Planck's constant and $p$ is the momentum of the particle. Thus $p=h / \\lambda_{\\mathrm{dB}}=h /(2 L)$, and the minimal possible energy of the particle is\n\n$$\nE_{\\min }=\\frac{p^{2}}{2 m}=\\frac{h^{2}}{8 m L^{2}}\n$$""]"	['$E_{\\min }=\\frac{h^{2}}{8 m L^{2}}$']	"Particles and Waves
+
+Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.
+
+ Part A. Quantum particle in a box
+
+Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by
+
+$$
+V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
+\tag{1}
+$$
+
+While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+105	A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ).	"[""The potential well should fit an integer number of the de Broglie half-wavelengths: $L=\\frac{1}{2} \\lambda_{\\mathrm{dB}}^{(n)} \\cdot n$, $n=1,2, \\ldots$ Therefore, particle's momentum, corresponding to the de Broglie wavelength $\\lambda_{\\mathrm{dB}}^{(n)}$ is\n\n$$\np_{n}=\\frac{h}{\\lambda_{\\mathrm{dB}}^{(n)}}=\\frac{h n}{2 L}\n$$\n\nand the corresponding energy is\n\n$$\nE_{n}=\\frac{p_{n}^{2}}{2 m}=\\frac{h^{2} n^{2}}{8 m L^{2}}, \\quad n=1,2,3, \\ldots\n\\tag{1}\n$$""]"	['$E_{n}=\\frac{h^{2} n^{2}}{8 m L^{2}}$']	"Particles and Waves
+
+Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.
+
+ Part A. Quantum particle in a box
+
+Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by
+
+$$
+V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
+\tag{1}
+$$
+
+While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.
+Context question:
+A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.
+Context answer:
+\boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+106	A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$.	"[""The energy of the emitted photon, $E=h c / \\lambda$ (here $c$ is the speed of light and $\\lambda$ is the photon's wavelength) should be equal to the energy difference $\\Delta E=E_{2}-E_{1}$, therefore\n\n$$\n\\lambda_{21}=\\frac{h c}{E_{2}-E_{1}}=\\frac{8 m c L^{2}}{3 h}\n$$""]"	['$\\lambda_{21}=\\frac{8 m c L^{2}}{3 h}$']	"Particles and Waves
+
+Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.
+
+ Part A. Quantum particle in a box
+
+Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by
+
+$$
+V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
+\tag{1}
+$$
+
+While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.
+Context question:
+A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.
+Context answer:
+\boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state.
+Context question:
+A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ).
+Context answer:
+\boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+107	C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants.	['At temperature $T$, the average kinetic energy of translational motion is $\\frac{3}{2} k_{\\mathrm{B}} T$. Equating this result to $p^{2} /(2 m)$, we obtain typical momentum $p=\\sqrt{3 m k_{\\mathrm{B}} T}$ and the de Broglie wavelength\n\n$$\n\\lambda_{\\mathrm{dB}}=\\frac{h}{p}=\\frac{h}{\\sqrt{3 m k_{\\mathrm{B}} T}}\n$$']	['$p=\\sqrt{3 m k_{\\mathrm{B}} T}$ , $\\lambda_{\\mathrm{dB}}=\\frac{h}{\\sqrt{3 m k_{\\mathrm{B}} T}}$']	"Particles and Waves
+
+Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.
+
+ Part A. Quantum particle in a box
+
+Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by
+
+$$
+V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
+\tag{1}
+$$
+
+While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.
+Context question:
+A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.
+Context answer:
+\boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state.
+Context question:
+A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ).
+Context answer:
+\boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$}
+
+
+Context question:
+A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$.
+Context answer:
+\boxed{$\lambda_{21}=\frac{8 m c L^{2}}{3 h}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part C. Bose-Einstein condensation
+
+This part is not directly related to Parts A and B. Here, we will study the collective behaviour of bosonic particles. Bosons do not respect the Pauli exclusion principle, and-at low temperatures or high densitiesexperience a dramatic phenomenon known as the Bose-Einstein condensation (BEC). This is a phase transition to an intriguing collective quantum state: a large number of identical particles 'condense' into a single quantum state and start behaving as a single wave. The transition is typically reached by cooling a fixed number of particles below the critical temperature. In principle, it can also be induced by keeping the temperature fixed and driving the particle density past its critical value.
+
+We begin by exploring the relation between the temperature and the particle density at the transition. As it turns out, estimates of their critical values can be deduced from a simple observation: Bose-Einstein condensation takes place when the de Broglie wavelength corresponding to the mean square speed of the particles is equal to the characteristic distance between the particles in a gas."	[]	Text-only	Competition	True		Expression		Open-ended	Modern Physics	Physics	English
+108	C.2 Calculate the typical distance between the particles in a gas, $\ell$, as a function of particle density $n$. Hence deduce the critical temperature $T_{c}$ as a function of atom's mass, their density and physical constants.	['The volume per particle $V / N$ is a good estimate for $\\ell^{3}$. We obtain $\\ell=n^{-1 / 3}$, with $n=N / V$ and equate $\\ell=\\lambda_{\\mathrm{dB}}$ to express $T_{c}=h^{2} n^{2 / 3} /\\left(3 m k_{\\mathrm{B}}\\right)$.']	['$\\ell=n^{-1 / 3}$ , $T_{c}=\\frac{h^{2} n^{2 / 3}}{3 m k_{\\mathrm{B}}}$']	"Particles and Waves
+
+Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.
+
+ Part A. Quantum particle in a box
+
+Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by
+
+$$
+V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
+\tag{1}
+$$
+
+While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.
+Context question:
+A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.
+Context answer:
+\boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state.
+Context question:
+A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ).
+Context answer:
+\boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$}
+
+
+Context question:
+A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$.
+Context answer:
+\boxed{$\lambda_{21}=\frac{8 m c L^{2}}{3 h}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part C. Bose-Einstein condensation
+
+This part is not directly related to Parts A and B. Here, we will study the collective behaviour of bosonic particles. Bosons do not respect the Pauli exclusion principle, and-at low temperatures or high densitiesexperience a dramatic phenomenon known as the Bose-Einstein condensation (BEC). This is a phase transition to an intriguing collective quantum state: a large number of identical particles 'condense' into a single quantum state and start behaving as a single wave. The transition is typically reached by cooling a fixed number of particles below the critical temperature. In principle, it can also be induced by keeping the temperature fixed and driving the particle density past its critical value.
+
+We begin by exploring the relation between the temperature and the particle density at the transition. As it turns out, estimates of their critical values can be deduced from a simple observation: Bose-Einstein condensation takes place when the de Broglie wavelength corresponding to the mean square speed of the particles is equal to the characteristic distance between the particles in a gas.
+Context question:
+C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants.
+Context answer:
+\boxed{$p=\sqrt{3 m k_{\mathrm{B}} T}$ , $\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}$}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Modern Physics	Physics	English
+109	A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.	"[""Stefan's law: $L_{\\odot}=\\left(4 \\pi R_{\\odot}^{2}\\right)\\left(\\sigma T_{\\mathrm{s}}^{4}\\right)$\n\n$$\nT_{\\mathrm{s}}=\\left(\\frac{L_{\\odot}}{4 \\pi R_{\\odot}^{2} \\sigma}\\right)^{1 / 4}=5.76 \\times 10^{3} \\mathrm{~K}\n$$""]"	['$5.76 \\times 10^{3} $']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :"	[]	Text-only	Competition	False	K	Numerical	1e1	Open-ended	Modern Physics	Physics	English
+110	A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$	['$$\nP_{\\text {in }}=\\int_{0}^{\\infty} u(\\nu) d \\nu=\\int_{0}^{\\infty} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi h}{c^{2}} \\nu^{3} \\exp \\left(-h \\nu / k_{\\mathrm{B}} T_{\\mathrm{s}}\\right) d \\nu\n$$\n\nLet $x=\\frac{h \\nu}{k_{\\mathrm{B}} T_{\\mathrm{s}}}$. Then, $\\nu=\\frac{k_{\\mathrm{B}} T_{\\mathrm{s}}}{h} x \\quad d \\nu=\\frac{k_{\\mathrm{B}} T_{\\mathrm{s}}}{h} d x$.\n\n$$\nP_{\\mathrm{in}}=\\frac{2 \\pi h A R_{\\odot}^{2}}{c^{2} d_{\\odot}^{2}} \\frac{\\left(k_{\\mathrm{B}} T_{\\mathrm{s}}\\right)^{4}}{h^{4}} \\int_{0}^{\\infty} x^{3} e^{-x} d x=\\frac{2 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\cdot 6=\\frac{12 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}}\n$$']	['$P_{\\mathrm{in}}=\\frac{12 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+111	A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.	['$$\n\\begin{aligned}\nn_{\\gamma}(\\nu) & =\\frac{u(\\nu)}{h \\nu} \\\\\n& =A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi}{c^{2}} \\nu^{2} \\exp \\left(-h \\nu / k_{\\mathrm{B}} T_{\\mathrm{s}}\\right)\n\\end{aligned}\n$$']	['$n_{\\gamma}(\\nu)=A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi}{c^{2}} \\nu^{2} e^ {\\frac{-h \\nu }{k_{\\mathrm{B}} T_{\\mathrm{s}}}}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+112	A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.	['The useful power output is the useful energy quantum per photon, $E_{\\mathrm{g}} \\equiv h \\nu_{\\mathrm{g}}$, multiplied by the number of photons with energy, $E \\geq E_{\\mathrm{g}}$.\n\n$$\n\\begin{aligned}\nP_{\\text {out }} & =h \\nu_{\\mathrm{g}} \\int_{\\nu_{\\mathrm{g}}}^{\\infty} n_{\\gamma}(\\nu) d \\nu \\\\\n& =h \\nu_{\\mathrm{g}} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi}{c^{2}} \\int_{\\nu_{\\mathrm{g}}}^{\\infty} \\nu^{2} \\exp \\left(-h \\nu / k_{\\mathrm{B}} T_{\\mathrm{s}}\\right) d \\nu \\\\\n& =k_{\\mathrm{B}} T_{\\mathrm{s}} x_{\\mathrm{g}} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi}{c^{2}}\\left(\\frac{k_{\\mathrm{B}} T_{\\mathrm{s}}}{h}\\right)^{3} \\int_{x_{\\mathrm{g}}}^{\\infty} x^{2} e^{-x} d x \\\\\n& =\\frac{2 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} x_{\\mathrm{g}}\\left(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}\n\\end{aligned}\n$$']	['$P_{\\text {out }}=\\frac{2 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} x_{\\mathrm{g}}(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2) e^{-x_{\\mathrm{g}}}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy)."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+113	A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.	['Efficiency $\\eta=\\frac{P_{\\text {out }}}{P_{\\text {in }}}=\\frac{x_{\\mathrm{g}}}{6}\\left(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}$']	['$\\eta=\\frac{x_{\\mathrm{g}}}{6}\\left(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+114	A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?	['$$\n\\eta=\\frac{1}{6}\\left(x_{\\mathrm{g}}^{3}+2 x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}\\right) e^{-x_{\\mathrm{g}}}\n$$\n\nPut limiting values, $\\eta(0)=0 \\quad \\eta(\\infty)=0$.\n\nSince the polynomial has all positive coefficients, it increases monotonically; the exponential function decreases monotonically. Therefore, $\\eta$ has only one maximum.\n\n$$\n\\begin{aligned}\n& \\frac{\\mathrm{d} \\eta}{\\mathrm{d} x_{\\mathrm{g}}}=\\frac{1}{6}\\left(-x_{\\mathrm{g}}^{3}+x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}} \\\\\n& \\left.\\frac{\\mathrm{d} \\eta}{\\mathrm{d} x_{\\mathrm{g}}}\\right|_{x_{\\mathrm{g}}=0}=\\left.\\frac{1}{3} \\quad \\frac{\\mathrm{d} \\eta}{\\mathrm{d} x_{\\mathrm{g}}}\\right|_{x_{\\mathrm{g}} \\rightarrow \\infty}=0\n\\end{aligned}\n$$\n\n<img_4462>']	['$\\frac{1}{3}$ , 0']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+"	[]	Text-only	Competition	True		Numerical	0	Open-ended	Modern Physics	Physics	English
+115	A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.	['The maximum will be for $\\frac{\\mathrm{d} \\eta}{\\mathrm{d} x_{\\mathrm{g}}}=\\frac{1}{6}\\left(-x_{\\mathrm{g}}^{3}+x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}=0$\n\n$$\n\\Rightarrow p\\left(x_{\\mathrm{g}}\\right) \\equiv x_{\\mathrm{g}}^{3}-x_{\\mathrm{g}}^{2}-2 x_{\\mathrm{g}}-2=0\n$$\n\nA Numerical Solution by the Bisection Method:\n\nNow,\n\n$$\n\\begin{aligned}\np(0) & =-2 \\\\\np(1) & =-4 \\\\\np(2) & =-2 \\\\\np(3) & =10 \\quad \\Rightarrow \\\\\np(2.5) & =2.375 \\quad \\Rightarrow \\quad 2<x_{0}<3 \\\\\np(2.25) & =-0.171 \\quad \\Rightarrow \\quad 2.25<x_{0}<2.5\n\\end{aligned}\n$$\n\nThe approximate value of $x_{\\mathrm{g}}$ where $\\eta$ is maximum is $x_{0}=2.27$.\n\nAlternative methods leading to the same result are acceptable.\n\n$\\eta(2.27)=0.457$']	['$x_0=2.27$ , $\\eta(2.27)=0.457$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?
+Context answer:
+\boxed{$\frac{1}{3}$ , 0}
+"	[]	Text-only	Competition	True		Numerical	1e-2,1e-3	Open-ended	Modern Physics	Physics	English
+116	A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this <br> value.	['$$\n\\begin{gathered}\nx_{\\mathrm{g}}=\\frac{1.11 \\times 1.60 \\times 10^{-19}}{1.38 \\times 10^{-23} \\times 5763}=2.23 \\\\\n\\eta_{\\mathrm{Si}}=\\frac{x_{\\mathrm{g}}}{6}\\left(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}=0.457\n\\end{gathered}\n$$']	['0.457']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?
+Context answer:
+\boxed{$\frac{1}{3}$ , 0}
+
+
+Context question:
+A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.
+Context answer:
+\boxed{$x_0=2.27$ , $\eta(2.27)=0.457$}
+"	[]	Text-only	Competition	False		Numerical	1e-3	Open-ended	Modern Physics	Physics	English
+117	A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$.	['The total gravitational potential energy of the Sun: $\\Omega=-\\int_{0}^{M_{\\odot}} \\frac{G m \\mathrm{~d} m}{r}$\n\nFor constant density, $\\rho=\\frac{3 M_{\\odot}}{4 \\pi R_{\\odot}^{3}} \\quad m=\\frac{4}{3} \\pi r^{3} \\rho \\quad \\mathrm{d} m=4 \\pi r^{2} \\rho \\mathrm{d} r$\n\n$$\n\\Omega=-\\int_{0}^{R_{\\odot}} G\\left(\\frac{4}{3} \\pi r^{3} \\rho\\right)\\left(4 \\pi r^{2} \\rho\\right) \\frac{\\mathrm{d} r}{r}=-\\frac{16 \\pi^{2} G \\rho^{2}}{3} \\frac{R_{\\odot}^{5}}{5}=-\\frac{3}{5} \\frac{G M_{\\odot}^{2}}{R_{\\odot}}\n$$']	['$\\Omega=-\\frac{3}{5} \\frac{G M_{\\odot}^{2}}{R_{\\odot}}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?
+Context answer:
+\boxed{$\frac{1}{3}$ , 0}
+
+
+Context question:
+A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.
+Context answer:
+\boxed{$x_0=2.27$ , $\eta(2.27)=0.457$}
+
+
+Context question:
+A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this <br> value.
+Context answer:
+\boxed{0.457}
+
+
+Extra Supplementary Reading Materials:
+
+In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+118	A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period.	['$$\n\\begin{gathered}\n\\tau_{\\mathrm{KH}}=\\frac{-\\Omega}{L_{\\odot}} \\\\\n\\tau_{\\mathrm{KH}}=\\frac{3 G M_{\\odot}^{2}}{5 R_{\\odot} L_{\\odot}}=1.88 \\times 10^{7} \\text { years }\n\\end{gathered}\n$$']	['$1.88 \\times 10^{7}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?
+Context answer:
+\boxed{$\frac{1}{3}$ , 0}
+
+
+Context question:
+A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.
+Context answer:
+\boxed{$x_0=2.27$ , $\eta(2.27)=0.457$}
+
+
+Context question:
+A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this <br> value.
+Context answer:
+\boxed{0.457}
+
+
+Extra Supplementary Reading Materials:
+
+In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction.
+Context question:
+A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$.
+Context answer:
+\boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$}
+"	[]	Text-only	Competition	False	years	Numerical	1e5	Open-ended	Modern Physics	Physics	English
+119	B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction.	['$$\n\\begin{aligned}\n& \\text { Solution: } \\\\\n& \\qquad \\begin{array}{ll}\n4.0 \\times 10^{-12} \\mathrm{~J} \\leftrightarrow 2 \\nu \\\\\n\\Rightarrow \\Phi_{\\nu}=\\frac{L_{\\odot}}{4 \\pi d_{\\odot}^{2} \\delta E} \\times 2=\\frac{3.85 \\times 10^{26}}{4 \\pi \\times\\left(1.50 \\times 10^{11}\\right)^{2} \\times 4.0 \\times 10^{-12}} \\times 2=6.8 \\times 10^{14} \\mathrm{~m}^{-2} \\mathrm{~s}^{-1} .\n\\end{array}\n\\end{aligned}\n$$']	['$6.8 \\times 10^{14}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?
+Context answer:
+\boxed{$\frac{1}{3}$ , 0}
+
+
+Context question:
+A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.
+Context answer:
+\boxed{$x_0=2.27$ , $\eta(2.27)=0.457$}
+
+
+Context question:
+A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this <br> value.
+Context answer:
+\boxed{0.457}
+
+
+Extra Supplementary Reading Materials:
+
+In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction.
+Context question:
+A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$.
+Context answer:
+\boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$}
+
+
+Context question:
+A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period.
+Context answer:
+\boxed{$1.88 \times 10^{7}$}
+
+
+Extra Supplementary Reading Materials:
+
+The $\tau_{\mathrm{KH}}$ calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational.
+
+B Neutrinos from the Sun : 
+
+In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is:
+
+$$
+4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}}
+$$
+
+The ""electron neutrinos"", $v_{\mathrm{e}}$, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem."	[]	Text-only	Competition	False	$\mathrm{~m}^{-2} \mathrm{~s}^{-1}$	Numerical	1e13	Open-ended	Modern Physics	Physics	English
+120	B2 In terms of $N_{1}$ and $N_{2}$, calculate what fraction, $f$, of $v_{\mathrm{e}}$ is converted to $v_{\mathrm{x}}$.	['$$\n\\begin{aligned}\n& N_{1}=\\epsilon N_{0} \\\\\n& N_{e}=\\epsilon N_{0}(1-f) \\\\\n& N_{x}=\\epsilon N_{0} f / 6 \\\\\n& N_{2}=N_{e}+N_{x}\n\\end{aligned}\n$$\n\nOR\n\n$$\n\\begin{gathered}\n(1-f) N_{1}+\\frac{f}{6} N_{1}=N_{2} \\\\\n\\Rightarrow f=\\frac{6}{5}\\left(1-\\frac{N_{2}}{N_{1}}\\right)\n\\end{gathered}\n$$']	['$f=\\frac{6}{5}(1-\\frac{N_{2}}{N_{1}})$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?
+Context answer:
+\boxed{$\frac{1}{3}$ , 0}
+
+
+Context question:
+A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.
+Context answer:
+\boxed{$x_0=2.27$ , $\eta(2.27)=0.457$}
+
+
+Context question:
+A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this <br> value.
+Context answer:
+\boxed{0.457}
+
+
+Extra Supplementary Reading Materials:
+
+In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction.
+Context question:
+A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$.
+Context answer:
+\boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$}
+
+
+Context question:
+A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period.
+Context answer:
+\boxed{$1.88 \times 10^{7}$}
+
+
+Extra Supplementary Reading Materials:
+
+The $\tau_{\mathrm{KH}}$ calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational.
+
+B Neutrinos from the Sun : 
+
+In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is:
+
+$$
+4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}}
+$$
+
+The ""electron neutrinos"", $v_{\mathrm{e}}$, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem.
+Context question:
+B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction.
+Context answer:
+\boxed{$6.8 \times 10^{14}$}
+
+
+Extra Supplementary Reading Materials:
+
+Travelling from the core of the Sun to the Earth, some of the electron neutrinos, $v_{e}$, are converted to other types of neutrinos, $v_{\mathrm{x}}$. The efficiency of the detector for detecting $v_{\mathrm{x}}$ is $1 / 6$ of its efficiency for detecting $v_{\mathrm{e}}$. If there is no neutrino conversion, we expect to detect an average of $N_{1}$ neutrinos in a year. However, due to the conversion, an average of $N_{2}$ neutrinos ( $v_{\mathrm{e}}$ and $v_{\mathrm{x}}$ combined) are actually detected per year."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+121	B.3 Assume that an electron knocked out by a neutrino loses energy at a constant rate of $\alpha$ per unit time, while it travels through water. If this electron emits Cherenkov radiation for a time, $\Delta t$, determine the energy imparted to this electron ( $E_{\text {imparted }}$ ) by the neutrino, in terms of $\alpha, \Delta t, n, m_{\mathrm{e}}$ and $c$. (Assume the electron to be at rest before its interaction with the neutrino.)	['When the electron stops emitting Cherenkov radiation, its speed has reduced to $v_{\\text {stop }}=c / n$. Its total energy at this time is\n\n$$\nE_{\\text {stop }}=\\frac{m_{\\mathrm{e}} c^{2}}{\\sqrt{1-v_{\\text {stop }}^{2} / c^{2}}}=\\frac{n m_{\\mathrm{e}} c^{2}}{\\sqrt{n^{2}-1}}\n$$\n\nThe energy of the electron when it was knocked out is\n\n$$\nE_{\\text {start }}=\\alpha \\Delta t+\\frac{n m_{\\mathrm{e}} c^{2}}{\\sqrt{n^{2}-1}}\n$$\n\nBefore interacting, the energy of the electron was equal to $m_{\\mathrm{e}} c^{2}$.\n\nThus, the energy imparted by the neutrino is\n\n$$\nE_{\\text {imparted }}=E_{\\text {start }}-m_{\\mathrm{e}} c^{2}=\\alpha \\Delta t+\\left(\\frac{n}{\\sqrt{n^{2}-1}}-1\\right) m_{\\mathrm{e}} c^{2}\n$$']	['$E_{\\text {imparted }}=\\alpha \\Delta t+(\\frac{n}{\\sqrt{n^{2}-1}}-1) m_{\\mathrm{e}} c^{2}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?
+Context answer:
+\boxed{$\frac{1}{3}$ , 0}
+
+
+Context question:
+A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.
+Context answer:
+\boxed{$x_0=2.27$ , $\eta(2.27)=0.457$}
+
+
+Context question:
+A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this <br> value.
+Context answer:
+\boxed{0.457}
+
+
+Extra Supplementary Reading Materials:
+
+In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction.
+Context question:
+A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$.
+Context answer:
+\boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$}
+
+
+Context question:
+A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period.
+Context answer:
+\boxed{$1.88 \times 10^{7}$}
+
+
+Extra Supplementary Reading Materials:
+
+The $\tau_{\mathrm{KH}}$ calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational.
+
+B Neutrinos from the Sun : 
+
+In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is:
+
+$$
+4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}}
+$$
+
+The ""electron neutrinos"", $v_{\mathrm{e}}$, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem.
+Context question:
+B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction.
+Context answer:
+\boxed{$6.8 \times 10^{14}$}
+
+
+Extra Supplementary Reading Materials:
+
+Travelling from the core of the Sun to the Earth, some of the electron neutrinos, $v_{e}$, are converted to other types of neutrinos, $v_{\mathrm{x}}$. The efficiency of the detector for detecting $v_{\mathrm{x}}$ is $1 / 6$ of its efficiency for detecting $v_{\mathrm{e}}$. If there is no neutrino conversion, we expect to detect an average of $N_{1}$ neutrinos in a year. However, due to the conversion, an average of $N_{2}$ neutrinos ( $v_{\mathrm{e}}$ and $v_{\mathrm{x}}$ combined) are actually detected per year.
+Context question:
+B2 In terms of $N_{1}$ and $N_{2}$, calculate what fraction, $f$, of $v_{\mathrm{e}}$ is converted to $v_{\mathrm{x}}$.
+Context answer:
+\boxed{$f=\frac{6}{5}(1-\frac{N_{2}}{N_{1}})$}
+
+
+Extra Supplementary Reading Materials:
+
+In order to detect neutrinos, large detectors filled with water are constructed. Although the interactions of neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in the detector. These energetic electrons move through water at high speeds, emitting electromagnetic radiation in the process. As long as the speed of such an electron is greater than the speed of light in water (refractive index, $n$ ), this radiation, called Cherenkov radiation, is emitted in the shape of a cone."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+122	B.4 If $\Delta E_{r m s}=5.54 \times 10^{-17} \mathrm{~J}$, calculate the rms speed of the Be nuclei, $\mathrm{V}_{\mathrm{Be}}$, and hence estimate $T_{\mathrm{c}}$. (Hint: $\Delta E_{r m s}$ depends on the rms value of the component of velocity along the line of sight).	['Moving ${ }^{7} \\mathrm{Be}$ nuclei give rise to Doppler effect for neutrinos. Since the fractional change in energy ( $\\Delta E_{\\mathrm{rms}} / E_{\\nu} \\sim 10^{-4}$ ) is small, the Doppler shift may be considered in the nonrelativistic limit (a relativistic treatment gives almost same answer). Taking the line of sight along the $z$-direction,\n\n$$\n\\begin{aligned}\n\\frac{\\Delta E_{\\mathrm{rms}}}{E_{\\nu}} & =\\frac{v_{z, r m s}}{c} \\\\\n& =3.85 \\times 10^{-4} \\\\\n& =\\frac{1}{\\sqrt{3}} \\frac{V_{\\mathrm{Be}}}{c}\n\\end{aligned}\n$$\n\n$\\Rightarrow V_{\\mathrm{Be}}=\\sqrt{3} \\times 3.85 \\times 10^{-4} \\times 3.00 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}=2.01 \\times 10^{5} \\mathrm{~m} \\mathrm{~s}^{-1}$.\n\nThe average temperature is obtained by equating the average kinetic energy to the thermal energy.\n\n$$\n\\begin{gathered}\n\\frac{1}{2} m_{\\mathrm{Be}} V_{\\mathrm{Be}}^{2}=\\frac{3}{2} k_{\\mathrm{B}} T_{\\mathrm{c}} \\\\\n\\Rightarrow \\quad T_{\\mathrm{c}}=1.13 \\times 10^{7} \\mathrm{~K}\n\\end{gathered}\n$$\n\n']	['$V_{\\mathrm{Be}}=2.01 \\times 10^{5}$ , $T_{\\mathrm{c}}=1.13 \\times 10^{7}$']	"Particles from the Sun ${ }^{1}$ 
+
+Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.
+
+Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.
+
+Note:
+
+(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant
+
+(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant
+
+(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant
+
+A Radiation from the sun :
+Context question:
+A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
+Context answer:
+\boxed{$5.76 \times 10^{3} $}
+
+
+Extra Supplementary Reading Materials:
+
+The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by
+
+$$
+u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
+$$
+
+where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.
+
+Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
+Context question:
+A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
+Context answer:
+\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}
+
+
+Context question:
+A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$}
+
+
+Extra Supplementary Reading Materials:
+
+The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).
+Context question:
+A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.
+Context answer:
+\boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.
+Context answer:
+\boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$}
+
+
+Context question:
+A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?
+Context answer:
+\boxed{$\frac{1}{3}$ , 0}
+
+
+Context question:
+A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.
+Context answer:
+\boxed{$x_0=2.27$ , $\eta(2.27)=0.457$}
+
+
+Context question:
+A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this <br> value.
+Context answer:
+\boxed{0.457}
+
+
+Extra Supplementary Reading Materials:
+
+In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction.
+Context question:
+A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$.
+Context answer:
+\boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$}
+
+
+Context question:
+A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period.
+Context answer:
+\boxed{$1.88 \times 10^{7}$}
+
+
+Extra Supplementary Reading Materials:
+
+The $\tau_{\mathrm{KH}}$ calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational.
+
+B Neutrinos from the Sun : 
+
+In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is:
+
+$$
+4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}}
+$$
+
+The ""electron neutrinos"", $v_{\mathrm{e}}$, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem.
+Context question:
+B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction.
+Context answer:
+\boxed{$6.8 \times 10^{14}$}
+
+
+Extra Supplementary Reading Materials:
+
+Travelling from the core of the Sun to the Earth, some of the electron neutrinos, $v_{e}$, are converted to other types of neutrinos, $v_{\mathrm{x}}$. The efficiency of the detector for detecting $v_{\mathrm{x}}$ is $1 / 6$ of its efficiency for detecting $v_{\mathrm{e}}$. If there is no neutrino conversion, we expect to detect an average of $N_{1}$ neutrinos in a year. However, due to the conversion, an average of $N_{2}$ neutrinos ( $v_{\mathrm{e}}$ and $v_{\mathrm{x}}$ combined) are actually detected per year.
+Context question:
+B2 In terms of $N_{1}$ and $N_{2}$, calculate what fraction, $f$, of $v_{\mathrm{e}}$ is converted to $v_{\mathrm{x}}$.
+Context answer:
+\boxed{$f=\frac{6}{5}(1-\frac{N_{2}}{N_{1}})$}
+
+
+Extra Supplementary Reading Materials:
+
+In order to detect neutrinos, large detectors filled with water are constructed. Although the interactions of neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in the detector. These energetic electrons move through water at high speeds, emitting electromagnetic radiation in the process. As long as the speed of such an electron is greater than the speed of light in water (refractive index, $n$ ), this radiation, called Cherenkov radiation, is emitted in the shape of a cone.
+Context question:
+B.3 Assume that an electron knocked out by a neutrino loses energy at a constant rate of $\alpha$ per unit time, while it travels through water. If this electron emits Cherenkov radiation for a time, $\Delta t$, determine the energy imparted to this electron ( $E_{\text {imparted }}$ ) by the neutrino, in terms of $\alpha, \Delta t, n, m_{\mathrm{e}}$ and $c$. (Assume the electron to be at rest before its interaction with the neutrino.)
+Context answer:
+\boxed{$E_{\text {imparted }}=\alpha \Delta t+(\frac{n}{\sqrt{n^{2}-1}}-1) m_{\mathrm{e}} c^{2}$}
+
+
+Extra Supplementary Reading Materials:
+
+The fusion of $\mathrm{H}$ into He inside the Sun takes place in several steps. Nucleus of ${ }^{7} \mathrm{Be}$ (rest mass, $m_{\mathrm{Be}}$ ) is produced in one of these intermediate steps. Subsequently, it can absorb an electron, producing a ${ }^{7} \mathrm{Li}$ nucleus (rest mass, $m_{\mathrm{Li}}<m_{\mathrm{Be}}$ ) and emitting a $v_{\mathrm{e}}$. The corresponding nuclear reaction is:
+
+$$
+{ }^{7} \mathrm{Be}+\mathrm{e}^{-} \rightarrow{ }^{7} \mathrm{Li}+v_{\mathrm{e}}
+$$
+
+When a Be nucleus ( $m_{\mathrm{Be}}=11.65 \times 10^{-27} \mathrm{~kg}$ ) is at rest and absorbs an electron also at rest, the emitted neutrino has energy $E_{v}=1.44 \times 10^{-13} \mathrm{~J}$. However, the Be nuclei are in random thermal motion due to the temperature $T_{\mathrm{c}}$ at the core of the Sun, and act as moving neutrino sources. As a result, the energy of emitted neutrinos fluctuates with a root mean square (rms) value $\Delta E_{r m s}$."	[]	Text-only	Competition	True	$ \mathrm{~m} \mathrm{~s}^{-1}$, K	Numerical	1e3,1e5	Open-ended	Modern Physics	Physics	English
+123	A small puck of mass $m$ is carefully placed onto the inner surface of the thin hollow thin cylinder of mass $M$ and of radius $R$. Initially, the cylinder rests on the horizontal plane and the puck is located at the height $R$ above the plane as shown in the figure on the left. Find the interaction force $F$ between the puck and the cylinder at the moment when the puck passes the lowest point of its trajectory. Assume that the friction between the puck and the inner surface of the cylinder is absent, and the cylinder moves on the plane without slipping. The free fall acceleration is $g$.	['Consider the forces acting on the puck and the cylinder and depicted in the figure on the right. The puck is subject to the gravity force $m g$ and the reaction force from the cylinder $N$. The cylinder is subject to the gravity force $M g$, the reaction force from the plane $N_{1}$, the friction force $F_{f r}$ and the pressure force from the puck $N^{\\prime}=-N$. The idea is to write the horizontal projections of the equations of motion. It is written for the puck as follows\n\n$$\nm a_{x}=N \\sin \\alpha,\n\\tag{A.1}\n$$\n\nwhere $a_{x}$ is the horizontal projection of the puck acceleration.\n\nFor the cylinder the equation of motion with the acceleration $w$ is found as\n\n$$\nM w=N \\sin \\alpha-F_{f r} .\n\\tag{A.2}\n$$\n\nSince the cylinder moves along the plane without sliding its\n\n<img_4540>\nangular acceleration is obtained as\n\n$$\n\\varepsilon=w / R\n\\tag{A.3}\n$$\n\nThen the equation of rotational motion around the center of mass of the cylinder takes the form\n\n$$\nI \\varepsilon=F_{f r} R\n\\tag{A.4}\n$$\n\nwhere the inertia moment of the hollow cylinder is given by\n\n$$\nI=M R^{2}\n\\tag{A.5}\n$$\n\nSolving (A.2)-(A.5) yields\n\n$$\n2 M w=N \\sin \\alpha\n\\tag{A.6}\n$$\n\nFrom equations (A.1) and (A.6) it is easily concluded that\n\n$$\nm a_{x}=2 M w .\n\\tag{A.7}\n$$\n\nSince the initial velocities of the puck and of the cylinder are both equal to zero, then, it follows from (A.7) after integrating that\n\n$$\nm u=2 M v .\n\\tag{A.8}\n$$\n\nIt is obvious that the conservation law for the system is written as\n\n$$\nm g R=\\frac{m u^{2}}{2}+\\frac{M v^{2}}{2}+\\frac{I \\omega^{2}}{2},\n\\tag{A.9}\n$$\n\nwhere the angular velocity of the cylinder is found to be\n\n$$\n\\omega=\\frac{v}{R}\n\\tag{A.10}\n$$\n\nsince it does not slide over the plane.\n\nSolving (A.8)-(A.10) results in velocities at the lowest point of the puck trajectory written as\n\n$$\nu=2 \\sqrt{\\frac{M g R}{(2 M+m)}}\n\\tag{A.12}\n$$\n$$\nv=\\frac{m}{M} \\sqrt{\\frac{M g R}{(2 M+m)}}\n\\tag{A.13}\n$$\n\nIn the reference frame sliding progressively along with the cylinder axis, the puck moves in a circle of radius $R$ and, at the lowest point of its trajectory, have the velocity\n\nand the acceleration\n\n$$\nv_{r e l}=u+v\n\\tag{A.14}\n$$\n\n$$\na_{\\text {rel }}=\\frac{v_{\\text {rel }}^{2}}{R}\n\\tag{A.15}\n$$\n\nAt the lowest point of the puck trajectory the acceleration of the cylinder axis is equal to zero, therefore, the puck acceleration in the laboratory reference frame is also given by (A.15).\n\n$$\nF-m g=\\frac{m v_{r e l}^{2}}{R}\n\\tag{A.16}\n$$\n\nthen the interaction force between the puck and the cylinder is finally found as\n\n$$\nF=3 m g\\left(1+\\frac{m}{3 M}\\right) \\text {. }\n\\tag{A.17}\n$$']	['$F=3 m g(1+\\frac{m}{3 M})$']		[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+124	"1 Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 
+
+Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$."	['1) According to the first law of thermodynamics, the amount of heat transmitted $\\delta Q$ to the gas in the bubble is found as\n\n$$\n\\delta Q=v C_{V} d T+p d V\n\\tag{B.1}\n$$\n\nwhere the molar heat capacity at arbitrary process is as follows\n\n$$\nC=\\frac{1}{v} \\frac{\\delta Q}{d T}=C_{V}+\\frac{p}{v} \\frac{d V}{d T}\n\\tag{B.2}\n$$\n\nHere $C_{V}$ stands for the molar heat capacity of the gas at constant volume, $p$ designates its pressure, $v$ is the total amount of moles of gas in the bubble, $V$ and $T$ denote the volume and temperature of the gas, respectively.\n\nEvaluate the derivative standing on the right hand side of (B.2). According to the Laplace formula, the gas pressure inside the bubble is defined by\n\n$$\np=\\frac{4 \\sigma}{r}\n\\tag{B.3}\n$$\n\nthus, the equation of any equilibrium process with the gas in the bubble is a polytrope of the form\n\n$$\np^{3} V=\\text { const. }\n\\tag{B.4}\n$$\n\nThe equation of state of an ideal gas has the form\n\n$$\np V=v R T\n\\tag{B.5}\n$$\n\nand hence equation (B.4) can be rewritten as\n\n$$\nT^{3} V^{-2}=\\text { const. }\n\\tag{B.6}\n$$\n\nDifferentiating (B.6) the derivative with respect to temperature sought is found as\n\n$$\n\\frac{d V}{d T}=\\frac{3 V}{2 T}\n\\tag{B.7}\n$$\n\nTaking into account that the molar heat capacity of a diatomic gas at constant volume is\n\n$$\nC_{V}=\\frac{5}{2} R\n\\tag{B.8}\n$$\n\nand using (B.5) it is finally obtained that\n\n$$\nC=C_{V}+\\frac{3}{2} R=4 R=33.2 \\frac{\\mathrm{J}}{\\mathrm{mole} \\cdot \\mathrm{K}}\n\\tag{B.9}\n$$']	['$33.2 $']	A bubble of radius $r=5.00 \mathrm{~cm}$, containing a diatomic ideal gas, has the soap film of thickness $h=$ $10.0 \mu \mathrm{m}$ and is placed in vacuum. The soap film has the surface tension $\sigma=4.00 \cdot 10^{-2} \frac{\mathrm{N}}{\mathrm{m}}$ and the density $\rho=1.10 \frac{\mathrm{g}}{\mathrm{cm}^{3}} .	[]	Text-only	Competition	False	$\frac{\mathrm{J}}{\mathrm{mole} \cdot \mathrm{K}}$	Numerical	1e-1	Open-ended	Thermodynamics	Physics	English
+125	"2 Find formula for the frequency $\omega$ of the small radial oscillations of the bubble and evaluate it under the assumption that the heat capacity of the soap film is much greater than the heat capacity of the gas in the bubble. Assume that the thermal equilibrium inside the bubble is reached much faster than the period of oscillations.
