problem_id stringlengths 6 8 | discipline stringclasses 2
values | image stringlengths 17 19 ⌀ | problem dict | answer stringlengths 1 1.36k |
|---|---|---|---|---|
PHY-116 | Physics | images/PHY-116.png | {
"en": "As shown in the classic problem of a hound chasing a fox, the fox moves from point A to B at a constant speed v, while the hound moves toward the fox at a constant speed u, always aiming at the fox in the ground reference frame. At \\(t = 0\\), the two are located at points A and D, respectively, with a dist... | (1)Δt = $(\sqrt{1-u^2/c^2}-\sqrt{1-v^2/c^2})*\frac{uL}{u^2-v^2}$
(2)x(y) = $-\frac{y((L/y)**(v/u)-(y/L)**(v/u)))}{2\sqrt{1-v^2/c^2}}$ |
PHY-117 | Physics | images/PHY-117.png | {
"en": "As shown in a schematic diagram of a laser-cooling atomic experiment, a single-frequency laser beam propagating opposite to a collimated atomic beam scatters with atoms to decelerate them. We model the atom as a two-level system with energies \\(E_1\\) and \\(E_2\\), simplified to one electron orbiting a nuc... | (1)$ω_0 = \frac{me^4}{16 \pi^2 ε_0^2 \hbar^3}$
(2)
(2.1)$f_d = \frac{e^2}{6 \pi ε_0 c^3} \frac{da}{dt}$
(2.2) $\eta = \exp{-\frac{e^2 ω_0^2 t}{6 \pi ε_0 c^3 m}}$
(3)
(3.1) $\alpha = \frac{\pi e^2 ω}{ε_0 c V} \frac{\frac{e^2 ω^3}{6 \pi ε_0 c^3}}{(\frac{e^2 ω^3}{6 \pi ε_0 c^3})^2+m^2 (ω_0^2-ω^2)^2}$
(3.2)
$F = -(E_2-E_1... |
PHY-118 | Physics | images/PHY-118.png | {
"en": "As shown in a finite one-dimensional box, calculate \\(\\delta x \\cdot \\delta p\\), where \\(\\delta x\\) and \\(\\delta p\\) are the uncertainties in the particle's position and momentum, respectively. Assume the particle has mass m and the reduced Planck constant is \\(\\hbar\\). The particle is in an ei... |
$\hbar \sqrt{k^2 \pi^2/12-1/2}$,E = $\frac{k^2 \pi^2 \hbar^2}{2 m L^2}$
|
PHY-119 | Physics | images/PHY-119.png | {
"en": "The figure shows a system of N particles confined to a circle, where adjacent particles interact via a potential energy \\frac{\\lambda x^2}{2}, and no interaction exists between non-adjacent particles. Here, x represents the angular displacement between two particles. The system temperature is T, and the th... | (1)Z = $\frac{2 \pi^{N-1}}{(N-1)!}-N λ \frac{(2 N \pi)^{N+1}}{ k T (N+1)!}$
U = $\frac{4 \pi^2 λ}{-N-1+\frac{4 \pi^2 λ}{k T}}$
C_v = $\frac{16 \pi^4 λ^2}{k T^2(N+1-{\frac{4 \pi^2 λ}{k T}})^2}$
(2) Follows a normal (Gaussian) distribution with mean \frac{2\pi}{N} and variance \frac{(N-1)kT}{N \lambda}. |
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