+
+Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$."	"[""Since the heat capacity of the gas is much smaller than the heat capacity of the soap film, and there is heat exchange between them, the gas can be considered as isothermal since the soap film plays the role of thermostat. Consider the fragment of soap film, limited by the angle $\\alpha$ as shown in the figure. It's area is found as\n\n$$\nS=\\pi(\\alpha r)^{2}\n\\tag{B.10}\n$$\n\nand the corresponding mass is obtained as\n\n$$\nm=\\rho S h \\text {. }\n\\tag{B.11}\n$$\n\nLet $x$ be an increase in the radius of the bubble, then the Newton second law for the fragment of the soap film mentioned above takes the form\n\n$$\nm \\ddot{x}=p^{\\prime} S^{\\prime}-F_{\\text {surf }},\n\\tag{B.12}\n$$\n\nwhere $F_{\\text {surf }}$ denotes the projection of the resultant surface tension force acting in the radial direction, $p^{\\prime}$ stands for the gas pressure beneath the surface of the soap film and\n\n$$\nS^{\\prime}=S\\left(1+2 \\frac{x}{r}\\right)\n$$\n\n$F_{\\text {surf }}$ is easily found as\n\n$$\nF_{\\text {surf }}=F_{S T} \\alpha=\\sigma \\cdot 2 \\cdot 2 \\pi[(r+x) \\alpha] \\cdot \\alpha .\n\\tag{B.13}\n$$\n\nSince the gaseous process can be considered isothermal, it is\n\n<img_4410>\nwritten that\n\n$$\np^{\\prime} V^{\\prime}=p V \\text {. }\n\\tag{B.14}\n$$\n\nAssuming that the volume increase is quite small, (B.14) yields\n\n$$\np^{\\prime}=p \\frac{1}{\\left(1+\\frac{x}{r}\\right)^{3}} \\approx p \\frac{1}{\\left(1+\\frac{3 x}{r}\\right)} \\approx p\\left(1-\\frac{3 x}{r}\\right)\n\\tag{B.15}\n$$\n\nThus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap film is derived as\n\n$$\n\\rho h \\ddot{x}=-\\frac{8 \\sigma}{r^{2}} x\n\\tag{B.16}\n$$\n\n\n\nwith the frequency\n\n$$\n\\omega=\\sqrt{\\frac{8 \\sigma}{\\rho h r^{2}}}=108 \\mathrm{~s}^{-1}\n\\tag{B.17}\n$$""]"	['$108 $']	"A bubble of radius $r=5.00 \mathrm{~cm}$, containing a diatomic ideal gas, has the soap film of thickness $h=$ $10.0 \mu \mathrm{m}$ and is placed in vacuum. The soap film has the surface tension $\sigma=4.00 \cdot 10^{-2} \frac{\mathrm{N}}{\mathrm{m}}$ and the density $\rho=1.10 \frac{\mathrm{g}}{\mathrm{cm}^{3}} .
+Context question:
+1 Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 
+
+Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$.
+Context answer:
+\boxed{$33.2 $}
+"	[]	Text-only	Competition	False	$\mathrm{~s}^{-1}$	Numerical	1e0	Open-ended	Thermodynamics	Physics	English
+126	A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$.	['If $V=b$ is substituted into the equation of state, then the gas pressure turns infinite. It is obvious that this is the moment when all the molecules are tightly packed. Therefore, the parameter $b$ is approximately equal to the volume of all molecules, i.e.\n\n$$\nb=N_{A} d^{3}\n\\tag{A1.1}\n$$']	['$b=N_{A} d^{3}$']	"Van der Waals equation of state
+
+In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered.
+
+Part A. Non-ideal gas equation of state
+
+Taking into account the finite size of the molecules, the gaseous equation of state takes the form
+
+$$
+P(V-b)=R T
+\tag{1}
+$$
+
+where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+127	A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.	['Let us derive an equation describing the change of the electron number density with time. It is determined by the two processes; the generation of ion pairs by external ionizer and the recombination of electrons with ions. At ionization process electrons and ions are generated in pairs, and at recombination processthey disappear in pairs as well.Thus, their concentrations are alwaysequal at any given time, i.e.\n\n$$\nn(t)=n_{e}(t)=n_{i}(t)\n\\tag{A1.1}\n$$\n\nThen the equation describing the numberdensityevolution of electrons and ions in time can be written as\n\n$$\n\\frac{d n(t)}{d t}=Z_{e x t}-r n(t)^{2}\n\\tag{A1.2}\n$$\n\nIt is easy to show that at $t \\rightarrow 0$ the function $\\tanh b t \\rightarrow 0$, therefore, by virtue of the initial condition $n(0)=0$,one finds\n\n$$\nn_{0}=0\n\\tag{A1.3}\n$$\n\nSubstituting $n_{e}(t)=a \\tanh b t$ in (A1.2) and separating it in the independent functions (hyperbolic, or 1 and $\\left.e^{x}\\right)$, one gets\n\n$$\na=\\sqrt{\\frac{Z_{e x t}}{r}}\n\\tag{A1.4}\n$$\n$$\nb=\\sqrt{r Z_{e x t}}\n\\tag{A1.5}\n$$']	['$n_{0}=0$ , $a=\\sqrt{\\frac{Z_{e x t}}{r}}$ , $b=\\sqrt{r Z_{e x t}}$']	"Problem3. Simplest model of gas discharge
+
+An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.
+
+Part A. Non-self-sustained gas discharge 
+
+In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.
+
+When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by
+
+$$
+Z_{\mathrm{rec}}=r n_{e} n_{i},
+$$
+
+where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively.
+
+Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows:
+
+$$
+n_{e}(t)=n_{0}+a \tanh b t
+$$
+
+where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent."	[]	Text-only	Competition	True		Expression		Open-ended	Electromagnetism	Physics	English
+128	A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously.	['According to equation (A1.4) the number density of electronsat steady-state is expressed in terms of the external ionizer activity as\n\n$$\nn_{e 1}=\\sqrt{\\frac{Z_{e x t 1}}{r}}\n\\tag{A2.1}\n$$\n$$\nn_{e 2}=\\sqrt{\\frac{Z_{e x t 2}}{r}}\n\\tag{A2.2}\n$$\n$$\nn_{e}=\\sqrt{\\frac{Z_{e x t 1}+Z_{e x t 2}}{r}}\n\\tag{A2.3}\n$$\n\nThus,the following analogue of the Pythagorean theorem is obtained as\n\n$$\nn_{e}=\\sqrt{n_{e 1}^{2}+n_{e 2}^{2}}=20.0 \\cdot 10^{10} \\mathrm{~cm}^{-3}\n\\tag{A2.4}\n$$']	['$20.0 \\cdot 10^{10}$']	"Problem3. Simplest model of gas discharge
+
+An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.
+
+Part A. Non-self-sustained gas discharge 
+
+In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.
+
+When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by
+
+$$
+Z_{\mathrm{rec}}=r n_{e} n_{i},
+$$
+
+where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively.
+
+Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows:
+
+$$
+n_{e}(t)=n_{0}+a \tanh b t
+$$
+
+where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent.
+Context question:
+A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.
+Context answer:
+\boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$}
+
+
+Extra Supplementary Reading Materials:
+
+Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$."	[]	Text-only	Competition	False	$ \mathrm{~cm}^{-3}$	Numerical	1e9	Open-ended	Electromagnetism	Physics	English
+129	A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge.	['In the steady state, the balance equations of electrons and ions in the tube volume take the form\n\n$$\nZ_{\\text {ext }} S L=r n_{e} n_{i} S L+\\frac{I_{e}}{e}\n\\tag{A3.1}\n$$\n$$\nZ_{\\text {ext }} S L=r n_{e} n_{i} S L+\\frac{I_{i}}{e}\n\\tag{A3.2}\n$$\n\nIt follows from equations (A3.1) and (A3.2) that the ion and electron currents are equal, i.e.\n\n$$\nI_{e}=I_{i}\n\\tag{A3.3}\n$$\n\nAt the same time the total current in each tube section is the sum of the electron and ion currents\n\n$$\nI=I_{e}+I_{i}\n\\tag{A3.4}\n$$\n\nBy definition ofthe current density the following relations hold\n\n$$\nI_{e}=\\frac{I}{2}=e n_{e} v S=e \\beta n_{e} E S\n\\tag{A3.5}\n$$\n$$\nI_{i}=\\frac{I}{2}=e n_{i} v S=e \\beta n_{i} E S\n\\tag{A3.6}\n$$\n\nSubstituting (A3.5) and (A3.6) into (A3.1) and (A3.2), the following quadratic equation for the current is derived\n\n$$\nZ_{e x t} S L=r S L\\left(\\frac{I}{2 e \\beta E S}\\right)^{2}+\\frac{I}{2 e}\n\\tag{A3.7}\n$$\n\nThe electric field strength in the gas is equal to\n\n$$\nE=\\frac{U}{L}\n\\tag{A3.8}\n$$\n\nand solution to the quadratic equation (A3.7) takes the form\n\n$$\nI=\\frac{e \\beta^{2} U^{2} S}{r L^{3}}\\left(-1 \\pm \\sqrt{1+\\frac{4 r Z_{e x t} L^{4}}{\\beta^{2} U^{2}}}\\right)\n\\tag{A3.9}\n$$\n\n\n\nIt is obvious that only positive root does make sense, i.e.\n\n$$\nI=\\frac{e \\beta^{2} U^{2} S}{r L^{3}}\\left(\\sqrt{1+\\frac{4 r Z_{e x t} L^{4}}{\\beta^{2} U^{2}}}-1\\right)\n\\tag{A3.10}\n$$']	['$I=\\frac{e \\beta^{2} U^{2} S}{r L^{3}}(\\sqrt{1+\\frac{4 r Z_{e x t} L^{4}}{\\beta^{2} U^{2}}}-1)$']	"Problem3. Simplest model of gas discharge
+
+An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.
+
+Part A. Non-self-sustained gas discharge 
+
+In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.
+
+When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by
+
+$$
+Z_{\mathrm{rec}}=r n_{e} n_{i},
+$$
+
+where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively.
+
+Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows:
+
+$$
+n_{e}(t)=n_{0}+a \tanh b t
+$$
+
+where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent.
+Context question:
+A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.
+Context answer:
+\boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$}
+
+
+Extra Supplementary Reading Materials:
+
+Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$.
+Context question:
+A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously.
+Context answer:
+\boxed{$20.0 \cdot 10^{10}$}
+
+
+Extra Supplementary Reading Materials:
+
+Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers.
+
+Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube.
+
+Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as
+
+$$
+v=\beta E
+$$
+
+where bisa constant called charge mobility."	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+130	A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$	['At low voltages (A3.10) simplifies and gives the following expression\n\n$$\nI=2 U e \\beta \\sqrt{\\frac{Z_{e x t}}{r}} \\frac{S}{L}\n\\tag{A4.1}\n$$\n\nwhich is actually the Ohm law.\n\nUsing the well-known relation\n\n$$\nR=\\frac{U}{I}\n\\tag{A4.2}\n$$\n\ntogether with\n\n$$\nR=\\rho \\frac{L}{S}\n\\tag{A4.3}\n$$\n\none gets\n\n$$\n\\rho=\\frac{1}{2 e \\beta} \\sqrt{\\frac{r}{Z_{e x t}}}\n\\tag{A4.4}\n$$']	['$\\frac{1}{2 e \\beta} \\sqrt{\\frac{r}{Z_{e x t}}}$']	"Problem3. Simplest model of gas discharge
+
+An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning.
+
+Part A. Non-self-sustained gas discharge 
+
+In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume.
+
+When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by
+
+$$
+Z_{\mathrm{rec}}=r n_{e} n_{i},
+$$
+
+where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively.
+
+Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows:
+
+$$
+n_{e}(t)=n_{0}+a \tanh b t
+$$
+
+where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent.
+Context question:
+A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.
+Context answer:
+\boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$}
+
+
+Extra Supplementary Reading Materials:
+
+Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$.
+Context question:
+A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously.
+Context answer:
+\boxed{$20.0 \cdot 10^{10}$}
+
+
+Extra Supplementary Reading Materials:
+
+Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers.
+
+Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube.
+
+Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as
+
+$$
+v=\beta E
+$$
+
+where bisa constant called charge mobility.
+Context question:
+A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge.
+Context answer:
+\boxed{$I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+131	"i. By adjusting the launching angle for a ball thrown with a fixed initial speed $v_{0}$ from the origin, targets can be hit within the region given by
+
+$$
+z \leq z_{0}-k x^{2}
+$$
+
+You can use this fact without proving it. Find the constants $z_{0}$ and $k$."	['When the stone is thrown vertically upwards, it can reach the point $x=0, z=v_{0}^{2} / 2 g$ (as it follows from the energy conservation law). Comparing this with the inequality $z \\leq z_{0}-k x^{2}$ we conclude that\n\n$$\nz_{0}=v_{0}^{2} / 2 g\n$$\n\nLet us consider the asymptotics $z \\rightarrow-\\infty$; the trajectory of the stone is a parabola, and at this limit, the horizontal displacement (for the given $z$ ) is very sensitive with respect to the curvature of the parabola: the flatter the parabola, the larger the displacement. The parabola has the flattest shape when the stone is thrown horizontally, $x=v_{0} t$ and $z=-g t^{2} / 2$, i.e. its trajectory is given by $z=-g x^{2} / 2 v_{0}^{2}$. Now, let us recall that $z \\leq z_{0}-k x^{2}$, i.e. $-g x^{2} / 2 v_{0}^{2} \\leq z_{0}-k x^{2} \\Rightarrow k \\leq g / 2 v_{0}^{2}$. Note that $k<g / 2 v_{0}^{2}$ would imply that there is a gap between the parabolic region $z \\leq z_{0}-k x^{2}$ and the given trajectory $z=-g x^{2} / 2 v_{0}^{2}$. This trajectory is supposed to be optimal for hitting targets far below $(z \\rightarrow-\\infty)$, so there should be no such a gap, and hence, we can exclude the option $k<g / 2 v_{0}^{2}$. This leaves us with\n\n$$\nk=g / 2 v_{0}^{2}\n$$']	['$\\frac{v_{0}^{2}}{ 2 g}$ , $\\frac{g}{2 v_{0}^{2}}$']	"Problem T1. Focus on sketches
+
+Part A. Ballistics
+
+A ball, thrown with an initial speed $v_{0}$, moves in a homogeneous gravitational field in the $x-z$ plane, where the $x$-axis is horizontal, and the $z$-axis is vertical and antiparallel to the free fall acceleration $g$. Neglect the effect of air drag."	[]	Text-only	Competition	True		Expression		Open-ended	Mechanics	Physics	English
+132	i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.	['$$\n\\begin{aligned}\n& T=\\text { const } \\Longrightarrow p V=\\text { const } \\\\\n& V \\propto r^{3} \\\\\n& \\therefore p \\propto r^{-3} \\Longrightarrow \\frac{p\\left(r_{1}\\right)}{p\\left(r_{0}\\right)}=2^{3}=8 .\n\\end{aligned}\n$$']	['$n=8$']	"Problem T3. Protostar formation
+
+Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant."	[]	Text-only	Competition	False		Numerical	0	Open-ended	Modern Physics	Physics	English
+133	ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.	"[""During the period considered the pressure is negligible. Therefore the gas is in free fall. By Gauss' theorem and symmetry, the gravitational field at any point in the ball is equivalent to the one generated when all the mass closer to the center is compressed into the center. Moreover, while the ball has not yet shrunk much, the field strength on its surface does not change much either. The acceleration of the outermost layer stays approximately constant. Thus,\n\n$$\nt \\approx \\sqrt{\\frac{2\\left(r_{0}-r_{2}\\right)}{g}}\n$$\n\nwhere\n\n$$\n\\begin{aligned}\ng & \\approx \\frac{G m}{r_{0}^{2}} \\\\\n\\therefore t & \\approx \\sqrt{\\frac{2 r_{0}^{2}\\left(r_{0}-r_{2}\\right)}{G m}}=\\sqrt{\\frac{0.1 r_{0}^{3}}{G m}} .\n\\end{aligned}\n$$""]"	['$\\sqrt{\\frac{0.1 r_{0}^{3}}{G m}}$']	"Problem T3. Protostar formation
+
+Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.
+Context question:
+i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.
+Context answer:
+\boxed{$n=8$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+134	iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws.	"[""Gravitationally the outer layer of the ball is influenced by the rest just as the rest were compressed into a point mass. Therefore we have Keplerian motion: the fall of any part of the outer layer consists in a halfperiod of an ultraelliptical orbit. The ellipse is degenerate into a line; its foci are at the ends of the line; one focus is at the center of the ball (by Kepler's $1^{\\text {st }}$ law) and the other one is at $r_{0}$, see figure (instead of a degenerate ellipse, a strongly elliptical ellipse is depicted). The period of the orbit is determined by the longer semiaxis of the ellipse (by Kepler's $3^{\\text {rd }}$ law). The longer semiaxis is $r_{0} / 2$ and we are interested in half a period. Thus, the answer is equal to the halfperiod of a circular orbit of radius $r_{0} / 2$ :\n\n$$\n\\left(\\frac{2 \\pi}{2 t_{r \\rightarrow 0}}\\right)^{2} \\frac{r_{0}}{2}=\\frac{G m}{\\left(r_{0} / 2\\right)^{2}} \\Longrightarrow t_{r \\rightarrow 0}=\\pi \\sqrt{\\frac{r_{0}^{3}}{8 G m}}\n$$\n\ncentre of the cloud initial position of a certain parcel\n\nstrongly elliptical orbit of the gas parcel of gas area covered by the radius vector\n\nAlternatively, one may write the energy conservation law $\\frac{\\dot{r}^{2}}{2}-\\frac{G m}{r}=E$ (that in turn is obtainable from Newton's II law $\\left.\\ddot{r}=-\\frac{G m}{r^{2}}\\right)$ with $E=-\\frac{G m}{r_{0}}$, separate the variables $\\left(\\frac{d r}{d t}=-\\sqrt{2 E+\\frac{2 G m}{r}}\\right)$ and write the integral $t=-\\int \\frac{d r}{\\sqrt{2 E+\\frac{2 G m}{r}}}$ This integral is probably not calculable during the limitted time given during the Olympiad, but a possible approach can be sketched as follows. Substituting $\\sqrt{2 E+\\frac{2 G m}{r}}=\\xi$ and $\\sqrt{2 E}=v$, one gets\n\n$$\n\\begin{aligned}\n\\frac{t_{\\infty}}{4 G m} & =\\int_{0}^{\\infty} \\frac{d \\xi}{\\left(v^{2}-\\xi^{2}\\right)^{2}} \\\\\n& =\\frac{1}{4 v^{3}} \\int_{0}^{\\infty}\\left[\\frac{v}{(v-\\xi)^{2}}+\\frac{v}{(v+\\xi)^{2}}+\\frac{1}{v-\\xi}+\\frac{1}{v+\\xi}\\right] d \\xi\n\\end{aligned}\n$$\n\nHere (after shifting the variable) one can use $\\int \\frac{d \\xi}{\\xi}=\\ln \\xi$ and $\\int \\frac{d \\xi}{\\xi^{2}}=-\\frac{1}{\\xi}$, finally getting the same answer as by Kepler's laws.""]"	['$t_{r \\rightarrow 0}=\\pi \\sqrt{\\frac{r_{0}^{3}}{8 G m}}$']	"Problem T3. Protostar formation
+
+Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.
+Context question:
+i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.
+Context answer:
+\boxed{$n=8$}
+
+
+Context question:
+ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.
+Context answer:
+\boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+135	iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$.	['By Clapeyron-Mendeleyev law,\n\n$$\np=\\frac{m R T_{0}}{\\mu V}\n$$\n\nWork done by gravity to compress the ball is\n\n$$\nW=-\\int p d V=-\\frac{m R T_{0}}{\\mu} \\int_{\\frac{4}{3} \\pi r_{0}^{3}}^{\\frac{4}{3} \\pi r_{3}^{3}} \\frac{d V}{V}=\\frac{3 m R T_{0}}{\\mu} \\ln \\frac{r_{0}}{r_{3}}\n$$\n\nThe temperature stays constant, so the internal energy does not change; hence, according to the $1^{\\text {st }}$ law of thermodynamics, the compression work $W$ is the heat radiated.']	['$Q=\\frac{3 m R T_{0}}{\\mu} \\ln \\frac{r_{0}}{r_{3}}$']	"Problem T3. Protostar formation
+
+Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.
+Context question:
+i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.
+Context answer:
+\boxed{$n=8$}
+
+
+Context question:
+ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.
+Context answer:
+\boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$}
+
+
+Context question:
+iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws.
+Context answer:
+\boxed{$t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+136	v. For radii smaller than $r_{3}$ you may neglect heat loss due to radiation. Determine how the temperature $T$ of the ball depends on its radius for $r<r_{3}$.	['The collapse continues adiabatically.\n\n$$\n\\begin{aligned}\n& p V^{\\gamma}=\\text { const } \\Longrightarrow T V^{\\gamma-1}=\\text { const. } \\\\\n& \\therefore T \\propto V^{1-\\gamma} \\propto r^{3-3 \\gamma} \\\\\n& \\therefore T=T_{0}\\left(\\frac{r_{3}}{r}\\right)^{3 \\gamma-3} .\n\\end{aligned}\n$$']	['$T(r)=T_{0}\\left(\\frac{r_{3}}{r}\\right)^{3 \\gamma-3}$']	"Problem T3. Protostar formation
+
+Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.
+Context question:
+i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.
+Context answer:
+\boxed{$n=8$}
+
+
+Context question:
+ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.
+Context answer:
+\boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$}
+
+
+Context question:
+iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws.
+Context answer:
+\boxed{$t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$}
+
+
+Context question:
+iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$.
+Context answer:
+\boxed{$Q=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+137	vi. Eventually we cannot neglect the effect of the pressure on the dynamics of the gas and the collapse stops at $r=r_{4}$ (with $r_{4} \ll r_{3}$ ). However, the radiation loss can still be neglected and the temperature is not yet high enough to ignite nuclear fusion. The pressure of such a protostar is not uniform anymore, but rough estimates with inaccurate numerical prefactors can still be done. Estimate the final radius $r_{4}$ and the respective temperature $T_{4}$.	['During the collapse, the gravitational energy is converted into heat. Since $r_{3} \\gg r_{4}$, The released gravitational energy can be estimated as $\\Delta \\Pi=-G m^{2}\\left(r_{4}^{-1}-r_{3}^{-1}\\right) \\approx-G m^{2} / r_{4}$ (exact calculation by integration adds a prefactor $\\frac{3}{5}$ ); the terminal heat energy is estimated as $\\Delta Q=c_{V} \\frac{m}{\\mu}\\left(T_{4}-T_{0}\\right) \\approx$ $c_{V} \\frac{m}{\\mu} T_{4}$ (the approximation $T_{4} \\gg T_{0}$ follows from the result of the previous question, when combined with $r_{3} \\gg r_{4}$ ). So, $\\Delta Q=\\frac{R}{\\gamma-1} \\frac{m}{\\mu} T_{4} \\approx \\frac{m}{\\mu} R T_{4}$. For the temperature $T_{4}$, we can use the result of the previous question, $T_{4}=T_{0}\\left(\\frac{r_{3}}{r_{4}}\\right)^{3 \\gamma-3}$. Since initial full energy was approximately zero, $\\Delta Q+\\Delta \\Pi \\approx 0$, we obtain\n\n$$\n\\frac{G m^{2}}{r_{4}} \\approx \\frac{m}{\\mu} R T_{0}\\left(\\frac{r_{3}}{r_{4}}\\right)^{3 \\gamma-3} \\Longrightarrow r_{4} \\approx r_{3}\\left(\\frac{R T_{0} r_{3}}{\\mu m G}\\right)^{\\frac{1}{3 \\gamma-4}}\n$$\n\nTherefore,\n\n$$\nT_{4} \\approx T_{0}\\left(\\frac{R T_{0} r_{3}}{\\mu m G}\\right)^{\\frac{3 \\gamma-3}{4-3 \\gamma}}\n$$\n\nAlternatively, one can obtain the result by approximately equating the hydrostatic pressure $\\rho r_{4} \\frac{G m}{r_{4}^{2}}$ to the gas pressure $p_{4}=\\frac{\\rho}{\\mu} R T_{4}$; the result will be exactly the same as given above.']	['$r_{3}\\left(\\frac{R T_{0} r_{3}}{\\mu m G}\\right)^{\\frac{1}{3 \\gamma-4}}$,$T_{0}\\left(\\frac{R T_{0} r_{3}}{\\mu m G}\\right)^{\\frac{3 \\gamma-3}{4-3 \\gamma}}$']	"Problem T3. Protostar formation
+
+Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.
+Context question:
+i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.
+Context answer:
+\boxed{$n=8$}
+
+
+Context question:
+ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.
+Context answer:
+\boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$}
+
+
+Context question:
+iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws.
+Context answer:
+\boxed{$t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$}
+
+
+Context question:
+iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$.
+Context answer:
+\boxed{$Q=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}$}
+
+
+Context question:
+v. For radii smaller than $r_{3}$ you may neglect heat loss due to radiation. Determine how the temperature $T$ of the ball depends on its radius for $r<r_{3}$.
+Context answer:
+\boxed{$T(r)=T_{0}\left(\frac{r_{3}}{r}\right)^{3 \gamma-3}$}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Modern Physics	Physics	English
+138	A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.	['Conservation of energy:\n\n$$\nm_{p} \\cdot c^{2}+V \\cdot e=m_{p} \\cdot c^{2} \\cdot \\gamma=\\frac{m_{p} \\cdot c^{2}}{\\sqrt{1-v^{2} / c^{2}}}\n$$\n\nPenalties\n\n$$\n\\text { No or incorrect total energy }\n$$\n\nSolve for velocity:\n\n$$\nv=c \\cdot \\sqrt{1-\\left(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+V \\cdot e}\\right)^{2}}\n$$']	['$v=c \\cdot \\sqrt{1-(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+V \\cdot e})^{2}}$']	"Large Hadron Collider
+
+Please read the general instructions in the separate envelope before you start this problem.
+
+In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 .
+
+| LHC ring |  |
+| :--- | :--- |
+| Circumference of ring | $26659 \mathrm{~m}$ |
+| Number of bunches per proton beam | 2808 |
+| Number of protons per bunch | $1.15 \times 10^{11}$ |
+| Proton beams |  |
+| Energy of protons | $7.00 \mathrm{TeV}$ |
+| Centre of mass energy | $14.0 \mathrm{TeV}$ |
+
+Table 1: Typical numerical values of relevant LHC parameters.
+
+Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$.
+
+The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$.
+
+Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN.
+
+Part A.LHC accelerator
+
+Acceleration:
+
+Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+139	A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.	['velocity (from previous question):\n\n$$\nv=c \\cdot \\sqrt{1-\\left(\\frac{m_{e} \\cdot c^{2}}{m_{e} \\cdot c^{2}+V \\cdot e}\\right)^{2}} \\text { or } c \\cdot \\sqrt{1-\\left(\\frac{m_{e} \\cdot c^{2}}{V \\cdot e}\\right)^{2}}\n$$\n\nrelative difference:\n\n$$\n\\begin{gathered}\n\\Delta=\\frac{c-v}{c}=1-\\frac{v}{c} \\\\\n\\rightarrow \\Delta \\simeq \\frac{1}{2}\\left(\\frac{m_{e} \\cdot c^{2}}{m_{e} \\cdot c^{2}+V \\cdot e}\\right)^{2} \\text { or } \\frac{1}{2}\\left(\\frac{m_{e} \\cdot c^{2}}{V \\cdot e}\\right)^{2}\n\\end{gathered}\n$$\n\nrelative difference\n\n$$\n\\Delta=3.63 \\cdot 10^{-11}\n$$']	"['$\\frac{1}{2}\\left(\\frac{m_{e} \\cdot c^{2}}{m_{e} \\cdot c^{2}+V \\cdot e}\\right)^{2}$, $\\Delta=3.63 \\cdot 10^{-11}$'
+ '$\\frac{1}{2}\\left(\\frac{m_{e} \\cdot c^{2}}{V \\cdot e}\\right)^{2}$, $\\Delta=3.63 \\cdot 10^{-11}$']"	"Large Hadron Collider
+
+Please read the general instructions in the separate envelope before you start this problem.
+
+In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 .
+
+| LHC ring |  |
+| :--- | :--- |
+| Circumference of ring | $26659 \mathrm{~m}$ |
+| Number of bunches per proton beam | 2808 |
+| Number of protons per bunch | $1.15 \times 10^{11}$ |
+| Proton beams |  |
+| Energy of protons | $7.00 \mathrm{TeV}$ |
+| Centre of mass energy | $14.0 \mathrm{TeV}$ |
+
+Table 1: Typical numerical values of relevant LHC parameters.
+
+Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$.
+
+The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$.
+
+Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN.
+
+Part A.LHC accelerator
+
+Acceleration:
+
+Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles.
+Context question:
+A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.
+Context answer:
+\boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$."	[]	Text-only	Competition	True		Expression,Numerical	,1e-12	Open-ended	Modern Physics	Physics	English
+140	"A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits.
+
+Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons."	['Balance of forces:\n\n$$\n\\frac{\\gamma \\cdot m_{p} \\cdot v^{2}}{r}=\\frac{m_{p} \\cdot v^{2}}{r \\cdot \\sqrt{1-\\frac{v^{2}}{c^{2}}}}=e \\cdot v \\cdot B\n$$\n\nIn case of a mistake, partial points can be given for intermediate steps (up to max 0.2). Examples:\n\n$$\n\\begin{array}{ll}\n\\text { Example: } & \\text { Lorentz force } \\\\\n\\text { Example: } & \\frac{\\gamma \\cdot m_{p} \\cdot v^{2}}{r}\n\\end{array}\n$$\n\nEnergy:\n\n$$\nE=(\\gamma-1) \\cdot m_{p} \\cdot c^{2} \\simeq \\gamma \\cdot m_{p} \\cdot c^{2} \\rightarrow \\gamma=\\frac{E}{m_{p} c^{2}}\n$$\n\nTherefore:\n\n$$\n\\frac{E \\cdot v}{c^{2} \\cdot r}=e \\cdot B\n$$\n\nWith\n\n$$\nv \\simeq c \\text { and } r=\\frac{L}{2 \\pi}\n$$\n\nfollows:\n\n$$\n\\rightarrow B=\\frac{2 \\pi \\cdot E}{e \\cdot c \\cdot L}\n$$\n\nSolution:\n\n$$\nB=5.50 \\mathrm{~T}\n$$']	['$B=\\frac{2 \\pi \\cdot E}{e \\cdot c \\cdot L}$ , $B=5.50$']	"Large Hadron Collider
+
+Please read the general instructions in the separate envelope before you start this problem.
+
+In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 .
+
+| LHC ring |  |
+| :--- | :--- |
+| Circumference of ring | $26659 \mathrm{~m}$ |
+| Number of bunches per proton beam | 2808 |
+| Number of protons per bunch | $1.15 \times 10^{11}$ |
+| Proton beams |  |
+| Energy of protons | $7.00 \mathrm{TeV}$ |
+| Centre of mass energy | $14.0 \mathrm{TeV}$ |
+
+Table 1: Typical numerical values of relevant LHC parameters.
+
+Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$.
+
+The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$.
+
+Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN.
+
+Part A.LHC accelerator
+
+Acceleration:
+
+Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles.
+Context question:
+A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.
+Context answer:
+\boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$.
+Context question:
+A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.
+Context answer:
+\boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$}
+
+
+Extra Supplementary Reading Materials:
+
+We now return to the protons in the LHC. Assume that the beam pipe has a circular shape."	[]	Text-only	Competition	True	,$\mathrm{~T}$	Expression,Numerical	,1e-1	Open-ended	Modern Physics	Physics	English
+141	A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$.	['Ansatz:\n\n$$\nP_{\\text {rad }}=a^{\\alpha} \\cdot q^{\\beta} \\cdot c^{\\gamma} \\cdot \\epsilon_{0}^{\\delta}\n$$\n\nDimensions: $[\\mathrm{a}]=\\mathrm{ms}^{-2},[\\mathrm{q}]=\\mathrm{C}=\\mathrm{As},[\\mathrm{c}]=\\mathrm{ms}^{-1},\\left[\\epsilon_{0}\\right]=\\mathrm{As}(\\mathrm{Vm})^{-1}=\\mathrm{A}^{2} \\mathrm{~s}^{2}\\left(\\mathrm{Nm}^{2}\\right)^{-1}=\\mathrm{A}^{2} \\mathrm{~s}^{4}\\left(\\mathrm{kgm}^{3}\\right)^{-}$']	['$P_{\\text {rad }}=a^{\\alpha} \\cdot q^{\\beta} \\cdot c^{\\gamma} \\cdot \\epsilon_{0}^{\\delta}$']	"Large Hadron Collider
+
+Please read the general instructions in the separate envelope before you start this problem.
+
+In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 .
+
+| LHC ring |  |
+| :--- | :--- |
+| Circumference of ring | $26659 \mathrm{~m}$ |
+| Number of bunches per proton beam | 2808 |
+| Number of protons per bunch | $1.15 \times 10^{11}$ |
+| Proton beams |  |
+| Energy of protons | $7.00 \mathrm{TeV}$ |
+| Centre of mass energy | $14.0 \mathrm{TeV}$ |
+
+Table 1: Typical numerical values of relevant LHC parameters.
+
+Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$.
+
+The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$.
+
+Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN.
+
+Part A.LHC accelerator
+
+Acceleration:
+
+Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles.
+Context question:
+A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.
+Context answer:
+\boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$.
+Context question:
+A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.
+Context answer:
+\boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$}
+
+
+Extra Supplementary Reading Materials:
+
+We now return to the protons in the LHC. Assume that the beam pipe has a circular shape.
+Context question:
+A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits.
+
+Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons.
+Context answer:
+\boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$}
+
+
+Extra Supplementary Reading Materials:
+
+Radiated Power: 
+
+An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+142	A.5 Calculate $P_{\text {tot }}$, the total radiated power of the LHC, for a proton energy of E= 7.00 $\mathrm{TeV}$ (Note table 1). You may use suitable approximations.	['Radiated Power:\n\n$$\nP_{r a d}=\\frac{\\gamma^{4} \\cdot a^{2} \\cdot e^{2}}{6 \\pi \\cdot c^{3} \\cdot \\epsilon_{0}}\n$$\n\nEnergy:\n\n$$\nE=(\\gamma-1) m_{p} \\cdot c^{2} \\text { or equally valid } E \\simeq \\gamma \\cdot m_{p} \\cdot c^{2}\n$$\n\nAcceleration:\n\n$$\na \\simeq \\frac{c^{2}}{r} \\text { with } \\quad r=\\frac{L}{2 \\pi}\n$$\n\nTherefore:\n\n$$\n\\begin{gathered}\nP_{\\text {rad }}=\\left(\\frac{E}{m_{p} c^{2}}+1\\right)^{4} \\cdot \\frac{e^{2} \\cdot c}{6 \\pi \\epsilon_{0} \\cdot r^{2}} \\text { or }\\left(\\frac{E}{m_{p} c^{2}}\\right)^{4} \\cdot \\frac{e^{2} \\cdot c}{6 \\pi \\epsilon_{0} \\cdot r^{2}} \\\\\n\\left(\\text { not required } P_{\\text {rad }}=7.94 \\cdot 10^{-12} \\mathrm{~W}\\right)\n\\end{gathered}\n$$\n\nTotal radiated power:\n\n$$\nP_{t o t}=2 \\cdot 2808 \\cdot 1.15 \\cdot 10^{11} \\cdot P_{r a d}=5.13 \\mathrm{~kW}\n$$']	['5.13']	"Large Hadron Collider
+
+Please read the general instructions in the separate envelope before you start this problem.
+
+In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 .
+
+| LHC ring |  |
+| :--- | :--- |
+| Circumference of ring | $26659 \mathrm{~m}$ |
+| Number of bunches per proton beam | 2808 |
+| Number of protons per bunch | $1.15 \times 10^{11}$ |
+| Proton beams |  |
+| Energy of protons | $7.00 \mathrm{TeV}$ |
+| Centre of mass energy | $14.0 \mathrm{TeV}$ |
+
+Table 1: Typical numerical values of relevant LHC parameters.
+
+Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$.
+
+The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$.
+
+Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN.
+
+Part A.LHC accelerator
+
+Acceleration:
+
+Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles.
+Context question:
+A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.
+Context answer:
+\boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$.
+Context question:
+A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.
+Context answer:
+\boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$}
+
+
+Extra Supplementary Reading Materials:
+
+We now return to the protons in the LHC. Assume that the beam pipe has a circular shape.
+Context question:
+A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits.
+
+Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons.
+Context answer:
+\boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$}
+
+
+Extra Supplementary Reading Materials:
+
+Radiated Power: 
+
+An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$.
+Context question:
+A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$.
+Context answer:
+\boxed{$P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}$}
+
+
+Extra Supplementary Reading Materials:
+
+The real formula for the radiated power contains a factor $1 /(6 \pi)$; moreover, a full relativistic derivation gives an additional multiplicative factor $\gamma^{4}$, with $\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}$."	[]	Text-only	Competition	False	kW	Numerical	1e-1	Open-ended	Modern Physics	Physics	English
+143	"3.3
+
+Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles:
+
+3.3.A
+
+The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$.
+
+(ie. $\frac{1}{\mathrm{r}_{\mathrm{0}}} ; \mathrm{r}_{0}$ is given in the problem).
+
+3.3.B
+
+In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $\left(\mathrm{p}_{\mathrm{0}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{0}}\right)^{2} \approx(\hbar)^{2}$"	['3.3 \n\nDetermination of the energy of $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ :\n\npotential energy $=-\\frac{Z \\cdot e^{2}}{4 \\cdot \\pi \\cdot \\varepsilon_{0} \\cdot r_{0}}=-\\frac{Z \\cdot q^{2}}{r_{0}}$\n\nkinetic energy $=\\frac{p^{2}}{2 \\cdot m}$\n\nIf the motion of the electrons is confined within the $x-y$-plane, principles of uncertainty in 3.1 can be written as\n\n$r_{0}^{2}=(\\Delta \\mathrm{x})^{2}+(\\Delta \\mathrm{y})^{2}$\n\n$\\mathrm{p}_{0}^{2}=\\left(\\Delta \\mathrm{p}_{\\mathrm{x}}\\right)^{2}+\\left(\\Delta \\mathrm{p}_{\\mathrm{y}}\\right)^{2}$\n\n$\\mathrm{p}_{0}^{2}=\\frac{\\hbar^{2}}{4} \\cdot\\left[\\frac{1}{(\\Delta \\mathrm{x})^{2}}+\\frac{1}{(\\Delta \\mathrm{y})^{2}}\\right]=\\frac{\\hbar^{2}}{4} \\cdot\\left[\\frac{2}{\\mathrm{r}_{0}^{2}}+\\frac{2}{\\mathrm{r}_{0}^{2}}\\right]=\\frac{\\hbar^{2}}{4} \\cdot \\frac{4}{\\mathrm{r}_{0}^{2}}$\n\nthus\n\n$\\mathrm{p}_{0}^{2} \\cdot \\mathrm{r}_{0}^{2}=\\hbar^{2}$\n\n$E\\left[A^{(z-1)+}\\right]=\\frac{p_{0}^{2}}{2 \\cdot m_{e}}-\\frac{Z \\cdot q^{2}}{r_{0}}=\\frac{\\hbar^{2}}{2 \\cdot m_{e} \\cdot r_{e}}-\\frac{Z \\cdot q^{2}}{r_{0}}$\n\nEnergy minimum exists, when $\\frac{\\mathrm{dE}}{\\mathrm{dr}_{0}}=0$.\n\nHence\n\n$\\frac{\\mathrm{dE}}{\\mathrm{dr}_{0}}=-\\frac{\\hbar^{2}}{\\mathrm{~m}_{\\mathrm{e}} \\cdot \\mathrm{r}_{\\mathrm{e}}^{3}}+\\frac{\\mathrm{Z} \\cdot \\mathrm{q}^{2}}{\\mathrm{r}_{0}^{2}}=0$\n\nthis gives $\\frac{1}{r_{0}}=\\frac{Z \\cdot q^{2} \\cdot m_{e}}{\\hbar^{2}}$\n\nhence\n\n$$\n\\begin{gathered}\nE\\left[A^{(Z-1)+}\\right]=\\frac{\\hbar^{2}}{2 \\cdot m_{e}} \\cdot\\left(\\frac{Z \\cdot q^{2} \\cdot m_{e}}{\\hbar}\\right)^{2}-Z \\cdot q^{2} \\cdot \\frac{Z \\cdot q^{2} \\cdot m_{e}}{\\hbar^{2}}=-\\frac{m_{e}}{2} \\cdot\\left(\\frac{Z \\cdot q^{2}}{\\hbar}\\right)^{2}=-\\frac{q^{2} \\cdot Z^{2}}{2 \\cdot r_{B}}=-E_{R} \\cdot Z^{2} \\\\\nE\\left[A^{(Z-1)+}\\right]=-E_{R} \\cdot Z^{2}\n\\end{gathered}\n$$']	['$-E_{R} \\cdot Z^{2}$']	"Recombination of Positive and Negative Ions in Ionized Gas
+
+Introduction
+
+A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit).
+
+Let this ion be represented by the symbol $\mathrm{A}^{(\mathrm{Z}-1)+}$
+
+Constants:
+
+| electric field constant | $\varepsilon_{\mathrm{0}}=8.85 \cdot 10^{-12} \frac{\mathrm{A} \cdot \mathrm{s}}{\mathrm{V} \cdot \mathrm{m}}$ |
+| :--- | :--- |
+| elementary charge | $\mathrm{e}= \pm 1.602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{s}$ |
+|  | $\mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{\mathrm{0}}}=2.037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{m}$ |
+| Planck's constant | $\hbar=1.054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ |
+| (rest) mass of an electron | $\mathrm{m}_{\mathrm{e}}=9.108 \cdot 10^{-31} \mathrm{~kg}$ |
+| Bohr's atomic radius | $\mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5.92 \cdot 10^{-11} \mathrm{~m}$ |
+| Rydberg's energy | $\mathrm{E}_{\mathrm{R}}=\frac{\mathrm{q}^{2}}{2 \cdot \mathrm{r}_{\mathrm{B}}}=2.180 \cdot 10^{-18} \mathrm{~J}$ |
+| (rest) mass of a proton | $\mathrm{m}_{\mathrm{P}} \cdot \mathrm{c}^{2}=1.503 \cdot 10^{-10} \mathrm{~J}$ |
+Context question:
+3.1
+
+Assume that the ion which has just one electron left the shell.
+
+$\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state.
+
+In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-\mathrm{y}$ - and $\mathrm{z}$-axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$.
+
+Write inequality involving $\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2}$ in a complete form.
+Context answer:
+$\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$
+
+
+Context question:
+3.2 
+
+The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon.
+
+Write down an equation which is to be used for calculation the frequency of an emitted photon.
+Context answer:
+3.2
+
+$\left|\vec{v}_{\mathrm{e}}\right| \ldots . .$. speed of the external electron before the capture
+
+$\left|\vec{V}_{i}\right| \ldots . .$. speed of $A^{(Z-1)+}$ before capturing
+
+$\left|\vec{V}_{\mathrm{f}}\right| \ldots . .$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ after capturing
+
+$\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .$. energy of the emitted photon
+
+conservation of energy:
+
+$\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right]$
+
+where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively.
+
+conservation of momentum:
+
+$m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$
+
+where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+144	"3.4
+
+Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ taken to be in the ground state, using the same principle as the calculation of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$. Given the average distance of each of the two electrons in the outermost shell (same as $r_{0}$ given in 3.3) denoted by $r_{1}$ and $r_{2}$, assume the average distance between the two electrons is given by $r_{1}+r_{2}$ and the average value of the square of the momentum of each electron obeys the principle of uncertainty ie.
+
+$\mathrm{p}_{1}^{2} \cdot \mathrm{r}_{1}^{2} \approx \hbar^{2}$ and $\mathrm{p}_{2}^{2} \cdot \mathrm{r}_{2}^{2} \approx \hbar^{2}$
+
+hint: Make use of the information that in the ground state $r_{1}=r_{2}$"	['3.4 \n\nIn the case of $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ ion captures a second electron\n\npotential energy of both electrons $=-2 \\cdot \\frac{Z \\cdot q^{2}}{r_{0}}$\n\nkinetic energy of the two electrons $=2 \\cdot \\frac{p^{2}}{2 \\cdot m}=\\frac{\\hbar^{2}}{m_{e} \\cdot r_{0}^{2}}$\n\npotential energy due to interaction between the two electrons $=\\frac{q^{2}}{\\left|\\vec{r}_{1}-\\vec{r}_{2}\\right|}=\\frac{q^{2}}{2 \\cdot r_{0}}$\n\n$E\\left[A^{(z-2)+}\\right]=\\frac{\\hbar^{2}}{m_{e} \\cdot r_{0}^{2}}-\\frac{2 \\cdot Z \\cdot q^{2}}{r_{0}^{2}}+\\frac{q^{2}}{2 \\cdot r_{0}}$\n\ntotal energy is lowest when $\\frac{\\mathrm{dE}}{\\mathrm{dr}}=0$\n\nhence\n\n$0=-\\frac{2 \\cdot \\hbar^{2}}{m_{e} \\cdot r_{0}^{3}}+\\frac{2 \\cdot Z \\cdot q^{2}}{r_{0}^{3}}-\\frac{q^{2}}{2 \\cdot r_{0}^{2}}$\n\nhence\n\n$\\frac{1}{r_{0}}=\\frac{q^{2} \\cdot m_{e}}{2 \\cdot \\hbar^{2}} \\cdot\\left(2 \\cdot z-\\frac{1}{2}\\right)=\\frac{1}{r_{B}} \\cdot\\left(z-\\frac{1}{4}\\right)$\n\n$E\\left[A^{(Z-2)+}\\right]=\\frac{\\hbar^{2}}{m_{e}} \\cdot\\left(\\frac{q^{2} \\cdot m_{e}}{2 \\cdot \\hbar^{2}}\\right)^{2}-\\frac{q^{2} \\cdot\\left(2 \\cdot Z-\\frac{1}{2}\\right)}{\\hbar} \\cdot \\frac{q^{2} \\cdot m_{e} \\cdot\\left(2 \\cdot Z-\\frac{1}{2}\\right)}{2 \\cdot \\hbar}$\n\n$E\\left[A^{(Z-2)+}\\right]=-\\frac{m_{e}}{4} \\cdot\\left[\\frac{q^{2} \\cdot\\left(2 \\cdot Z-\\frac{1}{2}\\right)}{\\hbar}\\right]^{2}=-\\frac{m_{e} \\cdot\\left[q^{2} \\cdot\\left(Z-\\frac{1}{4}\\right)\\right]^{2}}{\\hbar^{2}}=-\\frac{q^{2} \\cdot\\left(Z-\\frac{1}{4}\\right)^{2}}{\\hbar^{2}}$\n\nthis gives\n\n$$\nE\\left[A^{(z-2)+}\\right]=-2 \\cdot E_{R} \\cdot\\left(z-\\frac{1}{4}\\right)^{2}\n$$']	['$-2 \\cdot E_{R} \\cdot(z-\\frac{1}{4})^{2}$']	"Recombination of Positive and Negative Ions in Ionized Gas
+
+Introduction
+
+A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit).
+
+Let this ion be represented by the symbol $\mathrm{A}^{(\mathrm{Z}-1)+}$
+
+Constants:
+
+| electric field constant | $\varepsilon_{\mathrm{0}}=8.85 \cdot 10^{-12} \frac{\mathrm{A} \cdot \mathrm{s}}{\mathrm{V} \cdot \mathrm{m}}$ |
+| :--- | :--- |
+| elementary charge | $\mathrm{e}= \pm 1.602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{s}$ |
+|  | $\mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{\mathrm{0}}}=2.037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{m}$ |
+| Planck's constant | $\hbar=1.054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ |
+| (rest) mass of an electron | $\mathrm{m}_{\mathrm{e}}=9.108 \cdot 10^{-31} \mathrm{~kg}$ |
+| Bohr's atomic radius | $\mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5.92 \cdot 10^{-11} \mathrm{~m}$ |
+| Rydberg's energy | $\mathrm{E}_{\mathrm{R}}=\frac{\mathrm{q}^{2}}{2 \cdot \mathrm{r}_{\mathrm{B}}}=2.180 \cdot 10^{-18} \mathrm{~J}$ |
+| (rest) mass of a proton | $\mathrm{m}_{\mathrm{P}} \cdot \mathrm{c}^{2}=1.503 \cdot 10^{-10} \mathrm{~J}$ |
+Context question:
+3.1
+
+Assume that the ion which has just one electron left the shell.
+
+$\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state.
+
+In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-\mathrm{y}$ - and $\mathrm{z}$-axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$.
+
+Write inequality involving $\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2}$ in a complete form.
+Context answer:
+$\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$
+
+
+Context question:
+3.2 
+
+The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon.
+
+Write down an equation which is to be used for calculation the frequency of an emitted photon.
+Context answer:
+3.2
+
+$\left|\vec{v}_{\mathrm{e}}\right| \ldots . .$. speed of the external electron before the capture
+
+$\left|\vec{V}_{i}\right| \ldots . .$. speed of $A^{(Z-1)+}$ before capturing
+
+$\left|\vec{V}_{\mathrm{f}}\right| \ldots . .$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ after capturing
+
+$\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .$. energy of the emitted photon
+
+conservation of energy:
+
+$\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right]$
+
+where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively.
+
+conservation of momentum:
+
+$m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$
+
+where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.
+
+
+Context question:
+3.3
+
+Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles:
+
+3.3.A
+
+The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$.
+
+(ie. $\frac{1}{\mathrm{r}_{\mathrm{0}}} ; \mathrm{r}_{0}$ is given in the problem).
+
+3.3.B
+
+In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $\left(\mathrm{p}_{\mathrm{0}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{0}}\right)^{2} \approx(\hbar)^{2}$
+Context answer:
+\boxed{$-E_{R} \cdot Z^{2}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+145	"3.5
+
+Consider in particular the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ is at rest in the ground state when capturing an additional electron and the captured electron is also at rest prior to the capturing. Determine the numerical value of $Z$, if the frequency of the emitted photon accompanying electron capturing is $2,057\times 10^{17} \mathrm{rad} / \mathrm{s}$. Identify the element which gives rise to the ion."	['3.5 \n\nThe ion $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ is at rest when it captures the second electron also at rest before capturing.\n\nFrom the information provided in the problem, the frequency of the photon emitted is given by\n\n$v=\\frac{\\omega}{2 \\cdot \\pi}=\\frac{2,057 \\cdot 10^{17}}{2 \\cdot \\pi} \\mathrm{Hz}$\n\nThe energy equation can be simplified to $\\quad E\\left[A^{(Z-1)+}\\right]-E\\left[A^{(Z-2)+}\\right]=\\hbar \\cdot \\omega=h \\cdot v$ that is\n\n$-E_{R} \\cdot Z^{2}-\\left[-2 \\cdot E_{R} \\cdot\\left(Z-\\frac{1}{4}\\right)^{2}\\right]=\\hbar \\cdot \\omega$\n\nputting in known numbers follows\n\n$2,180 \\cdot 10^{-18} \\cdot\\left[-Z^{2}+2 \\cdot\\left(Z-\\frac{1}{4}\\right)^{2}\\right]=1,05 \\cdot 10^{-34} \\cdot 2,607 \\cdot 10^{17}$\n\nthis gives\n\n$Z^{2}-Z-12,7=0$\n\nwith the physical sensuous result $\\quad Z=\\frac{1+\\sqrt{1+51}}{2}=4,1$\n\nThis implies $\\mathrm{Z}=4$, and that means Beryllium\n\n']	['Z=4, Beryllium']	"Recombination of Positive and Negative Ions in Ionized Gas
+
+Introduction
+
+A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit).
+
+Let this ion be represented by the symbol $\mathrm{A}^{(\mathrm{Z}-1)+}$
+
+Constants:
+
+| electric field constant | $\varepsilon_{\mathrm{0}}=8.85 \cdot 10^{-12} \frac{\mathrm{A} \cdot \mathrm{s}}{\mathrm{V} \cdot \mathrm{m}}$ |
+| :--- | :--- |
+| elementary charge | $\mathrm{e}= \pm 1.602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{s}$ |
+|  | $\mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{\mathrm{0}}}=2.037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{m}$ |
+| Planck's constant | $\hbar=1.054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ |
+| (rest) mass of an electron | $\mathrm{m}_{\mathrm{e}}=9.108 \cdot 10^{-31} \mathrm{~kg}$ |
+| Bohr's atomic radius | $\mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5.92 \cdot 10^{-11} \mathrm{~m}$ |
+| Rydberg's energy | $\mathrm{E}_{\mathrm{R}}=\frac{\mathrm{q}^{2}}{2 \cdot \mathrm{r}_{\mathrm{B}}}=2.180 \cdot 10^{-18} \mathrm{~J}$ |
+| (rest) mass of a proton | $\mathrm{m}_{\mathrm{P}} \cdot \mathrm{c}^{2}=1.503 \cdot 10^{-10} \mathrm{~J}$ |
+Context question:
+3.1
+
+Assume that the ion which has just one electron left the shell.
+
+$\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state.
+
+In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-\mathrm{y}$ - and $\mathrm{z}$-axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$.
+
+Write inequality involving $\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2}$ in a complete form.
+Context answer:
+$\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$
+
+
+Context question:
+3.2 
+
+The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon.
+
+Write down an equation which is to be used for calculation the frequency of an emitted photon.
+Context answer:
+3.2
+
+$\left|\vec{v}_{\mathrm{e}}\right| \ldots . .$. speed of the external electron before the capture
+
+$\left|\vec{V}_{i}\right| \ldots . .$. speed of $A^{(Z-1)+}$ before capturing
+
+$\left|\vec{V}_{\mathrm{f}}\right| \ldots . .$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ after capturing
+
+$\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .$. energy of the emitted photon
+
+conservation of energy:
+
+$\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right]$
+
+where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively.
+
+conservation of momentum:
+
+$m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$
+
+where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.
+
+
+Context question:
+3.3
+
+Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles:
+
+3.3.A
+
+The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$.
+
+(ie. $\frac{1}{\mathrm{r}_{\mathrm{0}}} ; \mathrm{r}_{0}$ is given in the problem).
+
+3.3.B
+
+In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $\left(\mathrm{p}_{\mathrm{0}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{0}}\right)^{2} \approx(\hbar)^{2}$
+Context answer:
+\boxed{$-E_{R} \cdot Z^{2}$}
+
+
+Context question:
+3.4
+
+Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ taken to be in the ground state, using the same principle as the calculation of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$. Given the average distance of each of the two electrons in the outermost shell (same as $r_{0}$ given in 3.3) denoted by $r_{1}$ and $r_{2}$, assume the average distance between the two electrons is given by $r_{1}+r_{2}$ and the average value of the square of the momentum of each electron obeys the principle of uncertainty ie.
+
+$\mathrm{p}_{1}^{2} \cdot \mathrm{r}_{1}^{2} \approx \hbar^{2}$ and $\mathrm{p}_{2}^{2} \cdot \mathrm{r}_{2}^{2} \approx \hbar^{2}$
+
+hint: Make use of the information that in the ground state $r_{1}=r_{2}$
+Context answer:
+\boxed{$-2 \cdot E_{R} \cdot(z-\frac{1}{4})^{2}$}"	[]	Text-only	Competition	True		Numerical		Open-ended	Modern Physics	Physics	English
+146	A. Determine the boiling temperatures of the liquids A and B under the pressure $p_{0}$.	['PART A\n\nThe liquid boils when the pressure of its saturated vapor is equal to the external pressure. Thus, in order to find the boiling temperature of the liquid $i(i-\\mathrm{A}$ or $\\mathrm{B}$ ), one should determine such a temperature $T_{b i}$ (or $t_{b i}$ ) for which $p_{i} / p_{0}=1$.\n\nThen $\\ln \\left(p_{i} / p_{0}\\right)=0$, and we have:\n\n$$\nT_{b i}=-\\frac{\\alpha_{i}}{\\beta_{i}}\n$$\n\n\n\nThe coefficients $\\alpha_{i}$ and $\\beta_{i}$ are not given explicitly. However, they can be calculated from the formula given in the text of the problem. For this purpose one should make use of the numerical data given in the Tab. 1.1.\n\nFor the liquid A, we have:\n\n$$\n\\begin{aligned}\n& \\ln 0.284=\\frac{\\alpha_{A}}{(40+273.15) \\mathrm{K}}+\\beta_{A}, \\\\\n& \\ln 1.476=\\frac{\\alpha_{A}}{(90+273.15) \\mathrm{K}}+\\beta_{A} .\n\\end{aligned}\n$$\n\nAfter subtraction of these equations, we get:\n\n$$\n\\begin{gathered}\n\\ln 0.284-\\ln 1.476=\\alpha_{A}\\left(\\frac{1}{40+273.15}-\\frac{1}{90+273.15}\\right) \\mathrm{K}^{-1} . \\\\\n\\alpha_{A}=\\frac{\\ln \\frac{0.284}{1.476}}{\\frac{1}{40+273.15}-\\frac{1}{90+273.15}} \\mathrm{~K} \\approx-3748.49 \\mathrm{~K}\n\\end{gathered}\n$$\n\nHence,\n\n$$\n\\beta_{A}=\\ln 0.284-\\frac{\\alpha_{A}}{(40+273.15) \\mathrm{K}} \\approx 10.711\n$$\n\nThus, the boiling temperature of the liquid $\\mathrm{A}$ is equal to\n\n$$\nT_{b A}=3748.49 \\mathrm{~K} / 10.711 \\approx 349.95 \\mathrm{~K} .\n$$\n\nIn the Celsius scale the boiling temperature of the liquid $\\mathrm{A}$ is\n\n$$\nt_{b A}=(349.95-273.15)^{\\circ} \\mathrm{C}=76.80^{\\circ} \\mathrm{C} \\approx 77^{\\circ} \\mathrm{C} .\n$$\n\nFor the liquid B, in the same way, we obtain:\n\n$$\n\\begin{aligned}\n\\alpha_{B} & \\approx-5121.64 \\mathrm{~K} \\\\\n\\beta_{B} & \\approx 13.735 \\\\\nT_{b B} & \\approx 372-89 \\mathrm{~K} \\\\\nt_{b B} & \\approx 99.74^{\\circ} \\mathrm{C} \\approx 100^{\\circ} \\mathrm{C} .\n\\end{aligned}\n$$']	['$76.8$ , $100$']	"Consider two liquids A and B insoluble in each other. The pressures $p_{i}(i=\mathrm{A}$ or $\mathrm{B})$ of their saturated vapors obey, to a good approximation, the formula:
+
+$$
+\ln \left(p_{i} / p_{o}\right)=\frac{\alpha_{i}}{T}+\beta_{i} ; \quad i=\mathrm{A} \text { or } \mathrm{B}
+$$
+
+where $p_{o}$ denotes the normal atmospheric pressure, $T$ - the absolute temperature of the vapor, and $\alpha_{i}$ and $\beta_{i}$ ( $i=\mathrm{A}$ or $\left.\mathrm{B}\right)$ - certain constants depending on the liquid. (The symbol $\ln$ denotes the natural logarithm, i.e. logarithm with base $e=2.7182818 \ldots)$
+
+The values of the ratio $p_{i} / p_{0}$ for the liquids $\mathrm{A}$ and $\mathrm{B}$ at the temperature $40^{\circ} \mathrm{C}$ and $90^{\circ} \mathrm{C}$ are given in Tab. 1.1.
+
+Table 1.1
+
+| $t\left[{ }^{\circ} \mathrm{C}\right]$ | $i=\mathrm{A}$ | $p_{i} / p_{0}$ |
+| :---: | :---: | :---: |
+|  | 0.284 | 0.07278 |
+| 40 | 1.476 | 0.6918 |
+| 90 |  |  |
+
+The errors of these values are negligible."	[]	Text-only	Competition	True	$^{\circ}\mathrm{C}$	Numerical	1e0	Open-ended	Thermodynamics	Physics	English
+147	"A. An electron after leaving a device, which accelerated it with the potential difference $U$, falls into a region with an inhomogeneous field $\boldsymbol{B}$ generated with a system of stationary coils $L_{1}, L_{2}, \ldots, L_{n}$. The known currents in the coils are $i_{1}, i_{2}, \ldots, i_{n}$, respectively.
+
+What should the currents $i_{1}$ ', $i_{2}$ ', $\ldots, i_{n}$ ' in the coils $L_{1}, L_{2}, \ldots, L_{n}$ be, in order to guide the proton (initially accelerated with the potential difference $-U$ ) along the same trajectory (and in the same direction) as that of the electron?
+
+HINT: The problem can be solved by finding a condition under which the equation describing the trajectory is the same in both cases. It may be helpful to use the relation:
+
+$$
+\boldsymbol{p} \frac{d}{d t} \boldsymbol{p}=\frac{1}{2} \frac{d}{d t} \boldsymbol{p}^{2}=\frac{1}{2} \frac{d}{d t} p^{2}
+$$"	"[""PART A\n\nAt the beginning one should notice that the kinetic energy of the electron accelerated with the potential difference $U=511 \\mathrm{kV}$ equals to its rest energy $E_{0}$. Therefore, at least in the case of the electron, the laws of the classical physics cannot be applied. It is necessary to use relativistic laws.\n\nThe relativistic equation of motion of a particle with the charge $e$ in the magnetic field B has the following form:\n\n$$\n\\frac{d}{d t} \\mathbf{p}=\\mathbf{F}_{L}\n$$\n\nwhere $\\mathbf{p}=m_{0} \\gamma \\mathbf{v}$ denotes the momentum of the particle (vector) and\n\n$$\n\\mathbf{F}_{L}=e \\mathbf{v} \\times \\mathbf{B}\n$$\n\nis the Lorentz force (its value is $e v B$ and its direction is determined with the right hand rule). $m_{0}$ denotes the (rest) mass of the particle and $v$ denotes the velocity of the particle. The quantity $\\gamma$ is given by the formula:\n\n$$\n\\gamma=\\frac{1}{\\sqrt{1-\\frac{v^{2}}{c^{2}}}}\n$$\n\nThe Lorentz force $\\mathbf{F}_{L}$ is perpendicular to the velocity $\\mathbf{v}$ of the particle and to its momentum $\\mathbf{p}=m_{0} \\boldsymbol{\\gamma}$. Hence,\n\n$$\n\\mathbf{F}_{L} \\cdot \\mathbf{v}=\\mathbf{F}_{L} \\cdot \\mathbf{p}=0\n$$\n\nMultiplying the equation of motion by $\\mathbf{p}$ and making use of the hint given in the text of the problem, we get:\n\n$$\n\\frac{1}{2} \\frac{d}{d t} p^{2}=0\n$$\n\n\n\nIt means that the value of the particle momentum (and the value of the velocity) is constant during the motion:\n\n$$\np=m_{0} v \\gamma=\\text { const; } \\quad v=\\text { const. }\n$$\n\nThe same result can be obtained without any formulae in the following way:\n\nThe Lorentz force $\\mathbf{F}_{L}$ is perpendicular to the velocity $\\mathbf{v}$ (and to the momentum $p$ as $\\left.\\mathbf{p}=m_{0} \\mathcal{W}\\right)$ and, as a consequence, to the trajectory of the particle. Therefore, there is no force that could change the component of the momentum tangent to the trajectory. Thus, this component, whose value is equal to the length of $\\mathbf{p}$, should be constant: $\\mathbf{p}=$ const. (The same refers to the component of the velocity tangent to the trajectory as $\\mathbf{p}=m_{0} \\mathcal{W}$ ).\n\nLet $s$ denotes the path passed by the particle along the trajectory. From the definition of the velocity, we have:\n\n$$\n\\frac{d s}{d t}=v\n$$\n\nUsing this formula, we can rewrite the equation of motion as follows:\n\n$$\n\\begin{gathered}\nv \\frac{d}{d s} \\mathbf{p}=\\frac{d s}{d t} \\frac{d}{d s} \\mathbf{p}=\\frac{d}{d t} \\mathbf{p}=\\mathbf{F}_{L}, \\\\\n\\frac{d}{d s} \\mathbf{p}=\\frac{\\mathbf{F}_{L}}{v} .\n\\end{gathered}\n$$\n\nDividing this equation by $p$ and making use of the fact that $p=$ const, we obtain:\n\n$$\nv \\frac{d}{d s} \\frac{\\mathbf{p}}{p}=\\frac{\\mathbf{F}_{L}}{v p}\n$$\n\nand hence\n\n$$\n\\frac{d}{d s} \\mathbf{t}=\\frac{\\mathbf{F}_{L}}{v p}\n$$\n\nwhere $\\mathbf{t}=\\mathbf{p} / p=\\mathbf{v} / v$ is the versor (unit vector) tangent to the trajectory. The above equation is exactly the same for both electrons and protons if and only if the vector quantity:\n\n$$\n\\frac{\\mathbf{F}_{L}}{v p}\n$$\n\nis the same in both cases.\n\nDenoting corresponding quantities for protons with the same symbols as for the electrons, but with primes, one gets that the condition, under which both electrons and protons can move along the same trajectory, is equivalent to the equality:\n\n$$\n\\frac{\\mathbf{F}_{L}}{v p}=\\frac{\\mathbf{F}_{L}^{\\prime}}{v^{\\prime} p^{\\prime}}\n$$\n\nHowever, the Lorentz force is proportional to the value of the velocity of the particle, and the directions of any two vectors of the following three: $\\mathbf{t}$ (or $\\mathbf{v}$ ), $\\mathbf{F}_{\\mathrm{L}}, \\mathbf{B}$ determine the direction of the third of them (right hand rule). Therefore, the above condition can be written in the following form:\n\n$$\n\\frac{e \\mathbf{B}}{p}=\\frac{e^{\\prime} \\mathbf{B}^{\\prime}}{p^{\\prime}}\n$$\n\n\n\nHence,\n\n$$\n\\mathbf{B}^{\\prime}=\\frac{e}{e^{\\prime}} \\frac{p^{\\prime}}{p} \\mathbf{B}=\\frac{p^{\\prime}}{p} \\mathbf{B}\n$$\n\nThis means that at any point the direction of the field $\\mathbf{B}$ should be conserved, its orientation should be changed into the opposite one, and the value of the field should be multiplied by the same factor $p^{\\prime} / p$. The magnetic field $\\mathbf{B}$ is a vector sum of the magnetic fields of the coils that are arbitrarily distributed in the space. Therefore, each of this fields should be scaled with the same factor $-p^{\\prime} / p$. However, the magnetic field of any coil is proportional to the current flowing in it. This means that the required scaling of the fields can only be achieved by the scaling of all the currents with the same factor $-p^{\\prime} / p$ :\n\n$$\ni_{n}^{\\prime}=-\\frac{p^{\\prime}}{p} i_{n}\n$$\n\nNow we shall determine the ratio $p^{\\prime} / p$. The kinetic energies of the particles in both cases are the same; they are equal to $E_{k}=e|U|=511 \\mathrm{keV}$. The general relativistic relation between the total energy $E$ of the particle with the rest energy $E_{0}$ and its momentum $p$ has the following form:\n\n$$\nE^{2}=E_{0}^{2}+p^{2} c^{2}\n$$\n\nwhere $\\mathrm{c}$ denotes the velocity of light. energies:\n\nThe total energy of considered particles is equal to the sum of their rest and kinetic\n\n$$\nE=E_{0}+E_{k}\n$$\n\nUsing these formulae and knowing that in our case $E_{k}=e|U|=E_{e}$, we determine the momenta of the electrons $(p)$ and the protons $(p ')$. We get:\n\na) electrons:\n\n$$\n\\begin{gathered}\n\\left(E_{e}+E_{e}\\right)^{2}=E_{e}^{2}+p^{2} c^{2}, \\\\\np=\\frac{E_{e}}{c} \\sqrt{3} .\n\\end{gathered}\n$$\n\nb) protons\n\n$$\n\\begin{gathered}\n\\left(E_{p}+E_{e}\\right)^{2}=E_{p}^{2}+p^{\\prime 2} c^{2}, \\\\\np^{\\prime}=\\frac{E_{e}}{c} \\sqrt{\\left(\\frac{E_{p}}{E_{e}}+1\\right)^{2}-\\left(\\frac{E_{p}}{E_{e}}\\right)^{2}} .\n\\end{gathered}\n$$\n\nHence,\n\n$$\n\\frac{p^{\\prime}}{p}=\\frac{1}{\\sqrt{3}} \\sqrt{\\left(\\frac{E_{p}}{E}+1\\right)^{2}-\\left(\\frac{E_{p}}{E_{e}}\\right)^{2}} \\approx 35.0\n$$\n\nand\n\n$$\ni_{n}^{\\prime}=-35.0 i_{n} \\text {. }\n$$\n\n\n\nIt is worthwhile to notice that our protons are 'almost classical', because their kinetic energy $E_{k}\\left(=E_{e}\\right)$ is small compared to the proton rest energy $E_{p}$. Thus, one can expect that the momentum of the proton can be determined, with a good accuracy, from the classical considerations. We have:\n\n$$\n\\begin{gathered}\nE_{e}=E_{k}=\\frac{p^{\\prime 2}}{2 m_{p}}=\\frac{p^{\\prime 2} c^{2}}{2 m_{p} c^{2}}=\\frac{p^{\\prime 2} c^{2}}{2 E_{p}}, \\\\\np^{\\prime}=\\frac{1}{c} \\sqrt{2 E_{e} E_{p}} .\n\\end{gathered}\n$$\n\nOn the other hand, the momentum of the proton determined from the relativistic formulae can be written in a simpler form since $E_{\\mathrm{p}} / E_{\\mathrm{e}} \\gg 1$. We get:\n\n$$\np^{\\prime}=\\frac{E_{e}}{c} \\sqrt{\\left(\\frac{E_{p}}{E_{e}}+1\\right)^{2}-\\left(\\frac{E_{p}}{E_{e}}\\right)^{2}}=\\frac{E_{e}}{c} \\sqrt{2 \\frac{E_{p}}{E_{e}}+1} \\approx \\frac{E_{e}}{c} \\sqrt{2 \\frac{E_{p}}{E_{e}}}=\\frac{1}{c} \\sqrt{2 E_{e} E_{p}} .\n$$\n\nIn accordance with our expectations, we have obtained the same result as above.""]"	['$i_{n}^{\\prime}=-35.0 i_{n}$']	The problem concerns investigation of transforming the electron microscope with magnetic guiding of the electron beam (which is accelerated with the potential difference $U=$ $511 \mathrm{kV}$ ) into a proton microscope (in which the proton beam is accelerated with the potential difference $-U$ ). For this purpose, solve the following two problems:	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+148	"B. How many times would the resolving power of the above microscope increase or decrease if the electron beam were replaced with the proton beam? Assume that the resolving power of the microscope (i.e. the smallest distance between two point objects whose circular images can be just separated) depends only on the wave properties of the particles.
+
+Assume that the velocities of the electrons and protons before their acceleration are zero, and that there is no interaction between own magnetic moment of either electrons or protons and the magnetic field. Assume also that the electromagnetic radiation emitted by the moving particles can be neglected.
+
+NOTE: Very often physicists use 1 electron-volt $(1 \mathrm{eV})$, and its derivatives such as 1 $\mathrm{keV}$ or $1 \mathrm{MeV}$, as a unit of energy. 1 electron-volt is the energy gained by the electron that passed the potential difference equal to $1 \mathrm{~V}$.
+
+Perform the calculations assuming the following data:
+
+$$
+\begin{array}{ll}
+\text { Rest energy of electron: } & E_{e}=m_{e} c^{2}=511 \mathrm{keV} \\
+\text { Rest energy of proton: } & E_{p}=m_{p} c^{2}=938 \mathrm{MeV}
+\end{array}
+$$"	['PART B\n\nThe resolving power of the microscope (in the meaning mentioned in the text of the problem) is proportional to the wavelength, in our case to the length of the de Broglie wave:\n\n$$\n\\lambda=\\frac{h}{p}\n$$\n\nwhere $h$ denotes the Planck constant and $p$ is the momentum of the particle. We see that $\\lambda$ is inversely proportional to the momentum of the particle. Therefore, after replacing the electron beam with the proton beam the resolving power will be changed by the factor $p / p^{\\prime} \\approx 1 / 35$. It means that our proton microscope would allow observation of the objects about 35 times smaller than the electron microscope.']	['35']	"The problem concerns investigation of transforming the electron microscope with magnetic guiding of the electron beam (which is accelerated with the potential difference $U=$ $511 \mathrm{kV}$ ) into a proton microscope (in which the proton beam is accelerated with the potential difference $-U$ ). For this purpose, solve the following two problems:
+Context question:
+A. An electron after leaving a device, which accelerated it with the potential difference $U$, falls into a region with an inhomogeneous field $\boldsymbol{B}$ generated with a system of stationary coils $L_{1}, L_{2}, \ldots, L_{n}$. The known currents in the coils are $i_{1}, i_{2}, \ldots, i_{n}$, respectively.
+
+What should the currents $i_{1}$ ', $i_{2}$ ', $\ldots, i_{n}$ ' in the coils $L_{1}, L_{2}, \ldots, L_{n}$ be, in order to guide the proton (initially accelerated with the potential difference $-U$ ) along the same trajectory (and in the same direction) as that of the electron?
+
+HINT: The problem can be solved by finding a condition under which the equation describing the trajectory is the same in both cases. It may be helpful to use the relation:
+
+$$
+\boldsymbol{p} \frac{d}{d t} \boldsymbol{p}=\frac{1}{2} \frac{d}{d t} \boldsymbol{p}^{2}=\frac{1}{2} \frac{d}{d t} p^{2}
+$$
+Context answer:
+\boxed{$i_{n}^{\prime}=-35.0 i_{n}$}
+"	[]	Text-only	Competition	False		Numerical	1e-1	Open-ended	Electromagnetism	Physics	English
+149	1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.	['For an altitude change $d z$, the atmospheric pressure change is :\n\n$$\nd p=-\\rho g d z\n\\tag{1}\n$$\n\nwhere $g$ is the acceleration of gravity, considered constant, $\\rho$ is the specific mass of air, which is considered as an ideal gas:\n\n$$\n\\rho=\\frac{m}{V}=\\frac{p \\mu}{R T}\n$$\n\nPut this expression in (1) :\n\n$$\n\\frac{d p}{p}=-\\frac{\\mu g}{R T} d z\n$$\n\nIf the air temperature is uniform and equals $T_{0}$, then\n\n$$\n\\frac{d p}{p}=-\\frac{\\mu g}{R T_{0}} d z\n$$\n\nAfter integration, we have :\n\n$$\np(z)=p(0) \\mathbf{e}^{-\\frac{\\mu g}{R T_{0}} z}\n\\tag{2}\n$$']	['$p(z)=p(0) e^{-\\frac{\\mu g}{R T_{0}} z}$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+150	1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.	['If\n\n$$\nT(z)=T(0)-\\Lambda z\n\\tag{3}\n$$\n\nthen\n\n$$\n\\frac{d p}{p}=-\\frac{\\mu g}{R[T(0)-\\Lambda z]} d z\n\\tag{4}\n$$\n\nKnowing that :\n\n$$\n\\int \\frac{d z}{T(0)-\\Lambda z}=-\\frac{1}{\\Lambda} \\int \\frac{d[T(0)-\\Lambda z]}{T(0)-\\Lambda z}=-\\frac{1}{\\Lambda} \\ln (T(0)-\\Lambda z)\n$$\n\nby integrating both members of (4), we obtain :\n\n$$\n\\ln \\frac{p(z)}{p(0)}=\\frac{\\mu g}{R \\Lambda} \\ln \\frac{T(0)-\\Lambda z}{T(0)}=\\frac{\\mu g}{R \\Lambda} \\ln \\left(1-\\frac{\\Lambda z}{T(0)}\\right)\n$$\n$$\np(z)=p(0)\\left(1-\\frac{\\Lambda z}{T(0)}\\right)^{\\frac{\\mu g}{R \\Lambda}}\n\\tag{5}\n$$']	['$p(z)=p(0)\\left(1-\\frac{\\Lambda z}{T(0)}\\right)^{\\frac{\\mu g}{R \\Lambda}}$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ )."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+151	1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?	['The free convection occurs if:\n\n$$\n\\frac{\\rho(z)}{\\rho(0)}>1\n$$\n\nThe ratio of specific masses can be expressed as follows:\n\n$$\n\\frac{\\rho(z)}{\\rho(0)}=\\frac{p(z)}{p(0)} \\frac{T(0)}{T(z)}=\\left(1-\\frac{\\Lambda z}{T(0)}\\right)^{\\frac{\\mu g}{R \\Lambda}-1}\n$$\n\nThe last term is larger than unity if its exponent is negative:\n\n$$\n\\frac{\\mu g}{R \\Lambda}-1<0\n$$\n\nThen :\n\n$$\n\\Lambda>\\frac{\\mu g}{R}=\\frac{0.029 \\times 9.81}{8.31}=0.034 \\frac{\\mathrm{K}}{\\mathrm{m}}\n$$']	['$0.034$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+"	[]	Text-only	Competition	False	$ \frac{\mathrm{K}}{\mathrm{m}}$	Numerical	1e-3	Open-ended	Thermodynamics	Physics	English
+152	2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.	['In vertical motion, the pressure of the parcel always equals that of the surrounding air, the latter depends on the altitude. The parcel temperature $T_{\\text {parcel }}$ depends on the pressure.\n\nWe can write:\n\n$$\n\\frac{d T_{\\text {parcel }}}{d z}=\\frac{d T_{\\text {parcel }}}{d p} \\frac{d p}{d z}\n$$\n\n$p$ is simultaneously the pressure of air in the parcel and that of the surrounding air.\n\nExpression for $\\frac{d T_{\\text {parcel }}}{d p}$\n\nBy using the equation for adiabatic processes $p V^{\\gamma}=$ const and equation of state, we can deduce the equation giving the change of pressure and temperature in a quasi-equilibrium adiabatic process of an air parcel:\n\n$$\nT_{\\text {parcel }} p^{\\frac{1-\\gamma}{\\gamma}}=\\text { const }\n\\tag{6}\n$$\n\n\n\nwhere $\\gamma=\\frac{c_{p}}{c_{V}}$ is the ratio of isobaric and isochoric thermal capacities of air. By logarithmic differentiation of the two members of (6), we have:\n\n$$\n\\frac{d T_{\\text {parcel }}}{T_{\\text {parcel }}}+\\frac{1-\\gamma}{\\gamma} \\frac{d p}{p}=0\n$$\n\nOr\n\n$$\n\\frac{d T_{\\text {parcel }}}{d p}=\\frac{T_{\\text {parcel }}}{p} \\frac{\\gamma-1}{\\gamma}\n\\tag{7}\n$$\n\nNote: we can use the first law of thermodynamic to calculate the heat received by the parcel in an elementary process: $d Q=\\frac{m}{\\mu} c_{V} d T_{\\text {parcel }}+p d V$, this heat equals zero in an adiabatic process. Furthermore, using the equation of state for air in the parcel $p V=\\frac{m}{\\mu} R T_{\\text {parcel }}$ we can derive (6)\n\nExpression for $\\frac{d p}{d z}$\n\nFrom (1) we can deduce:\n\n$$\n\\frac{d p}{d z}=-\\rho g=-\\frac{p g \\mu}{R T}\n$$\n\nwhere $T$ is the temperature of the surrounding air.\n\nOn the basis of these two expressions, we derive the expression for $d T_{\\text {parcel }} / d z$ :\n\n$$\n\\frac{d T_{\\text {parcel }}}{d z}=-\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R} \\frac{T_{\\text {parcel }}}{T}=-G\n\\tag{8}\n$$\n\nIn general, $G$ is not a constant.']	['$\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R} \\frac{T_{\\text {parcel }}}{T}$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+153	2.2.1. Derive the expression of $\Gamma$	"[""If at any altitude, $T=T_{\\text {parcel }}$, then instead of $G$ in (8), we have :\n\n$$\n\\Gamma=\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R}=\\mathrm{const}\n\\tag{9}\n$$\n\nor\n\n\n\n$$\n\\Gamma=\\frac{\\mu g}{c_{p}}\n\\tag{9'}\n$$""]"	['$\\Gamma=\\frac{\\mu g}{c_{p}}$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.
+Context question:
+2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.
+Context answer:
+\boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$}
+
+
+Extra Supplementary Reading Materials:
+
+2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+154	2.2.2. Calculate the numerical value of $\Gamma$.	['Numerical value:\n\n$$\n\\Gamma=\\frac{1.4-1}{1.4} \\frac{0.029 \\times 9.81}{8.31}=0.00978 \\frac{\\mathrm{K}}{\\mathrm{m}} \\approx 10^{-2} \\frac{\\mathrm{K}}{\\mathrm{m}}\n$$']	['$0.00978 $']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.
+Context question:
+2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.
+Context answer:
+\boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$}
+
+
+Extra Supplementary Reading Materials:
+
+2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate.
+Context question:
+2.2.1. Derive the expression of $\Gamma$
+Context answer:
+\boxed{$\Gamma=\frac{\mu g}{c_{p}}$}
+"	[]	Text-only	Competition	False	$\frac{\mathrm{K}}{\mathrm{m}}$	Numerical	1e-3	Open-ended	Thermodynamics	Physics	English
+155	2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude.	['Thus, the expression for the temperature at the altitude $z$ in this special atmosphere (called adiabatic atmosphere) is :\n\n$$\nT(z)=T(0)-\\Gamma z\n\\tag{10}\n$$']	['$T(z)=T(0)-\\Gamma z$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.
+Context question:
+2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.
+Context answer:
+\boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$}
+
+
+Extra Supplementary Reading Materials:
+
+2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate.
+Context question:
+2.2.1. Derive the expression of $\Gamma$
+Context answer:
+\boxed{$\Gamma=\frac{\mu g}{c_{p}}$}
+
+
+Context question:
+2.2.2. Calculate the numerical value of $\Gamma$.
+Context answer:
+\boxed{$0.00978 $}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+156	2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$.	['Search for the expression of $T_{\\text {parcel }}(z)$\n\nSubstitute $T$ in (7) by its expression given in (3), we have:\n\n$$\n\\frac{d T_{\\text {parcel }}}{T_{\\text {parcel }}}=-\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R} \\frac{d z}{T(0)-\\Lambda z}\n$$\n\nIntegration gives:\n\n$$\n\\ln \\frac{T_{\\text {parcel }}(z)}{T_{\\text {parcel }}(0)}=-\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R}\\left(-\\frac{1}{\\Lambda}\\right) \\ln \\frac{T(0)-\\Lambda z}{T(0)}\n$$\n\nFinally, we obtain:\n\n$$\nT_{\\text {parcel }}(z)=T_{\\text {parcel }}(0)\\left(\\frac{T(0)-\\Lambda z}{T(0)}\\right)^{\\frac{\\Gamma}{\\Lambda}}\n\\tag{11}\n$$']	['$T_{\\text {parcel }}(z)=T_{\\text {parcel }}(0)\\left(\\frac{T(0)-\\Lambda z}{T(0)}\\right)^{\\frac{\\Gamma}{\\Lambda}}$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.
+Context question:
+2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.
+Context answer:
+\boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$}
+
+
+Extra Supplementary Reading Materials:
+
+2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate.
+Context question:
+2.2.1. Derive the expression of $\Gamma$
+Context answer:
+\boxed{$\Gamma=\frac{\mu g}{c_{p}}$}
+
+
+Context question:
+2.2.2. Calculate the numerical value of $\Gamma$.
+Context answer:
+\boxed{$0.00978 $}
+
+
+Context question:
+2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude.
+Context answer:
+\boxed{$T(z)=T(0)-\Gamma z$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+157	2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<<T(0)$ and $T(0) \approx T_{\text {parcel }}(0)$.	['From (11) we obtain\n\n$$\nT_{\\text {parcel }}(z)=T_{\\text {parcel }}(0)\\left(1-\\frac{\\Lambda z}{T(0)}\\right)^{\\frac{\\Gamma}{\\Lambda}}\n$$\n\nIf $\\Lambda z<<T(0)$, then by putting $x=\\frac{-T(0)}{\\Lambda z}$, we obtain\n\n$$\n\\begin{aligned}\nT_{\\text {parcel }}(z) & =T_{\\text {parcel }}(0)\\left(\\left(1+\\frac{1}{x}\\right)^{x}\\right)^{-\\frac{\\Gamma z}{T(0)}} \\\\\n& \\approx T_{\\text {parcel }}(0) \\mathbf{e}^{-\\frac{\\Gamma z}{T(0)}} \\approx T_{\\text {parcel }}(0)\\left(1-\\frac{\\Gamma z}{T(0)}\\right) \\approx T_{\\text {parcel }}(0)-\\Gamma z\n\\end{aligned}\n$$\n\n\n\nhence,\n\n$$\nT_{\\text {parcel }}(z) \\approx T_{\\text {parcel }}(0)-\\Gamma z\n\\tag{12}\n$$']	['$T_{\\text {parcel }}(0)-\\Gamma z$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.
+Context question:
+2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.
+Context answer:
+\boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$}
+
+
+Extra Supplementary Reading Materials:
+
+2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate.
+Context question:
+2.2.1. Derive the expression of $\Gamma$
+Context answer:
+\boxed{$\Gamma=\frac{\mu g}{c_{p}}$}
+
+
+Context question:
+2.2.2. Calculate the numerical value of $\Gamma$.
+Context answer:
+\boxed{$0.00978 $}
+
+
+Context question:
+2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude.
+Context answer:
+\boxed{$T(z)=T(0)-\Gamma z$}
+
+
+Context question:
+2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$.
+Context answer:
+\boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+158	3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$.	['In a stable atmosphere, with $\\Lambda<\\Gamma$, a parcel, which on ground has temperature $T_{\\text {parcel }}(0)>T(0)$ and pressure $p(0)$ equal to that of the atmosphere, can rise and reach a maximal altitude $h$, where $T_{\\text {parcel }}(h)=T(h)$.\n\nIn vertical motion from the ground to the altitude $h$, the air parcel realizes an adiabatic quasi-static process, in which its temperature changes from $T_{\\text {parcel }}(0)$ to $T_{\\text {parcel }}(h)=T(h)$. Using (11), we can write:\n\n$$\n\\begin{aligned}\n& \\left(1-\\frac{\\Lambda h}{T(0)}\\right)^{-\\frac{\\Gamma}{\\Lambda}}=\\frac{T_{\\text {parcel }}(0)}{T(h)}=\\frac{T_{\\text {parcel }}(0)}{T(0)\\left(1-\\frac{\\Lambda h}{T(0)}\\right)} \\\\\n& \\left(1-\\frac{\\Lambda h}{T(0)}\\right)^{1-\\frac{\\Gamma}{\\Lambda}}=T_{\\text {parcel }}(0) \\times T^{-1}(0) \\\\\n& 1-\\frac{\\Lambda h}{T(0)}=T_{\\text {parcel }}^{\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0) \\times T^{-\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0) \\\\\n& h=\\frac{1}{\\Lambda} T(0)\\left[1-T_{\\text {parcel }}^{\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0) \\times T^{-\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0)\\right] \\\\\n& =\\frac{1}{\\Lambda}\\left[T(0)-T_{\\text {parcel }}^{-\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0) T^{\\frac{\\Gamma}{\\Gamma-\\Lambda}}(0)\\right]\n\\end{aligned}\n$$\n\nSo that the maximal altitude $h$ has the following expression:\n\n$$\nh=\\frac{1}{\\Lambda}\\left[T(0)-\\left(\\frac{(T(0))^{\\Gamma}}{\\left(T_{\\text {parcel }}(0)\\right)^{\\Lambda}}\\right)^{\\frac{1}{\\Gamma-\\Lambda}}\\right]\n\\tag{13}\n$$']	['$h=\\frac{1}{\\Lambda}\\left[T(0)-\\left(\\frac{(T(0))^{\\Gamma}}{\\left(T_{\\text {parcel }}(0)\\right)^{\\Lambda}}\\right)^{\\frac{1}{\\Gamma-\\Lambda}}\\right]$']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.
+Context question:
+2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.
+Context answer:
+\boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$}
+
+
+Extra Supplementary Reading Materials:
+
+2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate.
+Context question:
+2.2.1. Derive the expression of $\Gamma$
+Context answer:
+\boxed{$\Gamma=\frac{\mu g}{c_{p}}$}
+
+
+Context question:
+2.2.2. Calculate the numerical value of $\Gamma$.
+Context answer:
+\boxed{$0.00978 $}
+
+
+Context question:
+2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude.
+Context answer:
+\boxed{$T(z)=T(0)-\Gamma z$}
+
+
+Context question:
+2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$.
+Context answer:
+\boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$}
+
+
+Context question:
+2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<<T(0)$ and $T(0) \approx T_{\text {parcel }}(0)$.
+Context answer:
+\boxed{$T_{\text {parcel }}(0)-\Gamma z$}
+
+
+Extra Supplementary Reading Materials:
+
+3. The atmospheric stability. 
+
+In this part, we assume that $T$ changes linearly with altitude.
+Context question:
+3.1. Consider an air parcel initially in equilibrium with its surrounding air at altitude
+
+
+
+$z_{0}$, i.e. it has the same temperature $T\left(z_{0}\right)$ as that of the surrounding air. If the parcel is moved slightly up and down (e.g. by atmospheric turbulence), one of the three following cases may occur:
+
+- The air parcel finds its way back to the original altitude $z_{0}$, the equilibrium of the parcel is stable. The atmosphere is said to be stable.
+- The parcel keeps moving in the original direction, the equilibrium of the parcel is unstable. The atmosphere is unstable.
+- The air parcel remains at its new position, the equilibrium of the parcel is indifferent. The atmosphere is said to be neutral.
+
+What is the condition on $\Lambda$ for the atmosphere to be stable, unstable or neutral?
+Context answer:
+$\Lambda>\Gamma$ is unstable
+$\Lambda<\Gamma$ is stable
+$\Lambda=\Gamma$ is neutral
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+159	"4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula $T(z)=T(0)-\Lambda z$ with different lapse rates $\Lambda$ in the three layers $0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}$ and $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}$.
+
+Consider an air parcel with temperature $T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}$ ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of $96 \mathrm{~m}$ and $119 \mathrm{~m}$."	['Using data from the Table, we obtain the plot of $z$ versus $T$ shown in Figure 3.\n\n<img_4370>\n\nFigure 3\n\nWe can divide the atmosphere under 200m into three layers, corresponding to the following altitudes:\n\n1) $0<z<96 \\mathrm{~m}, \\quad \\Lambda_{1}=\\frac{21.5-20.1}{91}=15.4 \\times 10^{-3} \\frac{\\mathrm{K}}{\\mathrm{m}}$.\n2) $96 \\mathrm{~m}<\\mathrm{z}<119 \\mathrm{~m}, \\Lambda_{2}=0$, isothermal layer.\n3) $119 \\mathrm{~m}<\\mathrm{z}<215 \\mathrm{~m}, \\Lambda_{3}=-\\frac{22-20.1}{215-119}=-0.02 \\frac{\\mathrm{K}}{\\mathrm{m}}$.\n\nIn the layer 1), the parcel temperature can be calculated by using (11)\n\n$$\nT_{\\text {parcel }}(96 \\mathrm{~m})=294.04 \\mathrm{~K} \\approx 294.0 \\mathrm{~K} \\text { that is } 21.0^{\\circ} \\mathrm{C}\n$$\n\nIn the layer 2), the parcel temperature can be calculated by using its expression in isothermal atmosphere $T_{\\text {parcel }}(z)=T_{\\text {parcel }}(0) \\exp \\left[-\\frac{\\Gamma z}{T(0)}\\right]$.\n\n\n\nThe altitude $96 \\mathrm{~m}$ is used as origin, corresponding to $0 \\mathrm{~m}$. The altitude $119 \\mathrm{~m}$ corresponds to $23 \\mathrm{~m}$. We obtain the following value for parcel temperature:\n\n$$\nT_{\\text {parcel }}(119 \\mathrm{~m})=293.81 \\mathrm{~K} \\text { that is } 20.8^{\\circ} \\mathrm{C}\n$$']	['$294.04$, $293.81 $']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.
+Context question:
+2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.
+Context answer:
+\boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$}
+
+
+Extra Supplementary Reading Materials:
+
+2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate.
+Context question:
+2.2.1. Derive the expression of $\Gamma$
+Context answer:
+\boxed{$\Gamma=\frac{\mu g}{c_{p}}$}
+
+
+Context question:
+2.2.2. Calculate the numerical value of $\Gamma$.
+Context answer:
+\boxed{$0.00978 $}
+
+
+Context question:
+2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude.
+Context answer:
+\boxed{$T(z)=T(0)-\Gamma z$}
+
+
+Context question:
+2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$.
+Context answer:
+\boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$}
+
+
+Context question:
+2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<<T(0)$ and $T(0) \approx T_{\text {parcel }}(0)$.
+Context answer:
+\boxed{$T_{\text {parcel }}(0)-\Gamma z$}
+
+
+Extra Supplementary Reading Materials:
+
+3. The atmospheric stability. 
+
+In this part, we assume that $T$ changes linearly with altitude.
+Context question:
+3.1. Consider an air parcel initially in equilibrium with its surrounding air at altitude
+
+
+
+$z_{0}$, i.e. it has the same temperature $T\left(z_{0}\right)$ as that of the surrounding air. If the parcel is moved slightly up and down (e.g. by atmospheric turbulence), one of the three following cases may occur:
+
+- The air parcel finds its way back to the original altitude $z_{0}$, the equilibrium of the parcel is stable. The atmosphere is said to be stable.
+- The parcel keeps moving in the original direction, the equilibrium of the parcel is unstable. The atmosphere is unstable.
+- The air parcel remains at its new position, the equilibrium of the parcel is indifferent. The atmosphere is said to be neutral.
+
+What is the condition on $\Lambda$ for the atmosphere to be stable, unstable or neutral?
+Context answer:
+$\Lambda>\Gamma$ is unstable
+$\Lambda<\Gamma$ is stable
+$\Lambda=\Gamma$ is neutral
+
+
+Context question:
+3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$.
+Context answer:
+\boxed{$h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]$}
+
+
+Extra Supplementary Reading Materials:
+
+4. The mixing height 
+
+Table 1 
+
+Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi.
+
+| Altitude, m | Temperature, ${ }^{\circ} \mathrm{C}$ |
+| :---: | :---: |
+| 5 | 21.5 |
+| 60 | 20.6 |
+| 64 | 20.5 |
+| 69 | 20.5 |
+| 75 | 20.4 |
+| 81 | 20.3 |
+| 90 | 20.2 |
+| 96 | 20.1 |
+| 102 | 20.1 |
+| 109 | 20.1 |
+| 113 | 20.1 |
+| 119 | 20.1 |
+| 128 | 20.2 |
+| 136 | 20.3 |
+| 145 | 20.4 |
+| 153 | 20.5 |
+| 159 | 20.6 |
+| 168 | 20.8 |
+| 178 | 21.0 |
+| 189 | 21.5 |
+| 202 | 21.8 |
+| 215 | 22.0 |
+| 225 | 22.1 |
+| 234 | 22.2 |
+| 246 | 22.3 |
+| 257 | 22.3 |"	[]	Text-only	Competition	True	K	Numerical	1e-1	Open-ended	Thermodynamics	Physics	English
+160	"4.2. Determine the maximal elevation $H$ the parcel can reach, and the temperature $T_{\text {parcel }}(H)$ of the parcel.
+
+$H$ is called the mixing height. Air pollutants emitted from ground can mix with the air in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer."	['In the layer 3), starting from $119 \\mathrm{~m}$, by using (13) we find the maximal elevation $h=23 \\mathrm{~m}$, and the corresponding temperature $293.6 \\mathrm{~K}$ (or $20.6{ }^{\\circ} \\mathrm{C}$ ).\n\nFinally, the mixing height is\n\n$$\nH=119+23=142 \\mathrm{~m} .\n$$\n\nAnd\n\n$$\nT_{\\text {parcel }}(142 \\mathrm{~m})=293.6 \\mathrm{~K} \\text { that is } 20.6^{\\circ} \\mathrm{C}\n$$\n\nFrom this relation, we can find $T_{\\text {parcel }}(119 \\mathrm{~m}) \\approx 293.82 \\mathrm{~K}$ and $h=23 \\mathrm{~m}$.\n\nNote: By using approximate expression (12) we can easily find $T_{\\text {parcel }}(z)=294 \\mathrm{~K}$ and 293.8 K at elevations $96 \\mathrm{~m}$ and $119 \\mathrm{~m}$, respectively. At $119 \\mathrm{~m}$ elevation, the difference between parcel and surrounding air temperatures is $0.7 \\mathrm{~K}$ (= 293.8 - 293.1), so that the maximal distance the parcel will travel in the third layer is $0.7 /\\left(\\Gamma-\\Lambda_{3}\\right)=0.7 / 0.03=23 \\mathrm{~m}$.']	['142, 293.6']	"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 
+
+Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.
+
+We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.
+
+Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.
+
+Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.
+
+The student may use the following data if necessary:
+
+The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).
+
+The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$
+
+The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$
+
+The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.
+
+The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.
+
+
+
+Mathematical hints 
+
+a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$
+
+b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.
+
+c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$
+
+1. Change of pressure with altitude.
+Context question:
+1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.
+Context answer:
+\boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$}
+
+
+Extra Supplementary Reading Materials:
+
+1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation
+
+$$
+T(z)=T(0)-\Lambda z
+$$
+
+where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).
+Context question:
+1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.
+Context answer:
+\boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$}
+
+
+Context question:
+1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?
+Context answer:
+\boxed{$0.034$}
+
+
+Extra Supplementary Reading Materials:
+
+2. Change of the temperature of an air parcel in vertical motion
+
+Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.
+
+As the size of the parcel is not large, the atmospheric pressure at different points on
+
+
+
+the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.
+Context question:
+2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.
+Context answer:
+\boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$}
+
+
+Extra Supplementary Reading Materials:
+
+2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate.
+Context question:
+2.2.1. Derive the expression of $\Gamma$
+Context answer:
+\boxed{$\Gamma=\frac{\mu g}{c_{p}}$}
+
+
+Context question:
+2.2.2. Calculate the numerical value of $\Gamma$.
+Context answer:
+\boxed{$0.00978 $}
+
+
+Context question:
+2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude.
+Context answer:
+\boxed{$T(z)=T(0)-\Gamma z$}
+
+
+Context question:
+2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$.
+Context answer:
+\boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$}
+
+
+Context question:
+2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<<T(0)$ and $T(0) \approx T_{\text {parcel }}(0)$.
+Context answer:
+\boxed{$T_{\text {parcel }}(0)-\Gamma z$}
+
+
+Extra Supplementary Reading Materials:
+
+3. The atmospheric stability. 
+
+In this part, we assume that $T$ changes linearly with altitude.
+Context question:
+3.1. Consider an air parcel initially in equilibrium with its surrounding air at altitude
+
+
+
+$z_{0}$, i.e. it has the same temperature $T\left(z_{0}\right)$ as that of the surrounding air. If the parcel is moved slightly up and down (e.g. by atmospheric turbulence), one of the three following cases may occur:
+
+- The air parcel finds its way back to the original altitude $z_{0}$, the equilibrium of the parcel is stable. The atmosphere is said to be stable.
+- The parcel keeps moving in the original direction, the equilibrium of the parcel is unstable. The atmosphere is unstable.
+- The air parcel remains at its new position, the equilibrium of the parcel is indifferent. The atmosphere is said to be neutral.
+
+What is the condition on $\Lambda$ for the atmosphere to be stable, unstable or neutral?
+Context answer:
+$\Lambda>\Gamma$ is unstable
+$\Lambda<\Gamma$ is stable
+$\Lambda=\Gamma$ is neutral
+
+
+Context question:
+3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$.
+Context answer:
+\boxed{$h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]$}
+
+
+Extra Supplementary Reading Materials:
+
+4. The mixing height 
+
+Table 1 
+
+Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi.
+
+| Altitude, m | Temperature, ${ }^{\circ} \mathrm{C}$ |
+| :---: | :---: |
+| 5 | 21.5 |
+| 60 | 20.6 |
+| 64 | 20.5 |
+| 69 | 20.5 |
+| 75 | 20.4 |
+| 81 | 20.3 |
+| 90 | 20.2 |
+| 96 | 20.1 |
+| 102 | 20.1 |
+| 109 | 20.1 |
+| 113 | 20.1 |
+| 119 | 20.1 |
+| 128 | 20.2 |
+| 136 | 20.3 |
+| 145 | 20.4 |
+| 153 | 20.5 |
+| 159 | 20.6 |
+| 168 | 20.8 |
+| 178 | 21.0 |
+| 189 | 21.5 |
+| 202 | 21.8 |
+| 215 | 22.0 |
+| 225 | 22.1 |
+| 234 | 22.2 |
+| 246 | 22.3 |
+| 257 | 22.3 |
+Context question:
+4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula $T(z)=T(0)-\Lambda z$ with different lapse rates $\Lambda$ in the three layers $0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}$ and $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}$.
+
+Consider an air parcel with temperature $T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}$ ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of $96 \mathrm{~m}$ and $119 \mathrm{~m}$.
+Context answer:
+\boxed{$294.04$, $293.81 $}
+"	[]	Text-only	Competition	True	m, K	Numerical	1e-0,1e-1	Open-ended	Thermodynamics	Physics	English
+161	"A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,
+
+$$
+M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
+\tag{!}
+$$
+
+The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as
+
+$$
+\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
+\tag{2}
+$$
+
+where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$."	"[""Apply Newton's law to mass $M_{1}$ :\n\n$$\nM_{1} \\frac{\\mathrm{d}^{2} \\vec{r}_{1}}{\\mathrm{~d} t^{2}}=G \\frac{M_{1} M_{2}}{\\left|\\vec{r}_{2}-\\vec{r}_{1}\\right|^{2}} \\frac{\\vec{r}_{2}-\\vec{r}_{1}}{\\left|\\vec{r}_{2}-\\overrightarrow{r_{1}}\\right|}\n\\tag{1}\n$$\n\nUse, from eq. (1) of the question sheet\n\n$$\n\\vec{r}_{2}=-\\frac{M_{1}}{M_{2}} \\vec{r}_{1}\n\\tag{2}\n$$\n\nin eq. (1) above, to obtain\n\n$$\n\\frac{\\mathrm{d}^{2} \\vec{r}_{1}}{\\mathrm{~d} t^{2}}=-\\frac{G M_{2}^{3}}{\\left(M_{1}+M_{2}\\right)^{2} r_{1}^{2}} \\frac{\\vec{r}_{1}}{r_{1}} .\n\\tag{3}\n$$""]"	['$n=3$ , $\\alpha=\\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$']	"LIGO-GW150914
+
+In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.
+
+Part A: Newtonian (conservative) orbits"	[]	Text-only	Competition	True		Numerical,Expression	0,	Open-ended	Mechanics	Physics	English
+162	"A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:
+
+$$
+E=A(\mu, \Omega, L)-G \frac{M \mu}{L},
+\tag{3}
+$$
+
+where
+
+$$
+\mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2}
+\tag{4}
+$$
+
+are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$."	['The total energy of the system is the sum of the two kinetic energies plus the gravitational potential energy. For circular motions, the linear velocity of each of the masses reads\n\n$$\n\\left|\\vec{v}_{1}\\right|=r_{1} \\Omega, \\quad\\left|\\vec{v}_{2}\\right|=r_{2} \\Omega\n\\tag{4}\n$$\n\nThus, the total energy is\n\n$$\nE=\\frac{1}{2}\\left(M_{1} r_{1}^{2}+M_{2} r_{2}^{2}\\right) \\Omega^{2}-\\frac{G M_{1} M_{2}}{L}\n\\tag{5}\n$$\n\nNow,\n\n$$\n\\left(M_{1} r_{1}-M_{2} r_{2}\\right)^{2}=0 \\quad \\Rightarrow \\quad M_{1} r_{1}^{2}+M_{2} r_{2}^{2}=\\mu L^{2}\n\\tag{6}\n$$\n\nThus,\n\n$$\nE=\\frac{1}{2} \\mu L^{2} \\Omega^{2}-G \\frac{M \\mu}{L} .\n\\tag{7}\n$$']	['$A(\\mu, \\Omega, L)=\\frac{1}{2} \\mu L^{2} \\Omega^{2}$']	"LIGO-GW150914
+
+In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.
+
+Part A: Newtonian (conservative) orbits
+Context question:
+A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,
+
+$$
+M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
+\tag{!}
+$$
+
+The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as
+
+$$
+\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
+\tag{2}
+$$
+
+where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.
+Context answer:
+\boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+163	A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$.	"[""Energy (3) of the question sheet can be interpreted as describing a system of a mass $\\mu$ in a circular orbit with angular velocity $\\Omega$, radius $L$, around a mass $M$ (at rest). Equating the gravitational acceleration to the centripetal acceleration:\n\n$$\nG \\frac{M}{L^{2}}=\\Omega^{2} L\n\\tag{8}\n$$\n\nThis is indeed Kepler's third law (for circular orbits). Then, from 7,\n\n$$\nE=-\\frac{1}{2} G \\frac{M \\mu}{L}\n\\tag{9}\n$$""]"	['$\\beta=-\\frac{1}{2}$']	"LIGO-GW150914
+
+In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.
+
+Part A: Newtonian (conservative) orbits
+Context question:
+A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,
+
+$$
+M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
+\tag{!}
+$$
+
+The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as
+
+$$
+\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
+\tag{2}
+$$
+
+where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.
+Context answer:
+\boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$}
+
+
+Context question:
+A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:
+
+$$
+E=A(\mu, \Omega, L)-G \frac{M \mu}{L},
+\tag{3}
+$$
+
+where
+
+$$
+\mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2}
+\tag{4}
+$$
+
+are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$.
+Context answer:
+\boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$}
+"	[]	Text-only	Competition	False		Numerical	0	Open-ended	Mechanics	Physics	English
+164	"B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain:
+
+$$
+\mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6}
+\tag{9}
+$$
+
+What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.]"	['First take the derivatives:\n\n$$\n\\frac{\\mathrm{d}^{3} Q_{i j}}{\\mathrm{~d} t^{3}}=4 \\Omega^{3} \\mu L^{2}\\left(\\begin{array}{ccc}\n\\sin 2 \\Omega t & -\\cos 2 \\Omega t & 0 \\\\\n-\\cos 2 \\Omega t & -\\sin 2 \\Omega t & 0 \\\\\n0 & 0 & 0\n\\end{array}\\right)\n\\tag{13}\n$$\n\nThen perform the sum:\n\n$$\n\\frac{\\mathrm{d} E}{\\mathrm{~d} t}=\\frac{G}{5 c^{5}}\\left(4 \\Omega^{3} \\mu L^{2}\\right)^{2}\\left[2 \\sin ^{2}(2 \\Omega t)+2 \\cos ^{2}(2 \\Omega t)\\right]=\\frac{32}{5} \\frac{G}{c^{5}} \\mu^{2} L^{4} \\Omega^{6} .\n\\tag{14}\n$$']	['$\\xi=\\frac{32}{5}$']	"LIGO-GW150914
+
+In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.
+
+Part A: Newtonian (conservative) orbits
+Context question:
+A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,
+
+$$
+M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
+\tag{!}
+$$
+
+The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as
+
+$$
+\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
+\tag{2}
+$$
+
+where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.
+Context answer:
+\boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$}
+
+
+Context question:
+A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:
+
+$$
+E=A(\mu, \Omega, L)-G \frac{M \mu}{L},
+\tag{3}
+$$
+
+where
+
+$$
+\mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2}
+\tag{4}
+$$
+
+are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$.
+Context answer:
+\boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$}
+
+
+Context question:
+A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$.
+Context answer:
+\boxed{$\beta=-\frac{1}{2}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part B: Introducing relativistic dissipation
+
+The correct theory of gravity, General Relativity, was formulated by Einstein in 1915, and predicts that gravity travels with the speed of light. The messengers carrying information about the interaction are called GWs. GWs are emitted whenever masses are accelerated, making the system of masses lose energy.
+
+Consider a system of two point-like particles, isolated from the rest of the Universe. Einstein proved that for small enough velocities the emitted GWs: 1) have a frequency which is twice as large as the orbital frequency; 2 ) can be characterized by a luminosity, i.e. emitted power $\mathcal{P}$, which is dominated by Einstein's quadrupole formula,
+
+$$
+\mathcal{P}=\frac{G}{5 c^{5}} \sum_{i=1}^{3} \sum_{j=1}^{3}\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)
+\tag{5}
+$$
+
+Here, $c$ is the velocity of light $c \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. For a system of 2 pointlike particles orbiting on the $x-y$ plane, $Q_{i j}$ is the following table ( $i, j$ label the row/column number)
+
+$$
+Q_{11}=\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 x_{A}^{2}-y_{A}^{2}\right), \quad Q_{22} =\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 y_{A}^{2}-x_{A}^{2}\right), \quad Q_{33}=-\sum_{A=1}^{2} \frac{M_{A}}{3}\left(x_{A}^{2}+y_{A}^{2}\right),
+\tag{6}
+$$
+$$
+Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A},
+\tag{7}
+$$
+
+and $Q_{i j}=0$ for all other possibilities. Here, $\left(x_{A}, y_{A}\right)$ is the position of mass A in the center-of-mass frame.
+Context question:
+B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as:
+
+$$
+Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t .
+\tag{8}
+$$
+
+Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$.
+Context answer:
+$$
+k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0
+$$
+"	[]	Text-only	Competition	False		Numerical	0	Open-ended	Mechanics	Physics	English
+165	"B.3 In the absence of GW emission the two masses will orbit on a fixed circular orbit indefinitely. However, the emission of GWs causes the system to lose energy and to slowly evolve towards smaller circular orbits. Obtain that the rate of change $\frac{\mathrm{d} \Omega}{\mathrm{d} t}$ of the orbital angular velocity takes the form
+
+$$
+\left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=(3 \xi)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5}
+\tag{10}
+$$
+
+where $M_{\mathrm{c}}$ is called the chirp mass. Obtain $M_{\mathrm{c}}$ as a function of $M$ and $\mu$. This mass determines the increase in frequency during the orbital decay. [The name ""chirp"" is inspired by the high pitch sound (increasing frequency) produced by small birds.]"	"[""Now we assume a sequency of Keplerian orbits, with decreasing energy, which is being taken from the system by the GWs.\n\nFirst, from (9), differentiating with respect to time,\n\n$$\n\\frac{\\mathrm{d} E}{\\mathrm{~d} t}=\\frac{G M \\mu}{2 L^{2}} \\frac{\\mathrm{d} L}{\\mathrm{~d} t}\n\\tag{15}\n$$\n\nSince this loss of energy is due to GWs, we equate it with (minus) the luminosity of GWs, given by (14)\n\n$$\n\\frac{G M \\mu}{2 L^{2}} \\frac{\\mathrm{d} L}{\\mathrm{~d} t}=-\\frac{32}{5} \\frac{G}{c^{5}} \\mu^{2} L^{4} \\Omega^{6}\n\\tag{16}\n$$\n\nWe can eliminate the $L$ and $\\mathrm{d} L / \\mathrm{d} t$ dependence in this equation in terms of $\\Omega$ and $\\mathrm{d} \\Omega / \\mathrm{d} t$, by using Kepler's third law (8), which relates:\n\n$$\nL^{3}=G \\frac{M}{\\Omega^{2}}, \\quad \\frac{\\mathrm{d} L}{\\mathrm{~d} t}=-\\frac{2}{3} \\frac{L}{\\Omega} \\frac{\\mathrm{d} \\Omega}{\\mathrm{d} t}\n\\tag{17}\n$$\n\n\n\nSubstituting in (16), we obtain:\n\n$$\n\\left(\\frac{\\mathrm{d} \\Omega}{\\mathrm{d} t}\\right)^{3}=\\left(\\frac{96}{5}\\right)^{3} \\frac{\\Omega^{11}}{c^{15}} G^{5} \\mu^{3} M^{2} \\equiv\\left(\\frac{96}{5}\\right)^{3} \\frac{\\Omega^{11}}{c^{15}}\\left(G M_{\\mathrm{c}}\\right)^{5}\n\\tag{18}\n$$""]"	['$M_{\\mathrm{c}}=(\\mu^{3} M^{2})^{1 / 5}$']	"LIGO-GW150914
+
+In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.
+
+Part A: Newtonian (conservative) orbits
+Context question:
+A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,
+
+$$
+M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
+\tag{!}
+$$
+
+The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as
+
+$$
+\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
+\tag{2}
+$$
+
+where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.
+Context answer:
+\boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$}
+
+
+Context question:
+A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:
+
+$$
+E=A(\mu, \Omega, L)-G \frac{M \mu}{L},
+\tag{3}
+$$
+
+where
+
+$$
+\mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2}
+\tag{4}
+$$
+
+are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$.
+Context answer:
+\boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$}
+
+
+Context question:
+A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$.
+Context answer:
+\boxed{$\beta=-\frac{1}{2}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part B: Introducing relativistic dissipation
+
+The correct theory of gravity, General Relativity, was formulated by Einstein in 1915, and predicts that gravity travels with the speed of light. The messengers carrying information about the interaction are called GWs. GWs are emitted whenever masses are accelerated, making the system of masses lose energy.
+
+Consider a system of two point-like particles, isolated from the rest of the Universe. Einstein proved that for small enough velocities the emitted GWs: 1) have a frequency which is twice as large as the orbital frequency; 2 ) can be characterized by a luminosity, i.e. emitted power $\mathcal{P}$, which is dominated by Einstein's quadrupole formula,
+
+$$
+\mathcal{P}=\frac{G}{5 c^{5}} \sum_{i=1}^{3} \sum_{j=1}^{3}\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)
+\tag{5}
+$$
+
+Here, $c$ is the velocity of light $c \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. For a system of 2 pointlike particles orbiting on the $x-y$ plane, $Q_{i j}$ is the following table ( $i, j$ label the row/column number)
+
+$$
+Q_{11}=\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 x_{A}^{2}-y_{A}^{2}\right), \quad Q_{22} =\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 y_{A}^{2}-x_{A}^{2}\right), \quad Q_{33}=-\sum_{A=1}^{2} \frac{M_{A}}{3}\left(x_{A}^{2}+y_{A}^{2}\right),
+\tag{6}
+$$
+$$
+Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A},
+\tag{7}
+$$
+
+and $Q_{i j}=0$ for all other possibilities. Here, $\left(x_{A}, y_{A}\right)$ is the position of mass A in the center-of-mass frame.
+Context question:
+B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as:
+
+$$
+Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t .
+\tag{8}
+$$
+
+Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$.
+Context answer:
+$$
+k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0
+$$
+
+
+Context question:
+B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain:
+
+$$
+\mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6}
+\tag{9}
+$$
+
+What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.]
+Context answer:
+\boxed{$\xi=\frac{32}{5}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+166	"B.4 Using the information provided above, relate the orbital angular velocity $\Omega$ with the GW frequency $f_{\mathrm{GW}}$. Knowing that, for any smooth function $F(t)$ and $a \neq 1$,
+
+$$
+\frac{\mathrm{d} F(t)}{\mathrm{d} t}=\chi F(t)^{a} \quad \Rightarrow \quad F(t)^{1-a}=\chi(1-a)\left(t-t_{0}\right)
+\tag{11}
+$$
+
+where $\chi$ is a constant and $t_{0}$ is an integration constant, show that (10) implies that the GW frequency is
+
+$$
+f_{\mathrm{GW}}^{-8 / 3}=8 \pi^{8 / 3} \xi\left(\frac{G M_{c}}{c^{3}}\right)^{(2 / 3)+p}\left(t_{0}-t\right)^{2-p}
+\tag{12}
+$$
+
+and determine the constant $p$."	['Angular and cycle frequencies are related as $\\Omega=2 \\pi f$. From the information provided above: GWs have a frequency which is twice as large as the orbital frequency, we have\n\n$$\n\\frac{\\Omega}{2 \\pi}=\\frac{f_{\\mathrm{GW}}}{2}\n\\tag{19}\n$$\n\nFormula (10) of the question sheet has the form\n\n$$\n\\frac{\\mathrm{d} \\Omega}{\\mathrm{d} t}=\\chi \\Omega^{11 / 3}, \\quad \\chi \\equiv \\frac{96}{5} \\frac{\\left(G M_{\\mathrm{c}}\\right)^{5 / 3}}{c^{5}} .\n\\tag{20}\n$$\n\nThus, from (11) of the question sheet\n\n$$\n\\Omega(t)^{-8 / 3}=\\frac{8}{3} \\chi\\left(t_{0}-t\\right),\n\\tag{21}\n$$\n\nor, using (20) and the definition of $\\chi$ gives\n\n$$\nf_{\\mathrm{GW}}^{-8 / 3}(t)=\\frac{(8 \\pi)^{8 / 3}}{5}\\left(\\frac{G M_{\\mathrm{c}}}{c^{3}}\\right)^{5 / 3}\\left(t_{0}-t\\right)\n\\tag{22}\n$$']	['p=1']	"LIGO-GW150914
+
+In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.
+
+Part A: Newtonian (conservative) orbits
+Context question:
+A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,
+
+$$
+M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
+\tag{!}
+$$
+
+The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as
+
+$$
+\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
+\tag{2}
+$$
+
+where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.
+Context answer:
+\boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$}
+
+
+Context question:
+A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:
+
+$$
+E=A(\mu, \Omega, L)-G \frac{M \mu}{L},
+\tag{3}
+$$
+
+where
+
+$$
+\mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2}
+\tag{4}
+$$
+
+are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$.
+Context answer:
+\boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$}
+
+
+Context question:
+A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$.
+Context answer:
+\boxed{$\beta=-\frac{1}{2}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part B: Introducing relativistic dissipation
+
+The correct theory of gravity, General Relativity, was formulated by Einstein in 1915, and predicts that gravity travels with the speed of light. The messengers carrying information about the interaction are called GWs. GWs are emitted whenever masses are accelerated, making the system of masses lose energy.
+
+Consider a system of two point-like particles, isolated from the rest of the Universe. Einstein proved that for small enough velocities the emitted GWs: 1) have a frequency which is twice as large as the orbital frequency; 2 ) can be characterized by a luminosity, i.e. emitted power $\mathcal{P}$, which is dominated by Einstein's quadrupole formula,
+
+$$
+\mathcal{P}=\frac{G}{5 c^{5}} \sum_{i=1}^{3} \sum_{j=1}^{3}\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)
+\tag{5}
+$$
+
+Here, $c$ is the velocity of light $c \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. For a system of 2 pointlike particles orbiting on the $x-y$ plane, $Q_{i j}$ is the following table ( $i, j$ label the row/column number)
+
+$$
+Q_{11}=\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 x_{A}^{2}-y_{A}^{2}\right), \quad Q_{22} =\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 y_{A}^{2}-x_{A}^{2}\right), \quad Q_{33}=-\sum_{A=1}^{2} \frac{M_{A}}{3}\left(x_{A}^{2}+y_{A}^{2}\right),
+\tag{6}
+$$
+$$
+Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A},
+\tag{7}
+$$
+
+and $Q_{i j}=0$ for all other possibilities. Here, $\left(x_{A}, y_{A}\right)$ is the position of mass A in the center-of-mass frame.
+Context question:
+B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as:
+
+$$
+Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t .
+\tag{8}
+$$
+
+Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$.
+Context answer:
+$$
+k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0
+$$
+
+
+Context question:
+B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain:
+
+$$
+\mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6}
+\tag{9}
+$$
+
+What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.]
+Context answer:
+\boxed{$\xi=\frac{32}{5}$}
+
+
+Context question:
+B.3 In the absence of GW emission the two masses will orbit on a fixed circular orbit indefinitely. However, the emission of GWs causes the system to lose energy and to slowly evolve towards smaller circular orbits. Obtain that the rate of change $\frac{\mathrm{d} \Omega}{\mathrm{d} t}$ of the orbital angular velocity takes the form
+
+$$
+\left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=(3 \xi)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5}
+\tag{10}
+$$
+
+where $M_{\mathrm{c}}$ is called the chirp mass. Obtain $M_{\mathrm{c}}$ as a function of $M$ and $\mu$. This mass determines the increase in frequency during the orbital decay. [The name ""chirp"" is inspired by the high pitch sound (increasing frequency) produced by small birds.]
+Context answer:
+\boxed{$M_{\mathrm{c}}=(\mu^{3} M^{2})^{1 / 5}$}
+"	[]	Text-only	Competition	False		Numerical	0	Open-ended	Mechanics	Physics	English
+167	A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.	['The magnetic force is the centripetal force:\n\n$$\nm \\frac{v^{2}}{r}=e v B \\Rightarrow r=\\frac{m v}{e B}\n$$\n\nFirst express the velocity in terms of the kinetic energy,\n\n$$\nK=\\frac{1}{2} m v^{2} \\Rightarrow v=\\sqrt{\\frac{2 K}{m}},\n$$\n\nand then insert it in the expression above for the radius to get']	['$r=\\frac{\\sqrt{2 K m}}{e B}$']	"Where is the neutrino?
+
+When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting.
+
+Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$;
+
+Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$
+
+Part A.ATLAS Detector physics"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+168	A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$.	['The radius of the circular motion of a charged particle in the presence of a uniform magnetic field is given by,\n\n$$\nr=\\frac{m v}{e B}\n$$\n\nThis formula is valid in the relativistic scenario if the mass correction, $m \\rightarrow \\gamma m$ is included:\n\n$$\nr=\\frac{\\gamma m v}{e B}=\\frac{p}{e B} \\Rightarrow p=r e B\n$$\n\nNote that the radius of the circular motion is half the radius of the inner part of the detector. One obtains $\\left[1 \\mathrm{MeV} / c=5.34 \\times 10^{-22} \\mathrm{~m} \\mathrm{~kg} \\mathrm{~s}^{-1}\\right.$ ]']	['$p=330$']	"Where is the neutrino?
+
+When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting.
+
+Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$;
+
+Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$
+
+Part A.ATLAS Detector physics
+Context question:
+A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.
+Context answer:
+\boxed{$r=\frac{\sqrt{2 K m}}{e B}$}
+
+
+Extra Supplementary Reading Materials:
+
+Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered."	[]	Text-only	Competition	False	$ \mathrm{MeV} / c$	Numerical	1e0	Open-ended	Modern Physics	Physics	English
+169	"A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as:
+
+$$
+P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2}
+$$
+
+where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$."	['The acceleration for the particle is $a=\\frac{e v B}{\\gamma m} \\sim \\frac{e c B}{\\gamma m}$, in the ultrarelativistic limit. Then,\n\n$$\nP=\\frac{e^{4} c^{2} \\gamma^{4} B^{2}}{6 \\pi \\epsilon_{0} c^{3} \\gamma^{2} m^{2}}=\\frac{e^{4} \\gamma^{2} c^{4} B^{2}}{6 \\pi \\epsilon_{0} c^{5} m^{2}}\n$$\n\nSince $E=\\gamma m c^{2}$ we can obtain $\\gamma^{2} c^{4}=\\frac{E^{2}}{m^{2}}$ and, finally,\n\n\n\n$$\nP=\\frac{e^{4}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} E^{2} B^{2}\n$$']	['$\\xi=\\frac{1}{6 \\pi}, n=5$ , $k=4$']	"Where is the neutrino?
+
+When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting.
+
+Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$;
+
+Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$
+
+Part A.ATLAS Detector physics
+Context question:
+A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.
+Context answer:
+\boxed{$r=\frac{\sqrt{2 K m}}{e B}$}
+
+
+Extra Supplementary Reading Materials:
+
+Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered.
+Context question:
+A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$.
+Context answer:
+\boxed{$p=330$}
+
+
+Extra Supplementary Reading Materials:
+
+When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by
+
+$$
+P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}}
+$$
+
+where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$."	[]	Text-only	Competition	True		Numerical	0	Open-ended	Modern Physics	Physics	English
+170	"A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is:
+
+$$
+E(t)=\frac{E_{0}}{1+\alpha E_{0} t},
+$$
+
+where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$."	['The power emitted by the particle is given by,\n\n$$\nP=-\\frac{\\mathrm{d} E}{\\mathrm{~d} t}=\\frac{e^{4}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} E^{2} B^{2}\n$$\n\nThe energy of the particle as a function of time can be calculated from\n\n$$\n\\int_{E_{0}}^{E(t)} \\frac{1}{E^{2}} \\mathrm{~d} E=-\\int_{0}^{t} \\frac{e^{4}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} B^{2} \\mathrm{~d} t\n$$\n\nwhere $E(0)=E_{0}$. This leads to,\n\n$$\n\\frac{1}{E(t)}-\\frac{1}{E_{0}}=\\frac{e^{4} B^{2}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} t \\quad \\Rightarrow \\quad E(t)=\\frac{E_{0}}{1+\\alpha E_{0} t},\n$$']	['$\\alpha=\\frac{e^{4} B^{2}}{6 \\pi \\epsilon_{0} m^{4} c^{5}}$']	"Where is the neutrino?
+
+When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting.
+
+Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$;
+
+Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$
+
+Part A.ATLAS Detector physics
+Context question:
+A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.
+Context answer:
+\boxed{$r=\frac{\sqrt{2 K m}}{e B}$}
+
+
+Extra Supplementary Reading Materials:
+
+Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered.
+Context question:
+A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$.
+Context answer:
+\boxed{$p=330$}
+
+
+Extra Supplementary Reading Materials:
+
+When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by
+
+$$
+P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}}
+$$
+
+where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$.
+Context question:
+A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as:
+
+$$
+P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2}
+$$
+
+where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$.
+Context answer:
+\boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+171	A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV.	['If the initial energy of the electron is $100 \\mathrm{GeV}$, the radius of curvature is extremely large ( $r=\\frac{E}{e B c} \\approx 167 \\mathrm{~m}$ ). Therefore, in approximation, one can consider the electron is moving in the inner part of the ATLAS detector along a straight line. The time of flight of the electron is $t=R / c$, where $R=1.1 \\mathrm{~m}$ is the radius of the inner part of the detector. The total energy lost due to synchrotron radiation is,\n\n$$\n\\Delta E=E(R / c)-E_{0}=\\frac{E_{0}}{1+\\alpha E_{0} \\frac{R}{c}}-E_{0} \\approx-\\alpha E_{0}^{2} \\frac{R}{c}\n$$']	['-56']	"Where is the neutrino?
+
+When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting.
+
+Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$;
+
+Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$
+
+Part A.ATLAS Detector physics
+Context question:
+A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.
+Context answer:
+\boxed{$r=\frac{\sqrt{2 K m}}{e B}$}
+
+
+Extra Supplementary Reading Materials:
+
+Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered.
+Context question:
+A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$.
+Context answer:
+\boxed{$p=330$}
+
+
+Extra Supplementary Reading Materials:
+
+When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by
+
+$$
+P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}}
+$$
+
+where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$.
+Context question:
+A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as:
+
+$$
+P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2}
+$$
+
+where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$.
+Context answer:
+\boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$}
+
+
+Context question:
+A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is:
+
+$$
+E(t)=\frac{E_{0}}{1+\alpha E_{0} t},
+$$
+
+where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$.
+Context answer:
+\boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$}
+"	[]	Text-only	Competition	False	MeV	Numerical	1e0	Open-ended	Modern Physics	Physics	English
+172	A.6 Find an expression for the cyclotron frequency of the electron as a function of time in the ultrarelativistic limit.	['In the ultrarelativistic limit, $v \\approx c$ and $E \\approx p c$. The cyclotron frequency is,\n\n$$\n\\omega(t)=\\frac{c}{r(t)}=\\frac{e c B}{p(t)}=\\frac{e c^{2} B}{E(t)}\n$$']	['$\\omega(t)=\\frac{e c^{2} B}{E_{0}}(1+\\frac{e^{4} B^{2}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} E_{0} t)$']	"Where is the neutrino?
+
+When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting.
+
+Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$;
+
+Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$
+
+Part A.ATLAS Detector physics
+Context question:
+A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.
+Context answer:
+\boxed{$r=\frac{\sqrt{2 K m}}{e B}$}
+
+
+Extra Supplementary Reading Materials:
+
+Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered.
+Context question:
+A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$.
+Context answer:
+\boxed{$p=330$}
+
+
+Extra Supplementary Reading Materials:
+
+When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by
+
+$$
+P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}}
+$$
+
+where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$.
+Context question:
+A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as:
+
+$$
+P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2}
+$$
+
+where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$.
+Context answer:
+\boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$}
+
+
+Context question:
+A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is:
+
+$$
+E(t)=\frac{E_{0}}{1+\alpha E_{0} t},
+$$
+
+where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$.
+Context answer:
+\boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$}
+
+
+Context question:
+A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV.
+Context answer:
+\boxed{-56}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+173	a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.	['This is when the spring force equals the maximal static friction, $k s=\\mu_{0} m g$, so $s=\\mu_{0} m g / k$.']	['$s=\\mu_{0} m g / k$']	"## Toffee Pudding
+
+
+
+A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over."	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+174	b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving.	['The net stretching of the spring is $s+v_{0} t-x$, leading to a rightward force $k s$. When the box is moving, it is always moving to the right, so the kinetic friction force $\\mu m g$ is always in the leftward direction, which means\n\n\n\n$$\n\nm a=k\\left(s+v_{0} t-x\\right)-\\mu m g\n\n$$\n\n\n\nwhich means\n\n\n\n$$\n\na=\\frac{k}{m}\\left(s+v_{0} t-x\\right)-\\mu g .\n\n$$']	['$a=\\frac{k}{m}\\left(s+v_{0} t-x\\right)-\\mu g$']	"## Toffee Pudding
+
+
+
+A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over.
+Context question:
+a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.
+Context answer:
+\boxed{$s=\mu_{0} m g / k$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+175	c. Find the time $t_{0}$ when the box stops for the first time.	['Taking the derivative, the velocity of the box is\n\n\n\n$$\n\nv=v_{0}(1-\\cos \\omega t)+(1-r) s \\omega \\sin \\omega t .\n\n$$\n\n\n\nThe box stops when this is equal to zero for the first time. There are several ways to\n\n\n\n\n\n\n\nevaluate this condition. First, we can use half-angle identities to find\n\n\n\n$$\n\n0=2 v_{0} \\sin ^{2} \\frac{\\omega t}{2}+2(1-r) s \\omega \\sin \\frac{\\omega t}{2} \\cos \\frac{\\omega t}{2}\n\n$$\n\n\n\nAs a result, the box stops when\n\n\n\n$$\n\n\\tan \\frac{\\omega t}{2}=-\\frac{(1-r) s \\omega}{v_{0}}\n\n$$\n\n\n\nUsing a basic property of the tangent function,\n\n\n\n$$\n\n\\tan \\left(\\pi-\\frac{\\omega t}{2}\\right)=\\frac{(1-r) s \\omega}{v_{0}}\n\n$$\n\n\n\nSolving for $t$, we conclude that\n\n\n\n$$\n\nt_{0}=\\frac{2 \\pi-2 \\alpha}{\\omega}, \\quad \\alpha=\\tan ^{-1} \\frac{(1-r) s \\omega}{v_{0}}\n\n$$\n\n\n\nNote that we cancelled a factor of $\\sin (\\omega t / 2)$, which has a zero at $t=2 \\pi / \\omega$. However, this is a larger time than the one we just found, so it is irrelevant.\n\n\n\nAnother way to arrive at the answer is to rewrite the original condition as\n\n\n\n$$\n\n1=\\cos \\omega t-\\tan \\alpha \\sin \\omega t\n\n$$\n\n\n\nSquaring both sides and using some trigonometric identities gives\n\n\n\n$$\n\n\\tan \\omega t=-\\frac{2 \\tan \\alpha}{1-\\tan ^{2} \\alpha}\n\n$$\n\n\n\nThis can then be further simplified using the tangent half-angle identity, upon which we recover the same result as above.']	['$t_{0}=\\frac{2 \\pi-2 \\alpha}{\\omega}, \\quad \\alpha=\\tan ^{-1} \\frac{(1-r) s \\omega}{v_{0}}$']	"## Toffee Pudding
+
+
+
+A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over.
+Context question:
+a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.
+Context answer:
+\boxed{$s=\mu_{0} m g / k$}
+
+
+Context question:
+b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving.
+Context answer:
+\boxed{$a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g$}
+
+
+Extra Supplementary Reading Materials:
+
+The position of the box as a function of time $t$ as defined in part (b) is
+
+
+
+$$
+
+x(t)=\frac{v_{0}}{\omega}(\omega t-\sin \omega t)+(1-r) s(1-\cos \omega t),
+
+$$
+
+
+
+where $\omega=\sqrt{k / m}$ and $r=\mu / \mu_{0}$. This expression applies as long as the box is still moving, and you can use it in the parts below. Express all your answers in terms of $v_{0}, \omega, s$, and $r$."	[]	Text-only	Competition	True		Expression		Open-ended	Mechanics	Physics	English
+176	e. After the box stops, how long will it stay at rest before starting to move again?	"['Using a result we found in part (d), the stretch is\n\n\n\n$$\n\n\\Delta \\ell=r s+\\frac{v_{0}}{\\omega \\cos \\alpha} \\sin \\left(\\omega t_{0}+\\alpha\\right)\n\n$$\n\n\n\nwhen the box stops. Plugging in the value of $t_{0}$ found in part (c),\n\n\n\n$$\n\n\\Delta \\ell=r s+\\frac{v_{0}}{\\omega \\cos \\alpha} \\sin (2 \\pi-\\alpha)=r s-\\frac{v_{0}}{\\omega} \\tan \\alpha=(2 r-1) s\n\n$$\n\n\n\nThe box starts to move again when the stretch becomes $s$, so the time is\n\n\n\n$$\n\n\\frac{s-(2 r-1) s}{v_{0}}=\\frac{2(1-r) s}{v_{0}}\n\n$$\n\n\n\nThe pattern of motion investigated in this problem is known as ""stick-slip"" and occurs in many practical contexts.']"	['$\\frac{2(1-r) s}{v_{0}}$']	"## Toffee Pudding
+
+
+
+A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over.
+Context question:
+a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.
+Context answer:
+\boxed{$s=\mu_{0} m g / k$}
+
+
+Context question:
+b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving.
+Context answer:
+\boxed{$a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g$}
+
+
+Extra Supplementary Reading Materials:
+
+The position of the box as a function of time $t$ as defined in part (b) is
+
+
+
+$$
+
+x(t)=\frac{v_{0}}{\omega}(\omega t-\sin \omega t)+(1-r) s(1-\cos \omega t),
+
+$$
+
+
+
+where $\omega=\sqrt{k / m}$ and $r=\mu / \mu_{0}$. This expression applies as long as the box is still moving, and you can use it in the parts below. Express all your answers in terms of $v_{0}, \omega, s$, and $r$.
+Context question:
+c. Find the time $t_{0}$ when the box stops for the first time.
+Context answer:
+\boxed{$t_{0}=\frac{2 \pi-2 \alpha}{\omega}, \quad \alpha=\tan ^{-1} \frac{(1-r) s \omega}{v_{0}}$}
+
+
+Context question:
+d. For what values of $r$ will the spring always be at least as long as its rest length?
+Context answer:
+$r \geq \frac{1}{2}\left(1+\left(\frac{v_{0}}{s \omega}\right)^{2}\right)$
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+177	"a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral:
+
+
+
+$$
+
+\int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3}
+
+$$"	"[""A point on the sphere's equator moves at a speed $\\omega R$, where $\\omega$ is the angular velocity of rotation. Setting $\\omega R=c$ gives $\\omega=c / R$.\n\n\n\nThe spinning sphere can be thought of as a stack of infinitesimal current loops, all of which have a magnetic moment pointing in the same direction. Consider making a thin, circular slice of the sphere's surface, corresponding to polar angles in the range $(\\theta, \\theta+d \\theta)$. This slice has a radius $R \\sin \\theta$, so that the surface area of the slice is\n\n\n\n$$\n\nd s=2 \\pi R \\sin \\theta R d \\theta\n\n$$\n\n\n\nThe charge of the slice is\n\n\n\n$$\n\nd Q=-\\frac{q d s}{4 \\pi R^{2}}=-\\frac{q \\sin \\theta}{2}\n\n$$\n\n\n\nSince the charge $d Q$ moves around the rotation axis one time per period $T=2 \\pi / \\omega$, the corresponding current is\n\n\n\n$$\n\nd I=\\frac{d Q}{T}=-\\frac{\\omega q \\sin \\theta}{4 \\pi}\n\n$$\n\n\n\nThe magnitude of the magnetic moment of this slice is\n\n\n\n$$\n\nd \\mu=\\pi(R \\sin \\theta)^{2}|d I|=\\frac{1}{4} q \\omega R^{2} \\sin ^{3} \\theta d \\theta\n\n$$\n\n\n\nUsing the provided integral, the total magnetic moment is\n\n\n\n$$\n\n\\mu=\\int_{0}^{\\pi} \\frac{1}{4} q \\omega R^{2} \\sin ^{3} \\theta d \\theta=\\frac{1}{3} q c R\n\n$$\n\n\n\nIf you weren't able to do this, you could also have given the answer $\\mu \\sim q c R$, which can be derived by dimensional analysis, for partial credit.""
+ 'Note that for a uniformly charged ring of mass $d m$, charge $d q$, and radius $r$, rotating with angular velocity $\\omega$, the ratio of the magnetic moment and the\n\n\n\n\n\n\n\nangular momentum is\n\n\n\n$$\n\n\\frac{\\mu}{L}=\\frac{\\pi r^{2}(\\omega d q / 2 \\pi)}{\\left(r^{2} d m\\right) \\omega}=\\frac{1}{2} \\frac{d q}{d m}\n\n$$\n\n\n\nThe ratio is independent of $r$ and $\\omega$. Since the sphere can be decomposed into such rings, the total magnetic moment and total angular momentum must have the same ratio,\n\n\n\n$$\n\n\\frac{\\mu}{L}=\\frac{1}{2} \\frac{q}{m}\n\n$$\n\n\n\nFinally, we know that $L=(2 / 3) m R^{2} \\omega$ for a spherical shell. Plugging this in and using $\\omega=c / R$ gives $\\mu=q c R / 3$ as before, but with no integration required.']"	['$\\mu=q c R / 3$']	"## Electroneering
+
+
+
+An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron.
+
+
+
+Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center."	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+178	b. The electron's magnetic moment is known to be $\mu=q \hbar / 2 m$, where $\hbar$ is the reduced Planck constant. In this model, what is the minimum possible radius of the electron? Express your answer in terms of $m$ and fundamental constants.	['Since the magnetic moment is fixed, and we want the radius to be small, we want the electron to be spinning as fast as possible. Thus, the magnetic moment has the value found in part (a), and equating this to the known value gives\n\n\n\n$$\n\nR=\\frac{3}{2} \\frac{\\hbar}{m c}\n\n$$\n\n\n\nAgain, you can get $R \\sim \\hbar / m c$ by dimensional analysis.']	['$R=\\frac{3}{2} \\frac{\\hbar}{m c}$']	"## Electroneering
+
+
+
+An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron.
+
+
+
+Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center.
+Context question:
+a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral:
+
+
+
+$$
+
+\int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3}
+
+$$
+Context answer:
+\boxed{$\mu=q c R / 3$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+179	"c. Assuming the radius is the value you found in part (b), how much energy is stored in the electric field of the electron? Express your answer in terms of $E_{0}=m c^{2}$ and the fine structure constant,
+
+
+
+$$
+
+\alpha=\frac{q^{2}}{4 \pi \epsilon_{0} \hbar c} \approx \frac{1}{137}
+
+$$"	"[""For a collection of charges, the total energy stored in the electrostatic field is\n\n\n\n$$\n\nU_{E}=\\frac{1}{2} \\sum_{i} q_{i} V_{i}\n\n$$\n\n\n\nwhere $V_{i}$ is the electric potential at $q_{i}$. In this case, the total charge is $q$, and all of the charge is at potential $q / 4 \\pi \\epsilon_{0} R$, so\n\n\n\n$$\n\nU_{E}=\\frac{q^{2}}{8 \\pi \\epsilon_{0} R}\n\n$$\n\n\n\nUsing the result of part (b),\n\n\n\n$$\n\nU_{E}=\\frac{1}{3} \\alpha E_{0}\n\n$$\n\n\n\nNote that you can't get this answer by dimensional analysis alone, since $\\alpha$ is dimensionless. (However, if you found $R$ by dimensional analysis, and additionally reasoned that $U_{E}$ could\ndepend only on $q, \\epsilon_{0}$, and $R$, then you could derive $U_{E} \\sim \\alpha E_{0}$, for partial credit.)""]"	['$U_{E}=\\frac{1}{3} \\alpha E_{0}$']	"## Electroneering
+
+
+
+An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron.
+
+
+
+Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center.
+Context question:
+a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral:
+
+
+
+$$
+
+\int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3}
+
+$$
+Context answer:
+\boxed{$\mu=q c R / 3$}
+
+
+Context question:
+b. The electron's magnetic moment is known to be $\mu=q \hbar / 2 m$, where $\hbar$ is the reduced Planck constant. In this model, what is the minimum possible radius of the electron? Express your answer in terms of $m$ and fundamental constants.
+Context answer:
+\boxed{$R=\frac{3}{2} \frac{\hbar}{m c}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+180	d. Roughly estimate the total energy stored in the magnetic field of the electron, in terms of $E_{0}$ and $\alpha$. (Hint: one way to do this is to suppose the magnetic field has roughly constant magnitude inside the sphere and is negligible outside of it, then estimate the field inside the sphere.)	['Following the hint, we can estimate\n\n\n\n$$\n\nU_{B} \\sim \\frac{B_{0}^{2}}{2 \\mu_{0}}\\left(\\frac{4}{3} \\pi R^{3}\\right)\n\n$$\n\n\n\nwhere $B_{0}$ is the typical magnetic field inside the sphere. Actually finding the value of $B_{0}$ would require doing some complicated integrals. To get a rough estimate, note that if we replaced the sphere with a ring of charge, then at the center of the ring,\n\n\n\n$$\n\nB_{0} \\sim \\frac{\\mu_{0} I}{R} \\sim \\frac{\\mu_{0} q c}{R^{2}}\n\n$$\n\n\n\nThus, we have\n\n\n\n$$\n\nU_{B} \\sim \\frac{1}{\\mu_{0}}\\left(\\frac{\\mu_{0} q c}{R^{2}}\\right)^{2} R^{3} \\sim \\frac{\\mu_{0} q^{2} c^{2}}{R} \\sim \\frac{\\mu_{0} m c^{3} q^{2}}{\\hbar}\n\n$$\n\n\n\nTo get this in terms of the fine structure constant, we use $c^{2}=1 / \\mu_{0} \\epsilon_{0}$, giving\n\n\n\n$$\n\nU_{B} \\sim m c^{2} \\frac{q^{2}}{\\epsilon_{0} \\hbar c} \\sim \\alpha E_{0}\n\n$$\n\n\n\nAn even faster way to get this result is to note that in general, the energy stored in magnetic fields tends to be a factor of order $(v / c)^{2}$ smaller than the energy stored in electric fields, where $v$ is the speed of the charge. In this problem the charge is all moving relativistically, so we must have $U_{B} \\sim U_{E}$.']	['$\\alpha E_{0}$']	"## Electroneering
+
+
+
+An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron.
+
+
+
+Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center.
+Context question:
+a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral:
+
+
+
+$$
+
+\int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3}
+
+$$
+Context answer:
+\boxed{$\mu=q c R / 3$}
+
+
+Context question:
+b. The electron's magnetic moment is known to be $\mu=q \hbar / 2 m$, where $\hbar$ is the reduced Planck constant. In this model, what is the minimum possible radius of the electron? Express your answer in terms of $m$ and fundamental constants.
+Context answer:
+\boxed{$R=\frac{3}{2} \frac{\hbar}{m c}$}
+
+
+Context question:
+c. Assuming the radius is the value you found in part (b), how much energy is stored in the electric field of the electron? Express your answer in terms of $E_{0}=m c^{2}$ and the fine structure constant,
+
+
+
+$$
+
+\alpha=\frac{q^{2}}{4 \pi \epsilon_{0} \hbar c} \approx \frac{1}{137}
+
+$$
+Context answer:
+\boxed{$U_{E}=\frac{1}{3} \alpha E_{0}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+181	i. What is the minimum heat extraction $-d Q_{2, \text { min }}$ required by the laws of thermodynamics to heat up the room by $d Q_{1}$ ?	"['The second law of thermodynamics implies that, no matter what you do, you must have $\\mathrm{d} S_{\\text {universe }} \\geq 0$, and if your process is to be as efficient as possible, we can assume it is reversible, so\n\n\n\n$$\n\n\\mathrm{d} S_{\\text {universe; reversible }}=0\n\n$$\n\n\n\nIf we do extract any work while allowing heat to transfer between reservoirs, we will later use that work to transfer more heat. So in the entire process, there are only heat transfers, and by conservation of energy,\n\n\n\n$$\n\n\\mathrm{d} Q_{0}+\\mathrm{d} Q_{1}+\\mathrm{d} Q_{2}=0\n\n$$\n\n\n\nThe entropy change associated with each reversible heat transfer is $d S=d Q / T$, so our assumption of zero entropy production becomes\n\n\n\n$$\n\n\\frac{\\mathrm{d} Q_{0}}{T_{0}}+\\frac{\\mathrm{d} Q_{1}}{T_{1}}+\\frac{\\mathrm{d} Q_{2}}{T_{2}}=0\n\n$$\n\n\n\nBy combining these equations, we can eliminate $d Q_{0}$ and solve for $d Q_{2}$, giving\n\n\n\n$$\n\n-\\mathrm{d} Q_{2 ; \\min }=\\frac{T_{2}}{T_{1}} \\frac{T_{1}-T_{0}}{T_{2}-T_{0}} \\mathrm{~d} Q_{1}\n\n$$\n\n\n\nFor the provided numbers, this happens to be about $0.11 d Q_{1}$. That is, a heat pump can be much more efficient than direct heating. This problem was inspired by Jaynes, E. T, ""Note on thermal heating efficiency."", American Journal of Physics 71.2 (2003): 180-182. (You can also solve the problem by considering an explicit procedure using Carnot engines. But since Carnot engines are reversible, all such procedures will just give the same answer.)']"	['$\\frac{T_{2}}{T_{1}} \\frac{T_{1}-T_{0}}{T_{2}-T_{0}} \\mathrm{~d} Q_{1}$']	"## Hot Pocket
+a. It's winter and you want to keep warm. The temperature is $T_{0}=263 \mathrm{~K}$ outside and $T_{1}=290 \mathrm{~K}$ in your room. You have started a fire, which acts as a hot reservoir at temperature $T_{2}=1800 \mathrm{~K}$.
+
+
+
+You want to add a small amount of heat $d Q_{1}$ to your room. The simplest method would be to extract heat $-d Q_{2, \mathrm{dump}}=d Q_{1}$ from the fire and directly transfer it to your room. However, it is possible to heat your room more efficiently. Suppose that you can exchange heat between any pair of reservoirs. You cannot use any external source of work, such as the electrical grid, but the work extracted from running heat engines can be stored and used without dissipation."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+182	i. As a parcel of air moves upward, it accelerates. Find a rough estimate for the average speed $v_{0}$ during its upward motion.	['The temperature of the air is higher than its surroundings by a fractional amount of order $\\Delta T / T$. Thus, by the ideal gas law, the density is lower than its surroundings by a fraction of order $\\Delta T / T$, which means the upward acceleration due to the buoyant force is of order $a=g \\Delta T / T$. Since this is roughly uniformly accelerated motion, $v_{0}^{2} \\propto a h$,\n\n\n\n\n\n\n\nwhich implies\n\n\n\n$$\n\nv_{0} \\sim \\sqrt{g h \\frac{\\Delta T}{T_{0}}}\n\n$$\n\n\n\nNote that because $d P / d z=-\\rho g$ in hydrostatic equilibrium, the pressure of the surrounding air varies between the bottom and top of the container, by a fractional amount of order $\\rho_{0} g h / P_{0}$. But since we assumed $\\rho_{0} g h / P_{0} \\ll \\Delta T / T_{0}$, we can neglect this effect.']	['$\\sqrt{g h \\frac{\\Delta T}{T_{0}}}$']	"b. When the air at the bottom of a container is heated, it becomes less dense than the surrounding air and rises. Simultaneously, cooler air falls downward. This process of net upward heat transfer is known as convection.
+
+
+
+Consider a closed, rectangular box of height $h$ filled with air initially of uniform temperature $T_{0}$. Next, suppose the bottom of the box is heated so that the air there instantly reaches temperature $T_{0}+\Delta T$. The hot parcel of air at the bottom rises upward until it hits the top of the box, where its temperature is instantly reduced to $T_{0}$.
+
+
+
+You may neglect any heat transfer and friction between the parcel of air and the surrounding air, and assume that the temperature difference is not too large. In addition, you may assume the height $h$ is small enough so that the pressure $P_{0}$ and density $\rho_{0}$ of the surrounding air are very nearly constant throughout the container. More precisely, assume that $\rho_{0} g h / P_{0} \ll \Delta T / T_{0} \ll 1$. Express your answers in terms of $P_{0}, g, h, \Delta T$, and $T_{0}$."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+183	ii. In the steady state, warm parcels of air are continuously moving upward from the bottom, and cold parcels of air are continuously moving downward from the top. Find a rough estimate for the net rate of upward energy transfer per area.	['The extra energy carried by a parcel of gas is\n\n\n\n$$\n\nn C_{p} \\Delta T \\sim n R \\Delta T \\sim P_{0} V \\frac{\\Delta T}{T_{0}}\n\n$$\n\n\n\nwhere $V$ is the volume of the parcel. The net volume of warm air transported upward per unit time is of order $A v_{0}$, where $A$ is the cross-sectional area of the box. Thus, the average power per area is roughly\n\n\n\n$$\n\nP_{0} v_{0} \\frac{\\Delta T}{T_{0}} \\sim P_{0} \\sqrt{g h}\\left(\\frac{\\Delta T}{T_{0}}\\right)^{3 / 2}\n\n$$\n\n\n\nThis is a simplified version of the mixing length theory of convection, which is essential for modeling the interiors of stars.']	"['$P_{0} v_{0} \\frac{\\Delta T}{T_{0}}$'
+ '$ P_{0} \\sqrt{g h}\\left(\\frac{\\Delta T}{T_{0}}\\right)^{3 / 2}$']"	"b. When the air at the bottom of a container is heated, it becomes less dense than the surrounding air and rises. Simultaneously, cooler air falls downward. This process of net upward heat transfer is known as convection.
+
+
+
+Consider a closed, rectangular box of height $h$ filled with air initially of uniform temperature $T_{0}$. Next, suppose the bottom of the box is heated so that the air there instantly reaches temperature $T_{0}+\Delta T$. The hot parcel of air at the bottom rises upward until it hits the top of the box, where its temperature is instantly reduced to $T_{0}$.
+
+
+
+You may neglect any heat transfer and friction between the parcel of air and the surrounding air, and assume that the temperature difference is not too large. In addition, you may assume the height $h$ is small enough so that the pressure $P_{0}$ and density $\rho_{0}$ of the surrounding air are very nearly constant throughout the container. More precisely, assume that $\rho_{0} g h / P_{0} \ll \Delta T / T_{0} \ll 1$. Express your answers in terms of $P_{0}, g, h, \Delta T$, and $T_{0}$.
+Context question:
+i. As a parcel of air moves upward, it accelerates. Find a rough estimate for the average speed $v_{0}$ during its upward motion.
+Context answer:
+\boxed{$\sqrt{g h \frac{\Delta T}{T_{0}}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+184	a. How much time elapses on Carla's stopwatch with each revolution?	"[""From time dilation, her clock ticks slower by a factor $\\gamma$. Therefore, each revolution takes\n\n\n\n$$\n\n\\frac{2 \\pi R}{\\gamma v}=\\frac{2 \\pi R \\sqrt{1-v^{2} / c^{2}}}{v}\n\n$$\n\n\n\nwhen measured by Carla's stopwatch.""]"	['$\\frac{2 \\pi R \\sqrt{1-v^{2} / c^{2}}}{v}$']	"## Spin Cycle
+
+
+
+Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+185	b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ?	"[""Carla's motion is perpendicular to the displacement between Clock $A$ and Clock $B$ when they are synchronized in CIRF. Therefore, the simultaneous synchronization in CIRF is also simultaneous in the lab frame. Thus, the offset is 0 .\n\n\n\nTo understand why this offset is 0 , you can also imagine placing an lightbulb halfway between the two clocks and having it send a light pulse at some known time. In both Carla's frame and the lab frame, the light pulse reaches the two clocks simultaneously.""]"	['0']	"## Spin Cycle
+
+
+
+Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light.
+Context question:
+a. How much time elapses on Carla's stopwatch with each revolution?
+Context answer:
+\boxed{$\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$}
+
+
+Extra Supplementary Reading Materials:
+
+Carla decides to do a fun experiment during her training. She places two stationary clocks down: Clock A at the center of the race track, i.e. the origin; and Clock B at a point on the race track denoted as $(R, 0)$. She then begins her training.
+
+
+
+For parts (b) through (d), we define Carla's inertial reference frame (CIRF) as an inertial reference frame in which Carla is momentarily at rest, and which has the same origin of coordinates as the lab frame. Thus, CIRF is a new inertial frame each moment. The times on the clocks and stopwatch are all calibrated such that they all read 0 in CIRF when she passes by Clock $B$ for the first time."	[]	Text-only	Competition	False		Numerical	0	Open-ended	Modern Physics	Physics	English
+186	c. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock A measure in CIRF?	"[""By symmetry, the speed at which the center clock ticks according to CIRF cannot change. In one revolution, Carla's stopwatch measures $\\frac{2 \\pi R \\sqrt{1-v^{2} / c^{2}}}{v}$, while the center clock measures $\\frac{2 \\pi R}{v}$. Then,\n\n\n\n\n\n$$\n\nt_{A}(\\tau)=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}\n\n$$""]"	['$t_{A}(\\tau)=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}$']	"## Spin Cycle
+
+
+
+Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light.
+Context question:
+a. How much time elapses on Carla's stopwatch with each revolution?
+Context answer:
+\boxed{$\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$}
+
+
+Extra Supplementary Reading Materials:
+
+Carla decides to do a fun experiment during her training. She places two stationary clocks down: Clock A at the center of the race track, i.e. the origin; and Clock B at a point on the race track denoted as $(R, 0)$. She then begins her training.
+
+
+
+For parts (b) through (d), we define Carla's inertial reference frame (CIRF) as an inertial reference frame in which Carla is momentarily at rest, and which has the same origin of coordinates as the lab frame. Thus, CIRF is a new inertial frame each moment. The times on the clocks and stopwatch are all calibrated such that they all read 0 in CIRF when she passes by Clock $B$ for the first time.
+Context question:
+b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ?
+Context answer:
+\boxed{0}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+187	d. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock B measure in CIRF?	"[""The readings on Clock B and on Clock A are not necessarily identical once Carla moves through the circle (because her motion becomes more parallel with the displacement between the two clocks, and thus simultaneity is lost).\n\n\n\nSuppose Carla is at $(R \\cos \\theta, R \\sin \\theta)$, so her velocity is given by $(-v \\sin \\theta, v \\cos \\theta)$. Suppose we place a light bulb between the two clocks and having it propagate a light pulse. In the lab frame, the light pulse reaches the two clocks simultaneously. In CIRF, the math is a little more complicated.\n\n\n\nWe first rotate our lab coordinates so that $\\hat{\\mathbf{a}}=-\\sin \\theta \\hat{\\mathbf{x}}+\\cos \\theta \\hat{\\mathbf{y}}$, and $\\hat{\\mathbf{b}}=\\cos \\theta \\hat{\\mathbf{x}}+\\sin \\theta \\hat{\\mathbf{y}}$. We now give the coordinates of the clocks and bulb in the rotated lab frame: Clock A, $(a, b)=(0,0)$; Clock B, $(a, b)=(-R \\sin \\theta, R \\cos \\theta)$; bulb, $(a, b)=(-R \\sin \\theta, R \\cos \\theta) / 2$. In the lab frame, a light pulse is emitted at\n\n\n\n$$\n\nt=0, a=-(R / 2) \\sin \\theta, b=(R / 2) \\cos \\theta\n\n$$\n\n\n\nThe light pulse reaches Clock A at\n\n\n\n$$\n\nt=R / 2, a=0, b=0\n\n$$\n\n\n\nand Clock B at\n\n\n\n$$\n\nt=R / 2, a=-R \\sin \\theta, b=R \\cos \\theta .\n\n$$\n\n\n\nUnder a Lorentz tranformation from the lab frame to CIRF, we have that the light pulse reaches Clock $A$ at $t^{\\prime}=\\gamma R / 2$ and Clock $B$ at $t^{\\prime}=\\gamma R / 2+\\gamma v R \\sin \\theta$. Thus, Clock $B$ reads the same time as Clock $A$ with offset $\\gamma v R \\sin \\theta$ in the reference frame moving at $v_{a}=v$, $v_{b}=0$. Note that Clock $A$ ticks slower by a factor of $\\gamma$ in this frame. Therefore, the time on clock $B$ is $v R \\sin \\theta$ behind the time on clock $A$.\n\n\n\nThen,\n\n\n\n$$\n\nt_{B}(\\tau)=t_{A}(\\tau)-v R \\sin \\theta=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}-v R \\sin \\theta\n\n$$\n\n\n\n(This is the answer we expect from the rear clock ahead effect!) Finally, we use that $\\theta=\\omega \\tau$ and $\\omega=\\frac{2 \\pi}{T}$, where $T$ is the period in Carla's frame. Then,\n\n\n\n$$\n\nt_{B}(\\tau)=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}-\\frac{v R}{c^{2}} \\sin \\left(\\frac{v \\tau}{R \\sqrt{1-v^{2} / c^{2}}}\\right)\n\n$$""]"	['$t_{B}(\\tau)=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}-\\frac{v R}{c^{2}} \\sin \\left(\\frac{v \\tau}{R \\sqrt{1-v^{2} / c^{2}}}\\right)$']	"## Spin Cycle
+
+
+
+Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light.
+Context question:
+a. How much time elapses on Carla's stopwatch with each revolution?
+Context answer:
+\boxed{$\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$}
+
+
+Extra Supplementary Reading Materials:
+
+Carla decides to do a fun experiment during her training. She places two stationary clocks down: Clock A at the center of the race track, i.e. the origin; and Clock B at a point on the race track denoted as $(R, 0)$. She then begins her training.
+
+
+
+For parts (b) through (d), we define Carla's inertial reference frame (CIRF) as an inertial reference frame in which Carla is momentarily at rest, and which has the same origin of coordinates as the lab frame. Thus, CIRF is a new inertial frame each moment. The times on the clocks and stopwatch are all calibrated such that they all read 0 in CIRF when she passes by Clock $B$ for the first time.
+Context question:
+b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ?
+Context answer:
+\boxed{0}
+
+
+Context question:
+c. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock A measure in CIRF?
+Context answer:
+\boxed{$t_{A}(\tau)=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+188	"a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up.
+
+Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout."	['We can use energy conservation to answer this question. For a bit of water with mass $m$, the total energy $E$ is the sum of the kinetic and gravitational potential energies,\n\n\n\n$$\n\nE=\\frac{1}{2} m v^{2}+m g y \\text {. } \\tag{B1-1}\n\n$$\n\n\n\n(With this sign convention, $g \\approx 10 \\mathrm{~m} / \\mathrm{s}^{2}$ is positive. As $y$ decreases, so does the potential energy.)\n\n\n\nAs the bit of water falls, its energy remains constant, and is equal to the initial value of\n\n\n\n$$\n\nE=\\frac{1}{2} m v_{0}^{2}+m g y_{0} \\tag{B1-2}\n\n$$\n\n\n\nEquating eliminating $E$ from equations B1-1 and B1-2, we have\n\n\n\n$$\n\n\\frac{1}{2} m v^{2}+m g y=\\frac{1}{2} m v_{0}^{2}+m g y_{0},\n\n$$\n\n\n\nand solving for $v$, we get\n\n\n\n$$\n\nv=\\sqrt{v_{0}^{2}+2 g\\left(y_{0}-y\\right)}\n\n$$\n\n\n\nThe equation for $v^{2}$ has three terms, but we were asked to choose the zero of $y$ such that there is only one. Evidently, two of the terms must cancel, and these must be the two constant terms, since the final term varies with $y$.\n\n\n\nThat means we need\n\n\n\n$$\n\nv_{0}^{2}+2 g y_{0}=0\n\n$$\n\n\n\nSolving for $y_{0}$, the vertical position of the spout is\n\n\n\n$$\n\ny_{0}=\\frac{-v_{0}^{2}}{2 g}\n\n$$\n\n\n\n\n\n\n\nWith this choice of the zero of $y$, the equation for $v$ simplifies to\n\n\n\n$$\n\nv=\\sqrt{-2 g y} \\tag{B1-3}\n\n$$\n\n\n\nWe note that the result of this equation is real because $y<0$ at the spout, and decreases as the water falls, so this equation shows that $v$ is real and increases as the water falls.']	['$y_{0}=\\frac{-v_{0}^{2}}{2 g}$ ,$v=\\sqrt{-2 g y}$']	## String Cheese	[]	Text-only	Competition	True		Expression		Open-ended	Mechanics	Physics	English
+189	"b. Assume that the stream of water falling from the faucet is cylindrically symmetric about a vertical axis through the center of the stream. Also assume that the volume of water per unit time exiting the spout is constant, and that the shape of the stream of water is constant over time.
+In this case, the radius $r$ of the stream of water is a function of vertical position $y$. Let the radius at the faucet be $r_{0}$. Using your result from part (a), find $r(y)$.
+
+
+If $r(y)$ is not constant, it implies that the water has some radial velocity during its fall, in contradiction to our assumptions in part (a) that the motion is purely vertical. You may assume throughout the problem that any such radial velocity is negligibly small."	"[""The same volume of water must fall through any horizontal cross-section of the stream each second because water doesn't disappear during its fall, and its density if constant. That volume per unit time $Q$ is the cross-sectional area of the stream multiplied by the speed of the water in the vertical direction. As an equation,\n\n\n\n$$\n\nQ=v \\pi r^{2} \\tag{B1-4}\n\n$$\n\n\n\n$Q$ is the same at all $y$, and is equal to its initial value of\n\n\n\n$$\n\nQ=v_{0} \\pi r_{0}^{2} \\tag{B1-5}\n\n$$\n\n\n\nEliminating $Q$ from B1-4 and B1-5 and solving for $r$ gives\n\n\n\n$$\n\nr=r_{0} \\sqrt{\\frac{v_{0}}{v}}\n\n$$\n\n\n\nPlugging in our equation B1-3 for $v$,\n\n\n\n$$\n\nr=r_{0} \\sqrt[4]{\\frac{v_{0}^{2}}{-2 g y}}\n\n$$""]"	['$r=r_{0} \\sqrt[4]{\\frac{v_{0}^{2}}{-2 g y}}$']	"## String Cheese
+Context question:
+a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up.
+
+Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout.
+Context answer:
+\boxed{$y_{0}=\frac{-v_{0}^{2}}{2 g}$ ,$v=\sqrt{-2 g y}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+190	"a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by
+
+
+$$
+
+u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}
+
+$$
+
+
+
+where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$."	['The density of the gas is given by $N / V$. In a spherical shell of radius $r$ and thickness $\\Delta r$, there are $\\left(4 \\pi r^{2} \\Delta r\\right) N / V$ atoms. The potential energy is given by\n\n\n\n$$\n\n\\Delta U=-\\left(4 \\pi r^{2} \\Delta r\\right) N / V \\epsilon d^{6} / r^{6}\n\n$$\n\n\n\nThen, the total potential energy is given by\n\n\n\n$$\n\nU=\\int_{d}^{\\infty}-\\left(4 \\pi r^{2} \\mathrm{~d} r\\right) N / V \\epsilon d^{6} / r^{6}=-2 a^{\\prime} N / V\n\n$$']	"['$-2 a^{\\prime} N / V$'
+ '$-2\\cdot \\frac{2 \\pi d^{3} \\epsilon N}{3V}$']"	"## Real Expansion
+
+
+
+Consider a ""real"" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this ""real"" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+191	b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.	['Now only half of the shell of radius $r$ is full of gas, and the other half is outside of the box. This mean that the potential energy is lessened by a factor of two, to $-a^{\\prime} N / V$.']	['$-a^{\\prime} N / V$']	"## Real Expansion
+
+
+
+Consider a ""real"" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this ""real"" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
+Context question:
+a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by
+
+
+$$
+
+u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}
+
+$$
+
+
+
+where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
+Context answer:
+\boxed{$-2 a^{\prime} N / V$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+192	c. Using Bernoulli's law $P+U+\rho v^{2} / 2=$ constant, with pressure $P$, potential energy density $U$, mass density $\rho$ and fluid velocity $v$, what is the pressure at the boundary of the box? Assume the interior pressure is given by the ideal gas law.	['The potential energy density difference is $-a^{\\prime} \\frac{N^{2}}{V^{2}}$. Since there is no velocity difference, this is also the pressure difference. If the pressure on the interior is $\\frac{N k T}{V}$, then the pressure on the box is $\\frac{N k T}{V}-a^{\\prime} \\frac{N^{2}}{V^{2}}$']	['$\\frac{N k T}{V}-a^{\\prime} \\frac{N^{2}}{V^{2}}$']	"## Real Expansion
+
+
+
+Consider a ""real"" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this ""real"" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
+Context question:
+a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by
+
+
+$$
+
+u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}
+
+$$
+
+
+
+where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
+Context answer:
+\boxed{$-2 a^{\prime} N / V$}
+
+
+Context question:
+b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.
+Context answer:
+\boxed{$-a^{\prime} N / V$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+193	d. Assuming most atoms are in the interior of the box, what is the total energy of the atoms in the box?	['The total kinetic energy is $\\frac{3}{2} N k T$. The total potential energy is $-a^{\\prime} N^{2} / V$ (we drop a factor of two to avoid double-counting). So the total energy is $\\frac{3}{2} N k T-a^{\\prime} N^{2} / V$.']	['$\\frac{3}{2} N k T-a^{\\prime} N^{2} / V$']	"## Real Expansion
+
+
+
+Consider a ""real"" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this ""real"" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
+Context question:
+a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by
+
+
+$$
+
+u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}
+
+$$
+
+
+
+where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
+Context answer:
+\boxed{$-2 a^{\prime} N / V$}
+
+
+Context question:
+b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.
+Context answer:
+\boxed{$-a^{\prime} N / V$}
+
+
+Context question:
+c. Using Bernoulli's law $P+U+\rho v^{2} / 2=$ constant, with pressure $P$, potential energy density $U$, mass density $\rho$ and fluid velocity $v$, what is the pressure at the boundary of the box? Assume the interior pressure is given by the ideal gas law.
+Context answer:
+\boxed{$\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+194	e. What is the final temperature of the gas after the expansion?	['Naively, we might say that the total potential energy of the gas is $-2 a^{\\prime} N^{2} / V$, but to avoid double-counting, we divide by 2 and instead arrive at $-a^{\\prime} N^{2} / V$. Then, the quantity\n\n\n\n$$\n\nE=\\frac{3}{2} N k_{B} T-\\frac{a^{\\prime} N^{2}}{V}\n\n$$\n\n\n\nis conserved. Therefore,\n\n\n\n$$\n\nT^{\\prime}=T-\\frac{a^{\\prime} N}{3 k_{B} V}\n\n$$']	['$T^{\\prime}=T-\\frac{a^{\\prime} N}{3 k_{B} V}$']	"## Real Expansion
+
+
+
+Consider a ""real"" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this ""real"" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
+Context question:
+a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by
+
+
+$$
+
+u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}
+
+$$
+
+
+
+where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
+Context answer:
+\boxed{$-2 a^{\prime} N / V$}
+
+
+Context question:
+b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.
+Context answer:
+\boxed{$-a^{\prime} N / V$}
+
+
+Context question:
+c. Using Bernoulli's law $P+U+\rho v^{2} / 2=$ constant, with pressure $P$, potential energy density $U$, mass density $\rho$ and fluid velocity $v$, what is the pressure at the boundary of the box? Assume the interior pressure is given by the ideal gas law.
+Context answer:
+\boxed{$\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$}
+
+
+Context question:
+d. Assuming most atoms are in the interior of the box, what is the total energy of the atoms in the box?
+Context answer:
+\boxed{$\frac{3}{2} N k T-a^{\prime} N^{2} / V$}
+
+
+Extra Supplementary Reading Materials:
+
+Now consider an insulated partitioned container with two sections, each of volume $V$. We fill one side of the container with $N$ atoms of this ""real"" gas at temperature $T$, which the other side being a vacuum. We then quickly remove the partition and let the gas expand to fill the entirety of the partitioned container. During this expansion, the energy of the gas remains unchanged."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+195	f. What is the increase in the entropy of the universe as a result of the free expansion? Give your answer to first order in $\frac{a^{\prime} N}{V k_{B} T}$.	['The entropy of the surroundings do not increase as a result of the free expansion (no heat is dumped to the surroundings, and the surroundings remain in thermal equilibrium). However, the entropy of the gas does increase because the gas is momentarily not in equilibrium. Therefore, we just have to compute the increase in entropy of the gas.\n\n\n\nBecause entropy is a state function, we compute this change in entropy by constructing a reversible process between the initial and final states of the expansion, and computing the change in entropy for this process. Consider constant energy reversible expansion of this gas. For this process, the work done by the gas is equal to the heat the gas takes in. Therefore,\n\n\n\n$$\n\n\\mathrm{d} S=\\frac{p \\mathrm{~d} v}{t}\n\n$$\n\n\n\nwhere we use lowercase letters to denote the quantities during the reversible expansion.\n\n\n\nRecall that\n\n\n\n$$\n\np v+\\frac{a^{\\prime} N^{2}}{v}=N k_{B} t\n\n$$\n\n\n\n\n\n\n\nIf the energy of the system is $E$, then,\n\n\n\n$$\n\n\\frac{3}{2} p v+\\frac{3 a^{\\prime} N^{2}}{2 v}-\\frac{a^{\\prime} N^{2}}{v}=E\n\n$$\n\n\n\nThen,\n\n\n\n$$\n\np=\\frac{2 E}{3 v}-\\frac{a^{\\prime} N^{2}}{3 v^{2}}\n\n$$\n\n\n\nFrom our expression of energy,\n\n\n\n$$\n\nt=\\frac{2}{3} \\frac{E+a^{\\prime} N^{2} / v}{N k_{B}}\n\n$$\n\n\n\nThen,\n\n\n\n$$\n\n\\Delta S=\\int_{V}^{2 V} \\frac{E N k_{B}}{E v+a^{\\prime} N^{2}}-\\frac{a^{\\prime} N^{3} k_{B}}{2\\left(E v^{2}+a^{\\prime} N^{2} v\\right)} \\mathrm{d} v\n\n$$\n\n\n\nTaylor expanding gives us\n\n\n\n$$\n\n\\Delta S=\\int_{V}^{2 V} \\frac{N k_{B}}{v}-\\frac{3 a^{\\prime} N^{3} k_{B}}{2 E v^{2}} \\mathrm{~d} v\n\n$$\n\n\n\nIntegrating gives us\n\n\n\n$$\n\n\\Delta S=N k_{B} \\log 2-\\frac{3 a^{\\prime} N^{3} k_{B}}{4 E V}\n\n$$\n\n\n\nUsing that $E \\approx 3 / 2 N k_{B} T$, we arrive at\n\n\n\n$$\n\n\\Delta S=N k_{B} \\log 2-\\frac{a^{\\prime} N^{2}}{2 V T}\n\n$$']	['$\\Delta S=N k_{B} \\log 2-\\frac{a^{\\prime} N^{2}}{2 V T}$']	"## Real Expansion
+
+
+
+Consider a ""real"" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this ""real"" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
+Context question:
+a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by
+
+
+$$
+
+u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}
+
+$$
+
+
+
+where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
+Context answer:
+\boxed{$-2 a^{\prime} N / V$}
+
+
+Context question:
+b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.
+Context answer:
+\boxed{$-a^{\prime} N / V$}
+
+
+Context question:
+c. Using Bernoulli's law $P+U+\rho v^{2} / 2=$ constant, with pressure $P$, potential energy density $U$, mass density $\rho$ and fluid velocity $v$, what is the pressure at the boundary of the box? Assume the interior pressure is given by the ideal gas law.
+Context answer:
+\boxed{$\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$}
+
+
+Context question:
+d. Assuming most atoms are in the interior of the box, what is the total energy of the atoms in the box?
+Context answer:
+\boxed{$\frac{3}{2} N k T-a^{\prime} N^{2} / V$}
+
+
+Extra Supplementary Reading Materials:
+
+Now consider an insulated partitioned container with two sections, each of volume $V$. We fill one side of the container with $N$ atoms of this ""real"" gas at temperature $T$, which the other side being a vacuum. We then quickly remove the partition and let the gas expand to fill the entirety of the partitioned container. During this expansion, the energy of the gas remains unchanged.
+Context question:
+e. What is the final temperature of the gas after the expansion?
+Context answer:
+\boxed{$T^{\prime}=T-\frac{a^{\prime} N}{3 k_{B} V}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+196	a. Find the final temperature $T_{f}$ of the two objects, and the total work $W$ done by the engine.	['Since a Carnot engine is reversible, it produces no entropy,\n\n\n\n$$\n\nd S_{1}+d S_{2}=\\frac{d Q_{1}}{T_{1}}+\\frac{d Q_{2}}{T_{2}}=0\n\n$$\n\n\nBy the definition of heat capacity, $d Q_{i}=C d T_{i}$, so\n\n\n\n$$\n\n\\frac{d T_{1}}{T_{1}}=-\\frac{d T_{2}}{T_{2}}\n\n$$\n\n\n\nIntegrating this equation shows that $T_{1} T_{2}$ is constant, so the final temperature is\n\n\n\n$$\n\nT_{f}=\\sqrt{T_{1} T_{2}}\n\n$$\n\n\n\nThe change in thermal energy of the objects is\n\n\n\n$$\n\nC\\left(T_{f}-T_{1}\\right)+C\\left(T_{f}-T_{2}\\right)=C\\left[2 \\sqrt{T_{1} T_{2}}-T_{1}-T_{2}\\right]\n\n$$\n\n\n\nBy the First Law of Thermodynamics, the missing energy has been used to do work, so\n\n\n\n$$\n\nW=C\\left[T_{1}+T_{2}-2 \\sqrt{T_{1} T_{2}}\\right]\n\n$$']	['$T_{f}=\\sqrt{T_{1} T_{2}}$, $W=C\\left[T_{1}+T_{2}-2 \\sqrt{T_{1} T_{2}}\\right]$']	"$g=9.8 \mathrm{~N} / \mathrm{kg}$
+
+
+
+$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$
+
+
+
+$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$
+
+
+
+$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$
+
+
+
+$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$
+
+
+
+$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$
+
+
+
+$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$
+
+
+
+$$
+
+\begin{aligned}
+
+& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+& e=1.602 \times 10^{-19} \mathrm{C} \\
+
+& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+& \ln (1+x) \approx x \text { for }|x| \ll 1 \\
+
+& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{aligned}
+
+$$
+
+$$
+
+\begin{array}{lrl}
+
+\text { velocity parameter } & \beta & =\frac{v}{c} \\
+
+\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\
+
+\text { relativistic momentum } & p & =\gamma \beta m c \\
+
+\text { relativistic energy } & E & =\gamma m c^{2} \\
+
+\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Consider two objects with equal heat capacities $C$ and initial temperatures $T_{1}$ and $T_{2}$. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine."	[]	Text-only	Competition	True		Expression		Open-ended	Thermodynamics	Physics	English
+197	b. Find the minimum temperature $T_{L}$ to which the first object can be lowered.	['By the Second Law of Thermodynamics, we must have $T_{L}=100 \\mathrm{~K}$. Otherwise, we would have a process whose sole effect was a net transfer of heat from a cold body to a warm one.']	['$100$']	"$g=9.8 \mathrm{~N} / \mathrm{kg}$
+
+
+
+$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$
+
+
+
+$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$
+
+
+
+$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$
+
+
+
+$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$
+
+
+
+$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$
+
+
+
+$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$
+
+
+
+$$
+
+\begin{aligned}
+
+& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+& e=1.602 \times 10^{-19} \mathrm{C} \\
+
+& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+& \ln (1+x) \approx x \text { for }|x| \ll 1 \\
+
+& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{aligned}
+
+$$
+
+$$
+
+\begin{array}{lrl}
+
+\text { velocity parameter } & \beta & =\frac{v}{c} \\
+
+\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\
+
+\text { relativistic momentum } & p & =\gamma \beta m c \\
+
+\text { relativistic energy } & E & =\gamma m c^{2} \\
+
+\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Consider two objects with equal heat capacities $C$ and initial temperatures $T_{1}$ and $T_{2}$. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine.
+Context question:
+a. Find the final temperature $T_{f}$ of the two objects, and the total work $W$ done by the engine.
+Context answer:
+\boxed{$T_{f}=\sqrt{T_{1} T_{2}}$, $W=C\left[T_{1}+T_{2}-2 \sqrt{T_{1} T_{2}}\right]$}
+
+
+Extra Supplementary Reading Materials:
+
+Now consider three objects with equal and constant heat capacity at initial temperatures $T_{1}=100 \mathrm{~K}, T_{2}=300 \mathrm{~K}$, and $T_{3}=300 \mathrm{~K}$. Suppose we wish to raise the temperature of the third object.
+
+
+
+To do this, we could run a Carnot engine between the first and second objects, extracting work $W$. This work can then be dissipated as heat to raise the temperature of the third object. Even better, it can be stored and used to run a Carnot engine between the first and third object in reverse, which pumps heat into the third object.
+
+
+
+Assume that all work produced by running engines can be stored and used without dissipation."	[]	Text-only	Competition	False	K	Numerical	1e0	Open-ended	Thermodynamics	Physics	English
+198	c. Find the maximum temperature $T_{H}$ to which the third object can be raised.	['The entropy of an object with constant heat capacity is\n\n\n\n$$\n\nS=\\int \\frac{d Q}{T}=C \\int \\frac{d T}{T}=C \\ln T\n\n$$\n\n\n\n\n\n\n\nSince the total entropy remains constant, $T_{1} T_{2} T_{3}$ is constant; this is a direct generalization of the result for $T_{f}$ found in part (a). Energy is also conserved, as it makes no sense to leave stored energy unused, so $T_{1}+T_{2}+T_{3}$ is constant.\n\n\n\nWhen one object is at temperature $T_{H}$, the other two must be at the same lower temperature $T_{0}$, or else further work could be extracted from their temperature difference, so\n\n\n\n$$\n\nT_{1}+T_{2}+T_{3}=T_{H}+2 T_{0}, \\quad T_{1} T_{2} T_{3}=T_{H} T_{0}^{2}\n\n$$\n\n\n\nPlugging in temperatures with values divided by 100 for convenience, and eliminating $T_{0}$ gives\n\n\n\n$$\n\nT_{H}\\left(7-T_{H}\\right)^{2}=36\n\n$$\n\n\n\nWe know that $T_{H}=1$ is one (spurious) solution, since this is the minimum possible final temperature as found in part (b). The other roots are $T_{H}=4$ and $T_{H}=9$ by the quadratic formula. The solution $T_{H}=9$ is impossible by energy conservation, so\n\n\n\n$$\n\nT_{H}=400 \\mathrm{~K}\n\n$$\n\n\n\nIt is also possible to solve the problem more explicitly. For example, one can run a Carnot cycle between the first two objects until they are at the same temperature, then run a Carnot cycle in reverse between the last two objects using the stored work. At this point, the first two objects will no longer be at the same temperature, so we can repeat the procedure; this yields an infinite series for $T_{H}$. Some students did this, and took only the first term of the series. This yields a fairly good approximation of $T_{H} \\approx 395 \\mathrm{~K}$.\n\n\n\nAnother explicit method is to continuously switch between running one Carnot engine forward and another Carnot engine in reverse; this yields three differential equations for $T_{1}, T_{2}$, and $T_{3}$. Solving the equations and setting $T_{1}=T_{2}$ yields $T_{3}=T_{H}$.']	['395']	"$g=9.8 \mathrm{~N} / \mathrm{kg}$
+
+
+
+$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$
+
+
+
+$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$
+
+
+
+$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$
+
+
+
+$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$
+
+
+
+$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$
+
+
+
+$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$
+
+
+
+$$
+
+\begin{aligned}
+
+& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+& e=1.602 \times 10^{-19} \mathrm{C} \\
+
+& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+& \ln (1+x) \approx x \text { for }|x| \ll 1 \\
+
+& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{aligned}
+
+$$
+
+$$
+
+\begin{array}{lrl}
+
+\text { velocity parameter } & \beta & =\frac{v}{c} \\
+
+\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\
+
+\text { relativistic momentum } & p & =\gamma \beta m c \\
+
+\text { relativistic energy } & E & =\gamma m c^{2} \\
+
+\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Consider two objects with equal heat capacities $C$ and initial temperatures $T_{1}$ and $T_{2}$. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine.
+Context question:
+a. Find the final temperature $T_{f}$ of the two objects, and the total work $W$ done by the engine.
+Context answer:
+\boxed{$T_{f}=\sqrt{T_{1} T_{2}}$, $W=C\left[T_{1}+T_{2}-2 \sqrt{T_{1} T_{2}}\right]$}
+
+
+Extra Supplementary Reading Materials:
+
+Now consider three objects with equal and constant heat capacity at initial temperatures $T_{1}=100 \mathrm{~K}, T_{2}=300 \mathrm{~K}$, and $T_{3}=300 \mathrm{~K}$. Suppose we wish to raise the temperature of the third object.
+
+
+
+To do this, we could run a Carnot engine between the first and second objects, extracting work $W$. This work can then be dissipated as heat to raise the temperature of the third object. Even better, it can be stored and used to run a Carnot engine between the first and third object in reverse, which pumps heat into the third object.
+
+
+
+Assume that all work produced by running engines can be stored and used without dissipation.
+Context question:
+b. Find the minimum temperature $T_{L}$ to which the first object can be lowered.
+Context answer:
+\boxed{$100$}
+"	[]	Text-only	Competition	False	K	Numerical	5e0	Open-ended	Thermodynamics	Physics	English
+199	a. Derive an expression for the deviation of the compass, $\delta \theta$, from north as a function of $K_{b}$, $K_{s}$, and $\theta$.	"[""We add the fields to get the local field. The northward component is\n\n\n\n$$\n\nB_{\\text {north }}=B_{e}-B_{e} K_{b} \\cos \\theta \\cos \\theta-B_{e} K_{s} \\sin \\theta \\sin \\theta\n\n$$\n\n\n\nwhile the eastward component is\n\n\n\n$$\n\nB_{\\text {east }}=-B_{e} K_{b} \\sin \\theta \\cos \\theta+B_{e} K_{s} \\cos \\theta \\sin \\theta\n\n$$\n\n\n\nThe deviation is given by\n\n\n\n$$\n\n\\tan \\delta \\theta=\\left(K_{s}-K_{b}\\right) \\frac{\\sin \\theta \\cos \\theta}{1-K_{b} \\cos ^{2} \\theta-K_{s} \\sin ^{2} \\theta}\n\n$$\n\n\n\nThis form is particularly nice, because as we'll see below, $K_{b}$ and $K_{s}$ are small enough to ignore in the denominator.""]"	['$\\delta \\theta=\\arctan \\left(K_{s}-K_{b}\\right) \\frac{\\sin \\theta \\cos \\theta}{1-K_{b} \\cos ^{2} \\theta-K_{s} \\sin ^{2} \\theta}$']	"$g=9.8 \mathrm{~N} / \mathrm{kg}$
+
+
+
+$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$
+
+
+
+$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$
+
+
+
+$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$
+
+
+
+$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$
+
+
+
+$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$
+
+
+
+$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$
+
+
+
+$$
+
+\begin{aligned}
+
+& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+& e=1.602 \times 10^{-19} \mathrm{C} \\
+
+& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+& \ln (1+x) \approx x \text { for }|x| \ll 1 \\
+
+& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{aligned}
+
+$$
+
+$$
+
+\begin{array}{lrl}
+
+\text { velocity parameter } & \beta & =\frac{v}{c} \\
+
+\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\
+
+\text { relativistic momentum } & p & =\gamma \beta m c \\
+
+\text { relativistic energy } & E & =\gamma m c^{2} \\
+
+\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We want to examine the effect of the ship's field on the ship's compass, which will be located in the middle of the ship.
+
+
+Let the strength of the Earth's magnetic field near the ship be $B_{e}$, and the orientation of the field be horizontal, pointing directly toward true north.
+
+
+
+The Earth's magnetic field $B_{e}$ will magnetize the ship, which will then create a second magnetic field $B_{s}$ in the vicinity of the ship's compass given by
+
+
+
+$$
+
+\overrightarrow{\mathbf{B}}_{s}=B_{e}\left(-K_{b} \cos \theta \hat{\mathbf{b}}+K_{s} \sin \theta \hat{\mathbf{s}}\right)
+
+$$
+
+
+
+where $K_{b}$ and $K_{s}$ are positive constants, $\theta$ is the angle between the heading of the ship and magnetic north, measured clockwise, $\hat{\mathbf{b}}$ and $\hat{\mathbf{s}}$ are unit vectors pointing in the forward direction of the ship (bow) and directly right of the forward direction (starboard), respectively.
+
+
+
+Because of the ship's magnetic field, the ship's compass will no longer necessarily point North."	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+200	b. Assuming that $K_{b}$ and $K_{s}$ are both much smaller than one, at what heading(s) $\theta$ will the deviation $\delta \theta$ be largest?	"[""By inspection, $\\theta=45^{\\circ}$ will yield the largest deviation. It's also acceptable to list $45^{\\circ}, 135^{\\circ}$, $225^{\\circ}$, and $315^{\\circ}$.""]"	['$45$, $135$, $225$, $315$']	"$g=9.8 \mathrm{~N} / \mathrm{kg}$
+
+
+
+$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$
+
+
+
+$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$
+
+
+
+$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$
+
+
+
+$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$
+
+
+
+$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$
+
+
+
+$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$
+
+
+
+$$
+
+\begin{aligned}
+
+& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+& e=1.602 \times 10^{-19} \mathrm{C} \\
+
+& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+& \ln (1+x) \approx x \text { for }|x| \ll 1 \\
+
+& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{aligned}
+
+$$
+
+$$
+
+\begin{array}{lrl}
+
+\text { velocity parameter } & \beta & =\frac{v}{c} \\
+
+\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\
+
+\text { relativistic momentum } & p & =\gamma \beta m c \\
+
+\text { relativistic energy } & E & =\gamma m c^{2} \\
+
+\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We want to examine the effect of the ship's field on the ship's compass, which will be located in the middle of the ship.
+
+
+Let the strength of the Earth's magnetic field near the ship be $B_{e}$, and the orientation of the field be horizontal, pointing directly toward true north.
+
+
+
+The Earth's magnetic field $B_{e}$ will magnetize the ship, which will then create a second magnetic field $B_{s}$ in the vicinity of the ship's compass given by
+
+
+
+$$
+
+\overrightarrow{\mathbf{B}}_{s}=B_{e}\left(-K_{b} \cos \theta \hat{\mathbf{b}}+K_{s} \sin \theta \hat{\mathbf{s}}\right)
+
+$$
+
+
+
+where $K_{b}$ and $K_{s}$ are positive constants, $\theta$ is the angle between the heading of the ship and magnetic north, measured clockwise, $\hat{\mathbf{b}}$ and $\hat{\mathbf{s}}$ are unit vectors pointing in the forward direction of the ship (bow) and directly right of the forward direction (starboard), respectively.
+
+
+
+Because of the ship's magnetic field, the ship's compass will no longer necessarily point North.
+Context question:
+a. Derive an expression for the deviation of the compass, $\delta \theta$, from north as a function of $K_{b}$, $K_{s}$, and $\theta$.
+Context answer:
+\boxed{$\delta \theta=\arctan \left(K_{s}-K_{b}\right) \frac{\sin \theta \cos \theta}{1-K_{b} \cos ^{2} \theta-K_{s} \sin ^{2} \theta}$}
+"	[]	Text-only	Competition	True	$^{\circ}$	Numerical	0	Open-ended	Electromagnetism	Physics	English
+201	a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small.	"[""We can solve the problem exactly, approximating only in the last step. This certainly isn't necessary; we do this to illustrate a useful technique. We set $c=1$ throughout, and transform to an inertial frame where the proton is initially at rest. Before the collision,\n\n\n\n$$\n\nE_{p}=m_{p}, \\quad E_{\\gamma}=\\left|p_{\\gamma}\\right|\n\n$$\n\n\n\nFor the $\\Delta$ particle, we have the usual relativistic relation\n\n\n\n$$\n\nE_{\\Delta}^{2}=p_{\\Delta}^{2}+m_{\\Delta}^{2}\n\n$$\n\n\n\nBy energy-momentum conservation,\n\n\n\n$$\n\nE_{p}+E_{\\gamma}=E_{\\Delta}, \\quad p_{\\gamma}=p_{\\Delta} .\n\n$$\n\n\n\nCombining these results gives\n\n\n\n$$\n\n\\left(m_{p}+E_{\\gamma}\\right)^{2}=E_{\\gamma}^{2}+m_{\\Delta}^{2} \\Rightarrow E_{\\gamma}=\\frac{m_{\\Delta}^{2}-m_{p}^{2}}{2 m_{p}}\n\n$$\n\n\n\nNow we need to transform back to the original frame, where the energy of the photon is $E_{b}$. We can use the Lorentz transformation for this, but it's a little easier to realize that $E=h f$ for photons, and apply the Doppler shift. Then\n\n\n\n$$\n\n\\alpha \\equiv \\frac{E_{b}}{E_{\\gamma}}=\\sqrt{\\frac{1-\\beta}{1+\\beta}}\n\n$$\n\n\n\nwhere $\\beta$ is the velocity parameter of the proton in the inertial frame where the photon has energy $E_{b}$. Solving for $\\beta$ in terms of the energy ratio $\\alpha$,\n\n\n\n$$\n\n\\beta=\\frac{1-\\alpha^{2}}{1+\\alpha^{2}}\n\n$$\n\n\n\nTo calculate the proton energy, we need the Lorentz factor,\n\n\n\n$$\n\n\\gamma=\\frac{1}{\\sqrt{1-\\beta^{2}}}=\\frac{1+\\alpha^{2}}{2 \\alpha}\n\n$$\n\n\n\n\n\n\n\nThen the proton energy in the original frame is\n\n\n\n$$\n\nE_{p}=\\gamma m_{p}=\\frac{m_{p}}{2}\\left(\\alpha+\\frac{1}{\\alpha}\\right)=\\frac{m_{p}}{2}\\left(\\frac{2 m_{p} E_{b}}{m_{\\Delta}^{2}-m_{p}^{2}}+\\frac{m_{\\Delta}^{2}-m_{p}^{2}}{2 m_{p} E_{b}}\\right)\n\n$$\n\n\n\nwhich is the exact answer.\n\n\n\nAt this point, we can approximate. The second term is much larger than the first, so\n\n\n\n$$\n\nE_{p} \\approx \\frac{m_{\\Delta}^{2}-m_{p}^{2}}{4 E_{b}}\n\n$$\n\n\n\nwhich was the required answer for this problem.""
+ ""We now show another method that approximates throughout. We'll do everything in the lab frame, so the symbols in this solution don't mean the same things they did in solution 1. In the lab frame, energy-momentum conservation gives\n\n\n\n$$\n\np_{p}-p_{b}=p_{\\Delta}, \\quad E_{p}+E_{b}=E_{\\Delta}\n\n$$\n\n\n\nSquaring both expressions and dropping $E_{b}^{2}$ terms, since $E_{b}$ is small,\n\n\n\n$$\n\np_{p}^{2}-2 p_{p} p_{b} \\approx p_{\\Delta}^{2}, \\quad E_{p}^{2}+2 E_{p} E_{b} \\approx E_{\\Delta}^{2}\n\n$$\n\n\n\nSubtracting these equations gives\n\n\n\n$$\n\nm_{p}^{2}+2 E_{p} E_{b}+2 p_{p} E_{b}=m_{\\Delta}^{2} \\Rightarrow E_{p}+p_{p}=\\frac{m_{\\Delta}^{2}-m_{p}^{2}}{2 E_{b}}\n\n$$\n\n\n\nSince $E_{b}$ is small, this quantity must be large. But this means the protons are ultrarelativistic, so $E_{p} \\approx p_{p}$, giving\n\n\n\n$$\n\nE_{p} \\approx \\frac{m_{\\Delta}^{2}-m_{p}^{2}}{4 E_{b}}\n\n$$\n\n\n\nas desired.""]"	['$\\frac{m_{\\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$']	"$g=9.8 \mathrm{~N} / \mathrm{kg}$
+
+
+
+$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$
+
+
+
+$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$
+
+
+
+$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$
+
+
+
+$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$
+
+
+
+$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$
+
+
+
+$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$
+
+
+
+$$
+
+\begin{aligned}
+
+& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+& e=1.602 \times 10^{-19} \mathrm{C} \\
+
+& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+& \ln (1+x) \approx x \text { for }|x| \ll 1 \\
+
+& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{aligned}
+
+$$
+
+$$
+
+\begin{array}{lrl}
+
+\text { velocity parameter } & \beta & =\frac{v}{c} \\
+
+\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\
+
+\text { relativistic momentum } & p & =\gamma \beta m c \\
+
+\text { relativistic energy } & E & =\gamma m c^{2} \\
+
+\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Relativistic particles obey the mass energy relation
+
+
+
+$$
+
+E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}
+
+$$
+
+
+
+where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light.
+
+
+
+A proton with mass $m_{p}$ and energy $E_{p}$ collides head on with a photon which is massless and has energy $E_{b}$. The two combine and form a new particle with mass $m_{\Delta}$ called $\Delta$, or ""delta"". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+202	b. In this case, the photon energy $E_{b}$ is that of the cosmic background radiation, which is an EM wave with wavelength $1.06 \mathrm{~mm}$. Determine the energy of the photons, writing your answer in electron volts.	['Plugging in the numbers gives\n\n\n\n$$\n\nE=\\frac{h c}{\\lambda}=0.00112 \\mathrm{eV}\n\n$$']	['$0.00112$']	"$g=9.8 \mathrm{~N} / \mathrm{kg}$
+
+
+
+$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$
+
+
+
+$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$
+
+
+
+$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$
+
+
+
+$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$
+
+
+
+$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$
+
+
+
+$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$
+
+
+
+$$
+
+\begin{aligned}
+
+& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+& e=1.602 \times 10^{-19} \mathrm{C} \\
+
+& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+& \ln (1+x) \approx x \text { for }|x| \ll 1 \\
+
+& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{aligned}
+
+$$
+
+$$
+
+\begin{array}{lrl}
+
+\text { velocity parameter } & \beta & =\frac{v}{c} \\
+
+\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\
+
+\text { relativistic momentum } & p & =\gamma \beta m c \\
+
+\text { relativistic energy } & E & =\gamma m c^{2} \\
+
+\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Relativistic particles obey the mass energy relation
+
+
+
+$$
+
+E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}
+
+$$
+
+
+
+where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light.
+
+
+
+A proton with mass $m_{p}$ and energy $E_{p}$ collides head on with a photon which is massless and has energy $E_{b}$. The two combine and form a new particle with mass $m_{\Delta}$ called $\Delta$, or ""delta"". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum.
+Context question:
+a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small.
+Context answer:
+\boxed{$\frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$}
+"	[]	Text-only	Competition	False	eV	Numerical	2e-5	Open-ended	Modern Physics	Physics	English
+203	c. Assuming this value for $E_{b}$, what is the energy of the proton, in electron volts, that will allow the above reaction? This sets an upper limit on the energy of cosmic rays. The mass of the proton is given by $m_{p} c^{2}=938 \mathrm{MeV}$ and the mass of the $\Delta$ is given by $m_{\Delta} c^{2}=1232 \mathrm{MeV}$.	['Restoring the factors of $c$ and plugging in the numbers gives\n\n\n\n$$\n\nE_{p}=1.4 \\times 10^{20} \\mathrm{eV}\n\n$$\n\n\n\nThis is known as the GZK bound for cosmic rays.']	['$1.4 \\times 10^{20}$']	"$g=9.8 \mathrm{~N} / \mathrm{kg}$
+
+
+
+$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$
+
+
+
+$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$
+
+
+
+$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$
+
+
+
+$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$
+
+
+
+$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$
+
+
+
+$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$
+
+
+
+$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$
+
+
+
+$$
+
+\begin{aligned}
+
+& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+& e=1.602 \times 10^{-19} \mathrm{C} \\
+
+& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+& \ln (1+x) \approx x \text { for }|x| \ll 1 \\
+
+& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{aligned}
+
+$$
+
+$$
+
+\begin{array}{lrl}
+
+\text { velocity parameter } & \beta & =\frac{v}{c} \\
+
+\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\
+
+\text { relativistic momentum } & p & =\gamma \beta m c \\
+
+\text { relativistic energy } & E & =\gamma m c^{2} \\
+
+\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Relativistic particles obey the mass energy relation
+
+
+
+$$
+
+E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}
+
+$$
+
+
+
+where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light.
+
+
+
+A proton with mass $m_{p}$ and energy $E_{p}$ collides head on with a photon which is massless and has energy $E_{b}$. The two combine and form a new particle with mass $m_{\Delta}$ called $\Delta$, or ""delta"". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum.
+Context question:
+a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small.
+Context answer:
+\boxed{$\frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$}
+
+
+Context question:
+b. In this case, the photon energy $E_{b}$ is that of the cosmic background radiation, which is an EM wave with wavelength $1.06 \mathrm{~mm}$. Determine the energy of the photons, writing your answer in electron volts.
+Context answer:
+\boxed{$0.00112$}
+"	[]	Text-only	Competition	False	eV	Numerical	5e18	Open-ended	Modern Physics	Physics	English
+204	a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.)	"[""The intensity of solar radiation at the surface of the sun is $\\sigma T_{s}^{4}$, so the intensity at the planet's orbit radius is\n\n\n\n$$\n\nI=\\sigma T_{s}^{4} \\frac{R_{s}^{2}}{R_{0}^{2}}\n\n$$\n\n\n\nThe area subtended by the planet is $\\pi R^{2}$, so\n\n\n\n$$\n\nP=\\pi \\sigma T_{s}^{4} \\frac{R^{2} R_{s}^{2}}{R_{0}^{2}}\n\n$$""]"	['$P=\\pi \\sigma T_{s}^{4} \\frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Green Revolution ${ }^{1}$
+
+
+
+In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$, and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_{1}$, while the dark side has a constant uniform temperature $T_{2}$. The orbit radius of the planet is $R_{0}$, the Sun has temperature $T_{s}$, and the radius of the Sun is $R_{s}$. Assume that outer space has zero temperature, and treat all objects as ideal blackbodies."	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+205	c. Find the wind power $P_{w}$ in terms of $P$ and the temperature ratio $T_{2} / T_{1}$.	['Let heat be transferred from the bright side at a rate $Q_{1}$ and transferred to the dark side at a rate $Q_{2}$. Then by conservation of energy,\n\n\n\n$$\n\nQ_{1}=P_{w}+Q_{2}\n\n$$\n\n\n\nSince the two hemispheres have constant temperatures, energy balance for each gives\n\n\n\n$$\n\nP=Q_{1}+\\left(2 \\pi R^{2} \\sigma\\right) T_{1}^{4}, \\quad Q_{2}=A T_{2}^{4}\n\n$$\n\n\n\nFinally, since the engine is reversible,\n\n\n\n$$\n\n\\frac{Q_{1}}{T_{1}}=\\frac{Q_{2}}{T_{2}}\n\n$$\n\n\n\nBy combining the first three equations, and defining $x=T_{2} / T_{1}$, we have\n\n\n\n$$\n\nP_{w}=Q_{1}-Q_{2}=P-\\left(2 \\pi R^{2} \\sigma\\right)\\left(T_{1}^{4}+T_{2}^{4}\\right)=P-\\left(2 \\pi R^{2} \\sigma\\right) T_{1}^{4}\\left(1+x^{4}\\right) .\n\n$$\n\n\n\nThis is not yet in terms of $P$ and $x$, so now we use the reversibility condition,\n\n\n\n$$\n\n\\frac{P-A T_{1}^{4}}{T_{1}}=\\frac{A T_{2}^{4}}{T_{2}}\n\n$$\n\n\n\nwhich simplifies to\n\n\n\n$$\n\nP=\\left(2 \\pi R^{2} \\sigma\\right)\\left(T_{1}^{4}+T_{1} T_{2}^{3}\\right)=\\left(2 \\pi R^{2} \\sigma\\right) T_{1}^{4}\\left(1+x^{3}\\right)\n\n$$\n\n\n\nPlugging this in above, we find\n\n\n\n$$\n\nP_{w}=P-\\frac{P}{1+x^{3}}\\left(1+x^{4}\\right)=\\frac{x^{3}(1-x)}{1+x^{3}} P\n\n$$']	['$P_{w}=\\frac{x^{3}(1-x)}{1+x^{3}} P$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Green Revolution ${ }^{1}$
+
+
+
+In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$, and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_{1}$, while the dark side has a constant uniform temperature $T_{2}$. The orbit radius of the planet is $R_{0}$, the Sun has temperature $T_{s}$, and the radius of the Sun is $R_{s}$. Assume that outer space has zero temperature, and treat all objects as ideal blackbodies.
+Context question:
+a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.)
+Context answer:
+\boxed{$P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+In order to keep both $T_{1}$ and $T_{2}$ constant, heat must be continually transferred from the bright side to the dark side. By viewing the two hemispheres as the two reservoirs of a reversible heat engine, work can be performed from this temperature difference, which appears in the form of wind power. For simplicity, we assume all of this power is immediately captured and stored by windmills.
+Context question:
+b. The equilibrium temperature ratio $T_{2} / T_{1}$ depends on the heat transfer rate between the hemispheres. Find the minimum and maximum possible values of $T_{2} / T_{1}$. In each case, what is the wind power $P_{w}$ produced?
+Context answer:
+The minimum value is simply zero, $P_{w}=0$.
+The maximum value is $T_{2} / T_{1}=1$. $P_{w}=0$.
+"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+206	d. Estimate the maximum possible value of $P_{w}$ as a fraction of $P$, to one significant figure. Briefly explain how you obtained this estimate.	['There are many ways to get the required answer. For example, by sketching the function, one can see that there is a unique maximum at an intermediate value of $x$, and furthermore that this maximum is at $x>0.5$, because of the rapid rise of the $x^{3}$ factor. One could then compute $P_{w} / P$ with a calculator at a few trial values such as $x=0.5,0.7,0.9$, which are already enough to get the desired accuracy.\n\n\n\nThe optimum value is $x=0.69$, at which point\n\n\n\n$$\n\nP_{w}^{\\max }=0.077 P\n\n$$\n\n\n\n\n\n\n\nHence in this model, at most $7.7 \\%$ of solar energy can be converted into wind energy. Any answer within $15 \\%$ of this value was accepted.\n\n']	['$P_{w}^{\\max }=0.077 P$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Green Revolution ${ }^{1}$
+
+
+
+In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$, and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_{1}$, while the dark side has a constant uniform temperature $T_{2}$. The orbit radius of the planet is $R_{0}$, the Sun has temperature $T_{s}$, and the radius of the Sun is $R_{s}$. Assume that outer space has zero temperature, and treat all objects as ideal blackbodies.
+Context question:
+a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.)
+Context answer:
+\boxed{$P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$}
+
+
+Extra Supplementary Reading Materials:
+
+In order to keep both $T_{1}$ and $T_{2}$ constant, heat must be continually transferred from the bright side to the dark side. By viewing the two hemispheres as the two reservoirs of a reversible heat engine, work can be performed from this temperature difference, which appears in the form of wind power. For simplicity, we assume all of this power is immediately captured and stored by windmills.
+Context question:
+b. The equilibrium temperature ratio $T_{2} / T_{1}$ depends on the heat transfer rate between the hemispheres. Find the minimum and maximum possible values of $T_{2} / T_{1}$. In each case, what is the wind power $P_{w}$ produced?
+Context answer:
+The minimum value is simply zero, $P_{w}=0$.
+The maximum value is $T_{2} / T_{1}=1$. $P_{w}=0$.
+
+
+Context question:
+c. Find the wind power $P_{w}$ in terms of $P$ and the temperature ratio $T_{2} / T_{1}$.
+Context answer:
+\boxed{$P_{w}=\frac{x^{3}(1-x)}{1+x^{3}} P$}"	[]	Text-only	Competition	False		Expression		Open-ended	Thermodynamics	Physics	English
+207	"a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that
+
+
+
+$$
+
+v=\mu E
+
+$$
+
+
+
+where $E$ is the local electric field and $\mu$ is the charge mobility.
+
+i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives."	['In the steady state, the current is the same everywhere. Consider the region $(x, x+d x)$. The time it takes for the charge in the second region to leave is $\\frac{\\mathrm{d} x}{v(x)}$. The amount of charge that leaves is $\\rho A \\mathrm{~d} x$. The current is thus given by $\\rho A v$, so $\\rho v$ is constant. Alternatively, one can write this as\n\n\n\n$$\n\nv \\frac{\\mathrm{d} \\rho}{\\mathrm{d} x}+\\rho \\frac{\\mathrm{d} v}{\\mathrm{~d} x}=0\n\n$$\n\n\n\nBoth forms were accepted.\n\n']	['$v \\frac{\\mathrm{d} \\rho}{\\mathrm{d} x}+\\rho \\frac{\\mathrm{d} v}{\\mathrm{~d} x}=0$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Electric Slide
+
+
+
+Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium."	[]	Text-only	Competition	False		Equation		Open-ended	Electromagnetism	Physics	English
+208	iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.	"[""We have that\n\n\n\n$$\n\n\\rho \\frac{d v}{d x}+v \\frac{d \\rho}{d x}=0\n\n$$\n\n\n\nand now that $v=-\\mu \\frac{d V}{d x}$, so substituting in Poisson's equation gives us that\n\n\n\n$$\n\n\\left(\\frac{d^{2} V}{d x^{2}}\\right)^{2}+\\frac{d V}{d x}\\left(\\frac{d^{3} V}{d x^{3}}\\right)=0\n\n$$\n\n\n\nUsing $V(x)=-V_{0}(x / d)^{b}$ gives\n\n\n\n$$\n\nb(b-1) b(b-1)=-b b(b-1)(b-2)\n\n$$\n\n\n\nThe solution with $b=0$ cannot satisfy the boundary conditions, while $b=1$ has zero current. Assuming $b$ is neither of these values, we have $b-1=-(b-2)$, so $b=3 / 2$. Substituting gives\n\n\n\n$$\n\nv=-\\frac{3 V_{0} \\mu x^{1 / 2}}{2 d^{3 / 2}}\n\n$$\n\n\n\nand\n\n\n\n$$\n\n\\rho=-\\frac{3 V_{0} \\epsilon_{0}}{4 d^{3 / 2} x^{1 / 2}}\n\n$$\n\n\n\nso\n\n\n\n$$\n\nI=\\rho A v=\\frac{9 \\epsilon_{0} \\mu A V_{0}^{2}}{8 d^{3}}\n\n$$\n\n\n\nwith the current flowing from left to right.""]"	['$I=\\frac{9 \\epsilon_{0} \\mu A V_{0}^{2}}{8 d^{3}}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Electric Slide
+
+
+
+Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.
+Context question:
+a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that
+
+
+
+$$
+
+v=\mu E
+
+$$
+
+
+
+where $E$ is the local electric field and $\mu$ is the charge mobility.
+
+i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives.
+Context answer:
+$v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$
+
+
+Extra Supplementary Reading Materials:
+
+a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that
+
+
+
+$$
+
+v=\mu E
+
+$$
+
+
+
+where $E$ is the local electric field and $\mu$ is the charge mobility.
+Context question:
+ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.)
+Context answer:
+$$
+
+\frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}}
+
+$$
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+209	i. Assume that in the steady state, conditions have been established so that a nonzero, steady current flows, and the electric potential again satisfies $V(x) \propto x^{b^{\prime}}$, where $b^{\prime}$ is another exponent you must find. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.	"[""We again have that $V(x)=V_{0}(x / d)^{b}$. Note that from Poisson's equation, $\\frac{d \\rho}{d x}=-\\epsilon_{0} \\frac{d^{3} V}{d x^{3}}$, so we need $b=3$ for this expression to be constant. Therefore,\n\n\n\n$$\n\nI=\\frac{6 \\mu k_{B} T A \\epsilon_{0} V_{0}}{q d^{3}}\n\n$$""]"	['$I=\\frac{6 \\mu k_{B} T A \\epsilon_{0} V_{0}}{q d^{3}}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Electric Slide
+
+
+
+Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.
+Context question:
+a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that
+
+
+
+$$
+
+v=\mu E
+
+$$
+
+
+
+where $E$ is the local electric field and $\mu$ is the charge mobility.
+
+i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives.
+Context answer:
+$v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$
+
+
+Extra Supplementary Reading Materials:
+
+a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that
+
+
+
+$$
+
+v=\mu E
+
+$$
+
+
+
+where $E$ is the local electric field and $\mu$ is the charge mobility.
+Context question:
+ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.)
+Context answer:
+$$
+
+\frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}}
+
+$$
+
+
+Context question:
+iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.
+Context answer:
+\boxed{$I=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}$}
+
+
+Extra Supplementary Reading Materials:
+
+b. For small $V_{0}$, the positive charges move by diffusion. The current due to diffusion is given by Fick's Law,
+
+
+
+$$
+
+I=-A D \frac{\mathrm{d} \rho}{\mathrm{d} x}
+
+$$
+
+
+
+Here, $D$ is the diffusion constant, which you can assume to be described by the Einstein relation
+
+
+
+$$
+
+D=\frac{\mu k_{B} T}{q}
+
+$$
+
+
+
+where $T$ is the temperature of the system."	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+210	ii. At roughly what voltage $V_{0}$ does the system transition from this regime to the high voltage regime of the previous part?	['We find the crossover voltage by equating our answers in previous parts, to get\n\n\n\n$$\n\n\\frac{6 \\mu k_{B} T A \\epsilon_{0} V_{0}}{q d^{3}}=\\frac{9 \\epsilon_{0} \\mu A V_{0}^{2}}{8 d^{3}}\n\n$$\n\n\n\nor\n\n\n\n$$\n\nV_{0}=\\frac{16 k_{B} T}{3 q}\n\n$$']	['$V_{0}=\\frac{16 k_{B} T}{3 q}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Electric Slide
+
+
+
+Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.
+Context question:
+a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that
+
+
+
+$$
+
+v=\mu E
+
+$$
+
+
+
+where $E$ is the local electric field and $\mu$ is the charge mobility.
+
+i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives.
+Context answer:
+$v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$
+
+
+Extra Supplementary Reading Materials:
+
+a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that
+
+
+
+$$
+
+v=\mu E
+
+$$
+
+
+
+where $E$ is the local electric field and $\mu$ is the charge mobility.
+Context question:
+ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.)
+Context answer:
+$$
+
+\frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}}
+
+$$
+
+
+Context question:
+iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.
+Context answer:
+\boxed{$I=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}$}
+
+
+Extra Supplementary Reading Materials:
+
+b. For small $V_{0}$, the positive charges move by diffusion. The current due to diffusion is given by Fick's Law,
+
+
+
+$$
+
+I=-A D \frac{\mathrm{d} \rho}{\mathrm{d} x}
+
+$$
+
+
+
+Here, $D$ is the diffusion constant, which you can assume to be described by the Einstein relation
+
+
+
+$$
+
+D=\frac{\mu k_{B} T}{q}
+
+$$
+
+
+
+where $T$ is the temperature of the system.
+Context question:
+i. Assume that in the steady state, conditions have been established so that a nonzero, steady current flows, and the electric potential again satisfies $V(x) \propto x^{b^{\prime}}$, where $b^{\prime}$ is another exponent you must find. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.
+Context answer:
+\boxed{$I=\frac{6 \mu k_{B} T A \epsilon_{0} V_{0}}{q d^{3}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+211	"a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.
+
+
+i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?"	"[""During the collision, the block receives impulses from the normal force, friction, and gravity. Since the collision is very short, the impulse due to gravity is negligible. Let $p_{N}$ and $p_{F}$ be the magnitudes of the impulses from the normal force and friction force.\n\n\n\nSince the block stays on the ramp after the collision, its final momentum is parallel to the ramp. Then the normal force must completely eliminate the block's initial momentum perpendicular to the ramp, so\n\n\n\n$$\n\np_{N}=m v \\cos \\theta\n\n$$\n\n\n\nSince the block still moves after the collision,\n\n\n\n$$\n\np_{F}=\\mu p_{N}\n\n$$\n\n\n\nThe block's initial momentum parallel to the ramp is $m v \\sin \\theta$, so\n\n\n\n$$\n\nm v \\sin \\theta-p_{F}=m u\n\n$$\n\n\n\nwhere $u$ is the final speed of the block. Solving for $u$ gives\n\n\n\n$$\n\nu=v(\\sin \\theta-\\mu \\cos \\theta)\n\n$$""]"	['$u=v(\\sin \\theta-\\mu \\cos \\theta)$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+212	ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?	['We set $u=0$ to obtain\n\n\n\n$$\n\n\\mu=\\tan \\theta\n\n$$\n\n\n\nNote that this is simply the no-slip condition for a block resting on an inclined plane! This is because in both cases, equality is achieved when the normal force and maximal friction force sum to a purely vertical force.']	['$\\mu=\\tan \\theta$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Context question:
+a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.
+
+
+i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?
+Context answer:
+\boxed{$u=v(\sin \theta-\mu \cos \theta)$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+213	"b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$.
+
+
+i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case?"	"[""If the sphere immediately begins to roll without slipping, we can calculate the frictional impulse independently of the normal impulse. We have\n\n\n\n$$\n\nm v \\sin \\theta-p_{F}=m u\n\n$$\n\n\n\nThe frictional impulse is responsible for the sphere's rotation, so its angular momentum about its center of mass is $L=p_{F} R$. But we also know that\n\n\n\n$$\n\nL=\\beta m R^{2} \\omega=\\beta m R u\n\n$$\n\n\n\nThen\n\n\n\n$$\n\np_{F}=\\beta m u \\text {. }\n\n$$\n\n\n\nSubstituting into the previous expression gives\n\n\n\n$$\n\nm v \\sin \\theta=(1+\\beta) m u \\Rightarrow u=\\frac{v \\sin \\theta}{1+\\beta}\n\n$$""]"	['$u=\\frac{v \\sin \\theta}{1+\\beta}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Context question:
+a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.
+
+
+i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?
+Context answer:
+\boxed{$u=v(\sin \theta-\mu \cos \theta)$}
+
+
+Context question:
+ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?
+Context answer:
+\boxed{$\mu=\tan \theta$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+214	ii. What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision?	['As in part (a), the normal impulse is $p_{N}=m v \\cos \\theta$ and the maximal frictional impulse is $p_{F}=\\mu p_{N}$. From the previous part, we need\n\n\n\n$$\n\np_{F}=\\frac{\\beta m v \\sin \\theta}{1+\\beta}\n\n$$\n\n\n\nand equating these expressions gives\n\n\n\n$$\n\n\\mu=\\frac{\\beta \\tan \\theta}{1+\\beta}\n\n$$']	['$\\mu=\\frac{\\beta \\tan \\theta}{1+\\beta}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Context question:
+a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.
+
+
+i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?
+Context answer:
+\boxed{$u=v(\sin \theta-\mu \cos \theta)$}
+
+
+Context question:
+ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?
+Context answer:
+\boxed{$\mu=\tan \theta$}
+
+
+Context question:
+b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$.
+
+
+i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case?
+Context answer:
+\boxed{$u=\frac{v \sin \theta}{1+\beta}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Mechanics	Physics	English
+215	a. What is the electric potential at a corner of the same cube? Write your answer in terms of $\rho, a, \epsilon_{0}$, and any necessary numerical constants.	['By dimensional analysis, the answer takes the form\n\n\n\n$$\n\n\\Phi_{c}(a, \\rho) \\approx \\frac{C \\rho a^{2}}{\\epsilon_{0}}\n\n$$\n\n\n\nfor a dimensionless constant $C$. Note that a cube of side length $a$ consists of 8 cubes of side length $a / 2$, each with a corner at the center of the larger cube. Then\n\n\n\n$$\n\n\\frac{0.1894 \\rho a^{2}}{\\epsilon_{0}}=8 \\frac{C \\rho(a / 2)^{2}}{\\epsilon_{0}}\n\n$$\n\n\n\nso $C=0.1894 / 2=0.0947$.\n\n']	['$\\Phi_{c}(a, \\rho) \\approx \\frac{C \\rho a^{2}}{\\epsilon_{0}}$, $C=0.0947$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+The electric potential at the center of a cube with uniform charge density $\rho$ and side length $a$ is
+
+
+
+$$
+
+\Phi \approx \frac{0.1894 \rho a^{2}}{\epsilon_{0}}
+
+$$
+
+
+
+You do not need to derive this. ${ }^{1}$
+
+
+
+For the entirety of this problem, any computed numerical constants should be written to three significant figures."	[]	Text-only	Competition	False		Equation,Numerical	,1e-3	Open-ended	Electromagnetism	Physics	English
+216	a. Find the standard deviation $\Delta r$ of the distribution for $r$, the distance from the center of the telescope mirror to the point of reflection of the photon.	['The square of the standard deviation is the variance, so\n\n\n\n$$\n\n(\\Delta r)^{2}=\\left\\langle r^{2}\\right\\rangle-\\langle r\\rangle^{2}\n\n$$\n\n\n\nComputing the average value of $r^{2}$ is mathematically the exact same thing as computing the moment of inertia of a uniform disk; we have\n\n\n\n$$\n\n\\left\\langle r^{2}\\right\\rangle=\\frac{1}{\\pi R^{2}} \\int_{0}^{R} r^{2}(2 \\pi r d r)=\\frac{1}{2} R^{2}\n\n$$\n\n\n\nSimilarly, the average value of $r$ is\n\n\n\n$$\n\n\\langle r\\rangle=\\frac{1}{\\pi R^{2}} \\int_{0}^{R} r(2 \\pi r d r)=\\frac{2}{3} R\n\n$$\n\n\n\nThen we have\n\n\n\n$$\n\n\\Delta r=\\frac{R}{\\sqrt{18}}\n\n$$']	['$\\Delta r=\\frac{R}{\\sqrt{18}}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+In this problem, use a particle-like model of photons: they propagate in straight lines and obey the law of reflection, but are subject to the quantum uncertainty principle. You may use small-angle approximations throughout the problem.
+
+
+
+A photon with wavelength $\lambda$ has traveled from a distant star to a telescope mirror, which has a circular cross-section with radius $R$ and a focal length $f \gg R$. The path of the photon is nearly aligned to the axis of the mirror, but has some slight uncertainty $\Delta \theta$. The photon reflects off the mirror and travels to a detector, where it is absorbed by a particular pixel on a charge-coupled device (CCD).
+
+
+
+Suppose the telescope mirror is manufactured so that photons coming in parallel to each other are focused to the same pixel on the CCD, regardless of where they hit the mirror. Then all small cross-sectional areas of the mirror are equally likely to include the point of reflection for a photon."	[]	Text-only	Competition	False		Expression		Open-ended	Optics	Physics	English
+217	i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.	['We will assume that light from the sun is completely absorbed, and then re-radiated as blackbody isotropically. In that case,\n\n\n\n$$\n\nP=\\frac{I}{c}=\\frac{L \\odot}{4 \\pi R^{2} c}\n\n$$\n\n\n\nThe relationship between the pressure $P$ and the energy density $I$ can be derived from the equation $p=E / c$ for photons, or simply postulated by dimensional analysis, since there is no other relevant speed.\n\n\n\nAlternatively, this relation can be derived using classical electromagnetism, though this was not required. For a plane wave, the momentum density (i.e. pressure) of the electromagnetic field is $P=\\epsilon_{0}|\\mathbf{E} \\times \\mathbf{B}|$, while the energy density is $I=|\\mathbf{S}| / c$ where $\\mathbf{S}=\\mathbf{E} \\times \\mathbf{B} / \\mu_{0}$. Then $P=I / c$ since $c^{2}=1 / \\epsilon_{0} \\mu_{0}$.']	['$P=\\frac{L \\odot}{4 \\pi R^{2} c}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.
+
+
+a. The force of radiation on a spherical particle of radius $r$ is given by
+
+
+
+$$
+
+F=P Q \pi r^{2}
+
+$$
+
+
+
+where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+218	ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.	['We have\n\n\n\n$$\n\nF_{\\text {gravity }}=\\frac{G M_{\\odot}}{R^{2}} \\frac{4}{3} \\pi \\rho r^{3}, \\quad F_{\\text {radiation }}=\\frac{L_{\\odot}}{4 \\pi R^{2} c} Q \\pi r^{2}=\\frac{L_{\\odot}}{4 R^{2} c} Q r^{2}\n\n$$\n\n\n\nwhich gives\n\n\n\n$$\n\n\\frac{F_{\\text {radiation }}}{F_{\\text {gravity }}}=\\frac{3 L_{\\odot}}{16 \\pi G c M_{\\odot} \\rho} \\frac{Q}{r}\n\n$$']	['$\\frac{F_{\\text {radiation }}}{F_{\\text {gravity }}}=\\frac{3 L_{\\odot}}{16 \\pi G c M_{\\odot} \\rho} \\frac{Q}{r}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.
+
+
+a. The force of radiation on a spherical particle of radius $r$ is given by
+
+
+
+$$
+
+F=P Q \pi r^{2}
+
+$$
+
+
+
+where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables.
+Context question:
+i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.
+Context answer:
+\boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+219	i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants.	['Using the circular motion equation\n\n\n\n$$\n\n\\frac{G M_{\\odot}}{R^{2}}=\\frac{v^{2}}{R}\n\n$$\n\n\n\nwe have\n\n\n\n$$\n\nv=\\sqrt{\\frac{G M_{\\odot}}{R}}\n\n$$']	['$v=\\sqrt{\\frac{G M_{\\odot}}{R}}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.
+
+
+a. The force of radiation on a spherical particle of radius $r$ is given by
+
+
+
+$$
+
+F=P Q \pi r^{2}
+
+$$
+
+
+
+where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables.
+Context question:
+i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.
+Context answer:
+\boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$}
+
+
+Context question:
+ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.
+Context answer:
+\boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$}
+
+
+Context question:
+iii. The quality factor is given by one of the following
+
+
+
+- If $r \ll \lambda, Q \sim(r / \lambda)^{2}$
+
+- If $r \sim \lambda, Q \sim 1$.
+
+- If $r \gg \lambda, Q=1$
+
+
+
+Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure?
+Context answer:
+particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough.
+
+
+Extra Supplementary Reading Materials:
+
+b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+220	ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v \ll c$.	['Work in the reference frame of the particle. In this frame, the radiation hits the particle at an angle $\\theta=v / c$ from the radial direction. The particle then re-emits the radiation isotropically, contributing no additional radiation pressure. (We ignore relativistic effects because they occur at second order in $v / c$, while the effect we care about is first order.) The tangential component of the force is\n\n\n\nso\n\n\n\n$$\n\nF=\\frac{v}{c} \\frac{L \\odot}{4 \\pi R^{2} c} Q \\pi r^{2}=\\frac{v}{c} \\frac{L_{\\odot}}{4 R^{2} c} Q r^{2}\n\n$$\n\n\n\n$$\n\n\\tau=-\\frac{v}{c} \\frac{L \\odot}{4 R c} Q r^{2}\n\n$$\n\n\nwhere the negative sign is because this tends to decrease the angular momentum.\n\n\n\nThe problem can also be solved in the reference frame of the Sun. In this frame, the radiation hits the particle radially, but the particle does not re-emit the radiation isotropically since it is moving; instead the radiation is Doppler shifted. This eventually leads to the same result, after a much more complicated calculation.']	['$\\tau=-\\frac{v}{c} \\frac{L \\odot}{4 R c} Q r^{2}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.
+
+
+a. The force of radiation on a spherical particle of radius $r$ is given by
+
+
+
+$$
+
+F=P Q \pi r^{2}
+
+$$
+
+
+
+where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables.
+Context question:
+i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.
+Context answer:
+\boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$}
+
+
+Context question:
+ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.
+Context answer:
+\boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$}
+
+
+Context question:
+iii. The quality factor is given by one of the following
+
+
+
+- If $r \ll \lambda, Q \sim(r / \lambda)^{2}$
+
+- If $r \sim \lambda, Q \sim 1$.
+
+- If $r \gg \lambda, Q=1$
+
+
+
+Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure?
+Context answer:
+particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough.
+
+
+Extra Supplementary Reading Materials:
+
+b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system.
+Context question:
+i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants.
+Context answer:
+\boxed{$v=\sqrt{\frac{G M_{\odot}}{R}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+221	iii. Since $\tau=d L / d t$, the angular momentum $L$ of the particle changes with time. As such, develop a differential equation to find $d R / d t$, the rate of change of the radial location of the particle. You may assume the orbit is always quasi circular.	['The angular momentum is\n\n\n\n$$\n\nL=m v R=\\frac{4}{3} \\pi \\rho r^{3} \\sqrt{\\frac{G M_{\\odot}}{R}} R=\\frac{4}{3} \\pi \\rho r^{3} \\sqrt{G M_{\\odot} R}\n\n$$\n\n\n\nDifferentiating both sides with respect to time,\n\n\n\n$$\n\n-\\frac{v}{c} \\frac{L_{\\odot}}{4 R c} Q r^{2}=\\frac{4}{3} \\pi \\rho r^{3} \\sqrt{\\frac{G M_{\\odot}}{R}} \\frac{1}{2} \\frac{d R}{d t}\n\n$$\n\n\n\nwhich simplifies to\n\n\n\n$$\n\n-\\frac{1}{c^{2}} \\frac{L_{\\odot}}{R} Q=\\frac{8}{3} \\pi \\rho r \\frac{d R}{d t}\n\n$$']	['$-\\frac{1}{c^{2}} \\frac{L_{\\odot}}{R} Q=\\frac{8}{3} \\pi \\rho r \\frac{d R}{d t}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.
+
+
+a. The force of radiation on a spherical particle of radius $r$ is given by
+
+
+
+$$
+
+F=P Q \pi r^{2}
+
+$$
+
+
+
+where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables.
+Context question:
+i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.
+Context answer:
+\boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$}
+
+
+Context question:
+ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.
+Context answer:
+\boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$}
+
+
+Context question:
+iii. The quality factor is given by one of the following
+
+
+
+- If $r \ll \lambda, Q \sim(r / \lambda)^{2}$
+
+- If $r \sim \lambda, Q \sim 1$.
+
+- If $r \gg \lambda, Q=1$
+
+
+
+Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure?
+Context answer:
+particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough.
+
+
+Extra Supplementary Reading Materials:
+
+b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system.
+Context question:
+i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants.
+Context answer:
+\boxed{$v=\sqrt{\frac{G M_{\odot}}{R}}$}
+
+
+Context question:
+ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v \ll c$.
+Context answer:
+\boxed{$\tau=-\frac{v}{c} \frac{L \odot}{4 R c} Q r^{2}$}
+"	[]	Text-only	Competition	False		Equation		Open-ended	Modern Physics	Physics	English
+222	iv. Develop an expression for the time required to remove particles of size $r \approx 1 \mathrm{~cm}$ and density $\rho \approx 1000 \mathrm{~kg} / \mathrm{m}^{3}$ originally in circular orbits at a distance $R=R_{\text {earth }}$, and use the numbers below to simplify your expression.	['Integrating both sides,\n\n\n\n$$\n\nT \\frac{L_{\\odot} Q}{c^{2}}=\\frac{4}{3} \\pi \\rho r R^{2}\n\n$$\n\n\n\nSince $r \\gg \\lambda$, we have $Q=1$ and\n\n\n\n$$\n\nT=\\frac{4}{3} \\frac{\\pi c^{2}}{L \\odot} \\rho r R^{2} \\approx 2 \\times 10^{14} \\mathrm{~s} \\approx 7 \\times 10^{6} \\mathrm{y}\n\n$$']	['$2 \\times 10^{14}$']	"$$
+
+\begin{array}{ll}
+
+g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\
+
+k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\
+
+c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\
+
+N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\
+
+\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\
+
+1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\
+
+m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\
+
+\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1
+
+\end{array}
+
+$$
+
+
+$$
+
+\begin{array}{lrr}
+
+M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\
+
+L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\
+
+R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\
+
+\lambda_{\max } & = & 500 \mathrm{~nm}
+
+\end{array}
+
+$$
+
+Extra Supplementary Reading Materials:
+
+Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.
+
+
+a. The force of radiation on a spherical particle of radius $r$ is given by
+
+
+
+$$
+
+F=P Q \pi r^{2}
+
+$$
+
+
+
+where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables.
+Context question:
+i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.
+Context answer:
+\boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$}
+
+
+Context question:
+ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.
+Context answer:
+\boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$}
+
+
+Context question:
+iii. The quality factor is given by one of the following
+
+
+
+- If $r \ll \lambda, Q \sim(r / \lambda)^{2}$
+
+- If $r \sim \lambda, Q \sim 1$.
+
+- If $r \gg \lambda, Q=1$
+
+
+
+Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure?
+Context answer:
+particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough.
+
+
+Extra Supplementary Reading Materials:
+
+b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system.
+Context question:
+i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants.
+Context answer:
+\boxed{$v=\sqrt{\frac{G M_{\odot}}{R}}$}
+
+
+Context question:
+ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v \ll c$.
+Context answer:
+\boxed{$\tau=-\frac{v}{c} \frac{L \odot}{4 R c} Q r^{2}$}
+
+
+Context question:
+iii. Since $\tau=d L / d t$, the angular momentum $L$ of the particle changes with time. As such, develop a differential equation to find $d R / d t$, the rate of change of the radial location of the particle. You may assume the orbit is always quasi circular.
+Context answer:
+\boxed{$-\frac{1}{c^{2}} \frac{L_{\odot}}{R} Q=\frac{8}{3} \pi \rho r \frac{d R}{d t}$}
+"	[]	Text-only	Competition	False	s	Numerical	5e13	Open-ended	Modern Physics	Physics	English
+223	Find the period of the motion $T$. Neglect the viscosity of the liquid.	"['Denote the density of the liquid by $\\varrho$, so the density of the cylinder is $\\gamma \\varrho$. In equilibrium (i.e. when the net force acting on the cylinder is zero) the immersed part of the cylinder has height $\\gamma h$.\n\nConsider the system in a moment when the cylinder is displaced by distance $x_{1}$ downward and moves down with velocity $v_{1}$. As a result of the motion of cylinder the liquid level rises by some height $x_{2}$, and the liquid flows in the gap between the cylinder and beaker with some velocity $v_{2}$ upwards (see Fig. 1).\n\n<img_4468>\n\nFig. 1\n\nThe relation between the aforementioned displacements and velocities are given by the continuity law:\n\n$$\nx_{1} s=x_{2}(S-s), \\quad v_{1} s=v_{2}(S-s)\n$$\n\nIn the following we express the potential and kinetic energy of the system. Compared to the equilibrium position the cylinder of mass $\\gamma \\varrho s h$ sunk by $x_{1}$, while the potential energy change caused by the redistribution of liquid can be imagined as the center of mass of liquid with mass $\\varrho s x_{1}$ rises by distance $\\gamma h+$ $x_{1} / 2+x_{2} / 2$. Taken the potential energy in the equilibrium state to be zero, the potential energy in the state indicated in the right figure can be written as\n\n$$\nE_{\\mathrm{pot}}=-\\gamma \\varrho \\operatorname{shg} x_{1}+\\varrho s x_{1} g\\left(\\gamma h+\\frac{x_{1}+x_{2}}{2}\\right) .\n$$\n\nAfter opening the bracket the first two terms cancel each other:\n\n$$\nE_{\\mathrm{pot}}=\\frac{1}{2} \\varrho \\operatorname{sg} x_{1}\\left(x_{1}+x_{2}\\right)\n$$\n\nAfter expressing $x_{2}$ from continuity law and some simplification we get a quadratic expression for the potential energy:\n\n$$\nE_{\\mathrm{pot}}=\\frac{1}{2} \\varrho s g x_{1}\\left(x_{1}+\\frac{s}{S-s} x_{1}\\right)=\\frac{1}{2} \\varrho \\frac{s S}{S-s} g x_{1}^{2} .\n$$\n\nNow let us calculate the kinetic energy of the system. The contribution from the cylinder is straightforward, $\\gamma \\varrho s h v_{1}^{2} / 2$, but the motion of the liquid is more complicated.\n\nNote. We may notice that since $s /(S-s)=50$, the speed $v_{2}$ of the liquid in the narrow gap is 50 times larger than the typical speed of the liquid below the cylinder (which can be estimated to be in the range of $v_{1}$ ). And while the mass of the liquid below the cylinder is much larger than the mass of liquid inside the gap (the ratio is ca. 25 if the „few centimeters"" in the problem text is taken to be $3.5 \\mathrm{~cm}$ ), the kinetic energy is proportional to the square of the velocity, so the kinetic energy of the liquid inside the gap is roughly 100 times larger than the kinetic energy of the liquid below the cylinder.\n\nSince the kinetic energy of the liquid below the cylinder is negligible, we can write the total kinetic energy of the system as:\n\n$$\nE_{\\text {kin }}=\\underbrace{\\frac{1}{2} \\gamma \\varrho s h v_{1}^{2}}_{\\text {cylinder }}+\\underbrace{\\frac{1}{2} \\varrho(S-s)\\left(\\gamma h+x_{1}+x_{2}\\right) v_{2}^{2}}_{\\text {liquid }} .\n$$\n\nHere $x_{1}, x_{2} \\ll \\gamma h$, so we shall keep only the term containing $\\gamma h$ in the second bracket:\n\n$$\nE_{\\text {kin }}=\\frac{1}{2} \\gamma \\varrho s h v_{1}^{2}+\\frac{1}{2} \\varrho(S-s) \\gamma h v_{2}^{2}\n$$\n\nExpressing $v_{2}$ from continuity law gives the following:\n\n$$\nE_{\\text {kin }}=\\frac{1}{2} \\gamma \\varrho s h v_{1}^{2}+\\frac{1}{2} \\varrho \\gamma h \\frac{s^{2}}{S-s} v_{1}^{2}=\\frac{1}{2} \\varrho \\gamma h \\frac{s S}{S-s} v_{1}^{2} .\n$$\n\nThe potential and kinetic energies can be written in the form\n\n$$\nE_{\\mathrm{pot}}=\\frac{1}{2} k_{\\mathrm{eff}} x_{1}^{2}, \\quad E_{\\mathrm{kin}}=\\frac{1}{2} m_{\\mathrm{eff}} v_{1}^{2},\n$$\n\nwhere the effective spring constant and effective mass are given by\n\n$$\nk_{\\mathrm{eff}}=\\varrho \\frac{s S}{S-s} g, \\quad m_{\\mathrm{eff}}=\\varrho \\gamma h \\frac{s S}{S-s}\n$$\n\nSo the oscillation is indeed harmonic, thus the angular frequency and the period are:\n\n$$\n\\omega=\\sqrt{\\frac{k_{\\mathrm{eff}}}{m_{\\mathrm{eff}}}}=\\sqrt{\\frac{g}{\\gamma h}}, \\quad T=2 \\pi \\sqrt{\\frac{\\gamma h}{g}}=0.53 \\mathrm{~s} .\n$$\n\nNote. The static restoring force, acting on the cylinder is due to the change (relative to the equilibrium position) of the hydrostatic pressure at its lower base:\n\n$$\nF=-\\operatorname{seg}\\left(x_{1}+x_{2}\\right)=-\\frac{s S}{S-s} \\rho g x_{1}\n$$\n\nThis immediately gives effective stiffness of the system $k_{\\text {eff }}=\\frac{s S}{S-s} \\rho g$.\n\nAlternatively, one may wish to integrate $\\int F \\mathrm{~d} x_{1}$ to get the potential energy\n\n$$\nE_{\\mathrm{pot}}=\\frac{s S}{S-s} \\frac{\\rho g}{2} x_{1}^{2}\n$$'
+ ""When the cylinder is displaced from its equilibrium position downwards by distance $x_{1}$, the net restoring force (pointing up) can be calculated as the sum of the weight of the cylinder and the force from the difference of pressures at the top $\\left(p_{0}\\right)$ and bottom $(p)$ of the cylinder. As a result of the net force, the cylinder accelerates upwards with $a_{1}$, and at the same time, the liquid located in the gap between the cylinder and the wall of the beaker accelerates down with $a_{2}$. The relation between the magnitudes of $a_{1}$ and $a_{2}$ is given by the continuity law:\n\n$$\ns a_{1}=(S-s) a_{2}\n$$\n\n<img_4371>\n\nFig. 2\n\nIf the liquid in the gap was not accelerating, the pressure difference $p-p_{0}$ would be equal to the hydrostatic pressure of the liquid column in the gap. Due to the acceleration of the liquid, $p-p_{0}$ can be expressed from Newton's 2nd law applied for the liquid column of unit area located in the gap:\n\n$$\np_{0}-p+\\varrho g\\left(\\gamma h+x_{1}+x_{2}\\right)=\\varrho\\left(\\gamma h+x_{1}+x_{2}\\right) a_{2}\n$$\n\nwhere we used the notations of Solution I, and the downward direction was taken as positive.\n\nNewton's 2nd law for the cylinder reads as\n\n$$\n\\left(p-p_{0}\\right) s-\\gamma \\varrho s h g=\\gamma \\varrho s h a_{1}\n$$\n\nAfter expressing $p-p_{0}$ from the previous equation, and then substituting it here we get:\n\n$\\varrho g\\left(\\gamma h+x_{1}+x_{2}\\right) s-\\varrho\\left(\\gamma h+x_{1}+x_{2}\\right) a_{2} s-\\gamma \\varrho s h g=\\gamma \\varrho s h a_{1}$.\n\nSince the amplitude of the liquid level is small, the terms containing $a_{2} x_{1}$ and $a_{2} x_{2}$ can be neglected. After rearranging we get:\n\n$$\n\\varrho g s\\left(x_{1}+x_{2}\\right)=\\gamma \\varrho s h\\left(a_{1}+a_{2}\\right) .\n$$\n\nUsing the relations between the displacements and accelerations we finally get:\n\n$$\na_{1}=\\frac{g}{\\gamma h} x_{1}\n$$\n\nTaking into account the opposite directions of $x_{1}$ and $a_{1}$, this is the dynamical condition of a simple harmonic motion with angular frequency and period\n\n$$\n\\omega=\\sqrt{\\frac{g}{\\gamma h}}, \\quad T=2 \\pi \\sqrt{\\frac{\\gamma h}{g}}=0.53 \\mathrm{~s} .\n$$""]"	['0.53']	A solid, uniform cylinder of height $h=10 \mathrm{~cm}$ and base area $s=100 \mathrm{~cm}^{2}$ floats in a cylindrical beaker of height $H=20 \mathrm{~cm}$ and inner bottom area $S=102 \mathrm{~cm}^{2}$ filled with a liquid. The ratio between the density of the cylinder and that of the liquid is $\gamma=0.70$. The bottom of the cylinder is above the bottom of the beaker by a few centimeters. The cylinder is oscillating vertically, so that its axis always coincides with that of the beaker. The amplitude of the liquid level oscillations is $A=1 \mathrm{~mm}$.	[]	Text-only	Competition	False	s	Numerical	5e-2	Open-ended	Thermodynamics	Physics	English
+224	1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.	['Using Gauss law\n\n$$\n\\oint \\mathbf{E} \\cdot \\mathbf{d} \\mathbf{A}=\\frac{q}{\\epsilon_{0}} .\n\\tag{1}\n$$\n\nFrom symetry we know that the electric field only has radial component. Choose a cylinder (with a line charge as the axis) as the Gaussian surface, we obtain\n\n$$\nE .2 \\pi r l=\\frac{\\lambda l}{\\epsilon_{0}}\n$$\n\nSimplify to obtain\n\n$$\n\\mathbf{E}=\\hat{r} \\frac{\\lambda}{2 \\pi \\epsilon_{0} r}\n\\tag{2}\n$$\n\n']	['$\\mathbf{E}=\\hat{r} \\frac{\\lambda}{2 \\pi \\epsilon_{0} r}$']	A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+225	"2.  The potential due to the line charge could be written as
+
+$$
+V(r)=f(r)+K,
+$$
+
+where $K$ is a constant. Determine $f(r)$."	['The potential is given by\n\n$$\n\\begin{aligned}\nV & =-\\int_{\\mathrm{ref}}^{r} \\mathbf{E} \\cdot \\mathbf{d} \\mathbf{l} \\\\\n& =-\\int_{\\mathrm{ref}}^{r} E \\cdot d r \\\\\nV & =-\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln r+K,\n\\end{aligned}\n\\tag{3}\n$$\n\nso $f(r)=-\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln r$. where $K$ is a constant.']	['$f(r)=-\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln r$']	"A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.
+Context question:
+1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.
+Context answer:
+$\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$
+"	[]	Text-only	Competition	False		Expression		Open-ended	Electromagnetism	Physics	English
+226	1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).	['The equation of motion is given by\n\n$$\n\\begin{aligned}\nF & =\\frac{d}{d t}(\\gamma m v) \\\\\n& =\\frac{m c \\dot{\\beta}}{\\left(1-\\beta^{2}\\right)^{\\frac{3}{2}}}\n\\end{aligned}\n\\tag{1}\n$$\n\n$$\nF=\\gamma^{3} m a\n\\tag{2}\n$$\n\nwhere $\\gamma=\\frac{1}{\\sqrt{1-\\beta^{2}}}$ and $\\beta=\\frac{v}{c}$. So the acceleration is given by\n\n$$\na=\\frac{F}{\\gamma^{3} m}\n\\tag{3}\n$$']	['$a=\\frac{F}{\\gamma^{3} m}$']	"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.
+
+First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).
+
+By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.
+
+Some mathematics formulas that might be useful
+
+- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
+- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
+- $\tanh x=\frac{\sinh x}{\cosh x}$
+- $1+\sinh ^{2} x=\cosh ^{2} x$
+- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$
+
+
+
+- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
+- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$
+
+
+Part A. Single Accelerated Particle 
+
+Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$."	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+227	2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.	['Eq.(3) can be rewritten as\n\n$$\n\\begin{aligned}\nc \\frac{d \\beta}{d t} & =\\frac{F}{\\gamma^{3} m} \\\\\n\\int_{0}^{\\beta} \\frac{d \\beta}{\\left(1-\\beta^{2}\\right)^{\\frac{3}{2}}} & =\\frac{F}{m c} \\int_{0}^{t} d t \\\\\n\\frac{\\beta}{\\sqrt{1-\\beta^{2}}} & =\\frac{F t}{m c}\n\\end{aligned}\n\\tag{4}\n$$\n\n$$\n\\beta=\\frac{\\frac{F t}{m c}}{\\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}}\n\\tag{5}\n$$']	['$\\beta=\\frac{\\frac{F t}{m c}}{\\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}}$']	"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.
+
+First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).
+
+By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.
+
+Some mathematics formulas that might be useful
+
+- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
+- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
+- $\tanh x=\frac{\sinh x}{\cosh x}$
+- $1+\sinh ^{2} x=\cosh ^{2} x$
+- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$
+
+
+
+- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
+- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$
+
+
+Part A. Single Accelerated Particle 
+
+Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
+Context question:
+1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
+Context answer:
+\boxed{$a=\frac{F}{\gamma^{3} m}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+228	3.  Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.	['Using Eq.(5), we get\n\n$$\n\\begin{aligned}\n\\int_{0}^{x} d x & =\\int_{0}^{t} \\frac{F t d t}{m \\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}} \\\\\nx & =\\frac{m c^{2}}{F}\\left(\\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}-1\\right) .\n\\end{aligned}\n\\tag{6}\n$$']	['$x=\\frac{m c^{2}}{F}\\left(\\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}-1\\right)$']	"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.
+
+First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).
+
+By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.
+
+Some mathematics formulas that might be useful
+
+- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
+- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
+- $\tanh x=\frac{\sinh x}{\cosh x}$
+- $1+\sinh ^{2} x=\cosh ^{2} x$
+- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$
+
+
+
+- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
+- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$
+
+
+Part A. Single Accelerated Particle 
+
+Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
+Context question:
+1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
+Context answer:
+\boxed{$a=\frac{F}{\gamma^{3} m}$}
+
+
+Context question:
+2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
+Context answer:
+\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+229	5.  Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.	['Eq.(3) can also be rewritten as\n\n$$\nc \\frac{d \\beta}{\\gamma d \\tau}=\\frac{g}{\\gamma^{3}}\n\\tag{14}\n$$\n\n$$\n\\int_{0}^{\\beta} \\frac{d \\beta}{1-\\beta^{2}} =\\frac{g}{c} \\int_{0}^{\\tau} d \\tau\n$$\n$$\n\\ln \\left(\\frac{1}{\\sqrt{1-\\beta^{2}}}+\\frac{\\beta}{\\sqrt{1-\\beta^{2}}}\\right) =\\frac{g \\tau}{c}\n\\tag{15}\n$$\n$$\n\\sqrt{\\frac{1+\\beta}{1-\\beta}} =e^{\\frac{g \\tau}{c}}\n$$\n\n$$\n\\beta\\left(e^{\\frac{g \\tau}{c}}+e^{-\\frac{g \\tau}{c}}\\right) =e^{\\frac{g \\tau}{c}}-e^{-\\frac{g \\tau}{c}}\n$$\n\n$$\n\\beta =\\tanh \\frac{g \\tau}{c} .\n\\tag{16}\n$$']	['$\\beta=\\tanh \\frac{g \\tau}{c}$']	"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.
+
+First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).
+
+By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.
+
+Some mathematics formulas that might be useful
+
+- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
+- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
+- $\tanh x=\frac{\sinh x}{\cosh x}$
+- $1+\sinh ^{2} x=\cosh ^{2} x$
+- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$
+
+
+
+- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
+- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$
+
+
+Part A. Single Accelerated Particle 
+
+Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
+Context question:
+1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
+Context answer:
+\boxed{$a=\frac{F}{\gamma^{3} m}$}
+
+
+Context question:
+2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
+Context answer:
+\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}
+
+
+Context question:
+3.  Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.
+Context answer:
+\boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}
+
+
+Context question:
+4.  Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+230	6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.	['The time dilation relation is\n\n$$\nd t=\\gamma d \\tau\n\\tag{17}\n$$\n\nFrom eq.(16), we have\n\n$$\n\\gamma=\\frac{1}{\\sqrt{1-\\beta^{2}}}=\\cosh \\frac{g \\tau}{c}\n\\tag{18}\n$$\n\nCombining this equations, we get\n\n$$\n\\begin{aligned}\n\\int_{0}^{t} d t & =\\int_{0}^{\\tau} d \\tau \\cosh \\frac{g \\tau}{c} \\\\\nt & =\\frac{c}{g} \\sinh \\frac{g \\tau}{c} .\n\\end{aligned}\n\\tag{19}\n$$']	['$t=\\frac{c}{g} \\sinh \\frac{g \\tau}{c}$']	"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.
+
+First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).
+
+By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.
+
+Some mathematics formulas that might be useful
+
+- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
+- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
+- $\tanh x=\frac{\sinh x}{\cosh x}$
+- $1+\sinh ^{2} x=\cosh ^{2} x$
+- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$
+
+
+
+- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
+- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$
+
+
+Part A. Single Accelerated Particle 
+
+Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.
+Context question:
+1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).
+Context answer:
+\boxed{$a=\frac{F}{\gamma^{3} m}$}
+
+
+Context question:
+2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.
+Context answer:
+\boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}
+
+
+Context question:
+3.  Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.
+Context answer:
+\boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$}
+
+
+Context question:
+4.  Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+5.  Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.
+Context answer:
+\boxed{$\beta=\tanh \frac{g \tau}{c}$}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+231	2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.	['For a magnetic moment making an angle of $\\phi$ with $\\mathbf{B}$,\n\n$$\n\\begin{aligned}\n\\frac{d \\boldsymbol{\\mu}}{d t} & =-\\gamma \\boldsymbol{\\mu} \\times \\mathbf{B} \\\\\n|\\boldsymbol{\\mu}| \\sin \\phi \\frac{d \\theta}{d t} & =\\gamma|\\boldsymbol{\\mu}| B_{0} \\sin \\phi \\\\\n\\omega_{0}=\\frac{d \\theta}{d t} & =\\gamma B_{0}\n\\end{aligned}\n\\tag{5}\n$$']	['$\\omega_{0}=\\gamma B_{0}$']	"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.
+
+The classical torque equation of spin is given by
+
+$$
+\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
+$$
+
+In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation
+
+$$
+\boldsymbol{\mu}=-\gamma \boldsymbol{L}
+$$
+
+where $\gamma$ is the gyromagnetic ratio.
+
+In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.
+
+Part A. Larmor precession
+Context question:
+1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.
+
+(Hint: You can use properties of vector products.)
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+232	2.  For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?	['The new precession frequency as viewed on the rotating frame $S^{\\prime}$ is\n\n$$\n\\begin{aligned}\n\\vec{\\Delta} & =\\left(\\omega_{0}-\\omega\\right) \\mathbf{k}^{\\prime} \\\\\n\\Delta & =\\gamma B_{0}-\\omega .\n\\end{aligned}\n\\tag{7}\n$$']	['$\\Delta =\\gamma B_{0}-\\omega$']	"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.
+
+The classical torque equation of spin is given by
+
+$$
+\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
+$$
+
+In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation
+
+$$
+\boldsymbol{\mu}=-\gamma \boldsymbol{L}
+$$
+
+where $\gamma$ is the gyromagnetic ratio.
+
+In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.
+
+Part A. Larmor precession
+Context question:
+1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.
+
+(Hint: You can use properties of vector products.)
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.
+Context answer:
+\boxed{$\omega_{0}=\gamma B_{0}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part B. Rotating frame
+
+In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes
+
+$$
+\frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right)
+$$
+
+
+
+$$
+\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A})
+$$
+
+where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$.
+Context question:
+1.  Show that the time evolution of the magnetic moment follows the equation
+
+$$
+\left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f}
+$$
+
+where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field.
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+233	4.  Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )?	['In this case, the effective magnetic field becomes\n\n$$\n\\begin{aligned}\n\\mathbf{B}_{\\mathrm{eff}} & =\\mathbf{B}-\\omega / \\gamma \\mathbf{k}^{\\prime} \\\\\n& =\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b\\left(\\cos 2 \\omega t \\mathbf{i}^{\\prime}-\\sin 2 \\omega t \\mathbf{j}^{\\prime}\\right)\n\\end{aligned}\n\\tag{9}\n$$\n\nwhich has a time average of $\\overline{\\mathbf{B}_{\\mathrm{eff}}}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}$.']	['$\\mathbf{B}_{\\mathrm{eff}}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b\\left(\\cos 2 \\omega t \\mathbf{i}^{\\prime}-\\sin 2 \\omega t \\mathbf{j}^{\\prime}\\right)$ , $\\overline{\\mathbf{B}_{\\mathrm{eff}}}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}$']	"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.
+
+The classical torque equation of spin is given by
+
+$$
+\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
+$$
+
+In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation
+
+$$
+\boldsymbol{\mu}=-\gamma \boldsymbol{L}
+$$
+
+where $\gamma$ is the gyromagnetic ratio.
+
+In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.
+
+Part A. Larmor precession
+Context question:
+1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.
+
+(Hint: You can use properties of vector products.)
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.
+Context answer:
+\boxed{$\omega_{0}=\gamma B_{0}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part B. Rotating frame
+
+In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes
+
+$$
+\frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right)
+$$
+
+
+
+$$
+\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A})
+$$
+
+where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$.
+Context question:
+1.  Show that the time evolution of the magnetic moment follows the equation
+
+$$
+\left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f}
+$$
+
+where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field.
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+2.  For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?
+Context answer:
+\boxed{$\Delta =\gamma B_{0}-\omega$}
+
+
+Context question:
+3.  Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is
+
+$$
+\Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}}
+$$
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Modern Physics	Physics	English
+234	"1.  In the rotating frame $S^{\prime}$, show that the effective field can be approximated by
+
+$$
+\boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime},
+$$
+
+which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ?"	"['The oscillating field can be considered as a superposition of two oppositely rotating field:\n\n$$\n2 b \\cos \\omega_{0} t \\mathbf{i}=b\\left(\\cos \\omega_{0} t \\mathbf{i}+\\sin \\omega_{0} t \\mathbf{j}\\right)+b\\left(\\cos \\omega_{0} t \\mathbf{i}-\\sin \\omega_{0} t \\mathbf{j}\\right)\n$$\n\nwhich gives an effective field of (with $\\omega=\\omega_{0}=\\gamma B_{0}$ ):\n\n$$\n\\mathbf{B}_{\\mathrm{eff}}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b \\mathbf{i}^{\\prime}+b\\left(\\cos 2 \\omega_{0} t \\mathbf{i}^{\\prime}-\\sin 2 \\omega_{0} t \\mathbf{j}^{\\prime}\\right)\n$$\n\nSince $\\omega_{0} \\gg \\gamma b$, the rotation of the term $b\\left(\\cos 2 \\omega_{0} t \\mathbf{i}^{\\prime}-\\sin 2 \\omega_{0} t \\mathbf{j}^{\\prime}\\right)$ is so fast compared to the frequency $\\gamma b$. This means that we can take the approximation\n\n$$\n\\mathbf{B}_{\\mathrm{eff}} \\approx\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b \\mathbf{i}^{\\prime}=b \\mathbf{i}^{\\prime}\n\\tag{10}\n$$\n\nwhere the magnetic moment precesses with frequency $\\Omega=\\gamma b$.\n\nAs $\\Omega=\\gamma b \\ll \\omega_{0}$, the magnetic moment does not ""feel"" the rotating term $b\\left(\\cos 2 \\omega_{0} t \\mathbf{i}^{\\prime}-\\sin 2 \\omega_{0} t \\mathbf{j}^{\\prime}\\right)$ which averaged to zero.']"	['$\\Omega=\\gamma b$']	"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.
+
+The classical torque equation of spin is given by
+
+$$
+\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
+$$
+
+In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation
+
+$$
+\boldsymbol{\mu}=-\gamma \boldsymbol{L}
+$$
+
+where $\gamma$ is the gyromagnetic ratio.
+
+In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.
+
+Part A. Larmor precession
+Context question:
+1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.
+
+(Hint: You can use properties of vector products.)
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.
+Context answer:
+\boxed{$\omega_{0}=\gamma B_{0}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part B. Rotating frame
+
+In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes
+
+$$
+\frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right)
+$$
+
+
+
+$$
+\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A})
+$$
+
+where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$.
+Context question:
+1.  Show that the time evolution of the magnetic moment follows the equation
+
+$$
+\left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f}
+$$
+
+where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field.
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+2.  For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?
+Context answer:
+\boxed{$\Delta =\gamma B_{0}-\omega$}
+
+
+Context question:
+3.  Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is
+
+$$
+\Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}}
+$$
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+4.  Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )?
+Context answer:
+\boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part C. Rabi oscillation  
+
+For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: ""up"" and ""down"". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation
+
+$$
+N_{\uparrow}+N_{\downarrow}=N
+$$
+
+The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis:
+
+$$
+M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} .
+$$
+
+In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by
+
+$$
+\boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} .
+$$"	[]	Text-only	Competition	False		Expression		Open-ended	Modern Physics	Physics	English
+235	3.  Under the application of magnetic field described above, determine the fractional population of each spin up $P_{\uparrow}=N_{\uparrow} / N$ and spin down $P_{\downarrow}=N_{\downarrow} / N$ as a function of time. Plot $P_{\uparrow}(t)$ and $P_{\downarrow}(t)$ on the same graph vs. time $t$. The alternating spin up and spin down population as a function of time is called Rabi oscillation.	['From the relations\n\n$$\n\\begin{aligned}\nP_{\\uparrow}-P_{\\downarrow} & =\\frac{\\mu_{z}}{\\mu}=\\cos \\theta, \\\\\nP_{\\uparrow}+P_{\\downarrow} & =1,\n\\end{aligned}\n$$\n\n\n\nwe obtain the results $\\left(\\omega=\\omega_{0}\\right)$\n\n$$\n\\begin{aligned}\nP_{\\downarrow} & =\\frac{1-\\cos \\theta}{2} \\\\\n& =\\frac{1-\\cos ^{2} \\alpha-\\sin ^{2} \\alpha \\cos \\Omega t}{2} \\\\\n& =\\sin ^{2} \\alpha \\frac{1-\\cos \\Omega t}{2} \\\\\n& =\\frac{b^{2}}{\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right)^{2}+b^{2}} \\sin ^{2} \\frac{\\Omega t}{2} \\\\\n& =\\sin ^{2} \\frac{\\Omega t}{2},\n\\end{aligned}\n\\tag{13}\n$$\n\nand\n\n$$\nP_{\\uparrow}=\\frac{b^{2}}{\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right)^{2}+b^{2}} \\cos ^{2} \\frac{\\Omega t}{2}=\\cos ^{2} \\frac{\\Omega t}{2}\n\\tag{14}\n$$\n\n<img_4555>']	['$P_{\\downarrow}=\\sin ^{2} \\frac{\\Omega t}{2}$ , $P_{\\uparrow}=\\cos ^{2} \\frac{\\Omega t}{2}$']	"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue.
+
+The classical torque equation of spin is given by
+
+$$
+\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B}
+$$
+
+In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation
+
+$$
+\boldsymbol{\mu}=-\gamma \boldsymbol{L}
+$$
+
+where $\gamma$ is the gyromagnetic ratio.
+
+In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars.
+
+Part A. Larmor precession
+Context question:
+1.  Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant.
+
+(Hint: You can use properties of vector products.)
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.
+Context answer:
+\boxed{$\omega_{0}=\gamma B_{0}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part B. Rotating frame
+
+In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes
+
+$$
+\frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right)
+$$
+
+
+
+$$
+\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A})
+$$
+
+where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$.
+Context question:
+1.  Show that the time evolution of the magnetic moment follows the equation
+
+$$
+\left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f}
+$$
+
+where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field.
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+2.  For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?
+Context answer:
+\boxed{$\Delta =\gamma B_{0}-\omega$}
+
+
+Context question:
+3.  Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is
+
+$$
+\Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}}
+$$
+Context answer:
+\boxed{证明题}
+
+
+Context question:
+4.  Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )?
+Context answer:
+\boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$}
+
+
+Extra Supplementary Reading Materials:
+
+Part C. Rabi oscillation  
+
+For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: ""up"" and ""down"". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation
+
+$$
+N_{\uparrow}+N_{\downarrow}=N
+$$
+
+The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis:
+
+$$
+M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} .
+$$
+
+In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by
+
+$$
+\boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} .
+$$
+Context question:
+1.  In the rotating frame $S^{\prime}$, show that the effective field can be approximated by
+
+$$
+\boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime},
+$$
+
+which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ?
+Context answer:
+\boxed{$\Omega=\gamma b$}
+
+
+Context question:
+2.  Determine the angle $\alpha$ that $\boldsymbol{\mu}$ makes with $\boldsymbol{B}_{\text {eff }}$. Also, prove that the magnetization varies with time as
+
+$$
+M(t)=N \mu(\cos \Omega t) .
+$$
+Context answer:
+\boxed{证明题}
+"	[]	Text-only	Competition	True		Expression		Open-ended	Modern Physics	Physics	English