USAPhO 2007, Полуфинал — Задача 1. A1

AAPT AIP United States Physics Team 2007 — 2007 Semi-Final Exam · 25 т.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

  • Work Part A first. You have 90 minutes to complete all four problems.

  • After you have completed Part A, you may take a break.

  • Then work Part B. You have 90 minutes to complete both problems.

  • Show all your work. Partial credit will be given.

  • Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

    Doe, Jamie

    A1 – 1/3

  • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s, or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.

  • Questions with the same point value are not necessarily of the same difficulty.

  • Do not discuss the contents of this exam with anyone until after March 27th.

  • Good luck!

Possibly Useful Information - (Use for both part A and for part B)
Gravitational field at the Earth’s surfaceg=9.8 N/kgg = 9.8\ \mathrm{N/kg}
Newton’s gravitational constantG=6.67×10−11 N⋅m2/kg2G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}
Coulomb’s constantk=1/4πεo=8.99×109 N⋅m2/C2k = 1/4\pi\varepsilon_o = 8.99 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}
Biot-Savart constantkm=μo/4π=10−7 T⋅m/Ak_m = \mu_o/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}
Speed of light in a vacuumc=3.00×108 m/sc = 3.00 \times 10^{8}\ \mathrm{m/s}
Boltzmann’s constantkB=1.38×10−23 J/Kk_B = 1.38 \times 10^{-23}\ \mathrm{J/K}
Avogadro’s numberNA=6.02×1023 (mol)−1N_A = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}
Ideal gas constantR=NAkB=8.31 J/(mol⋅K)R = N_A k_B = 8.31\ \mathrm{J/(mol\cdot K)}
Stefan-Boltzmann constantσ=5.67×10−8 J/(s⋅m2⋅K4)\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}
Elementary chargee=1.602×10−19 Ce = 1.602 \times 10^{-19}\ \mathrm{C}
1 electron volt1 eV=1.602×10−19 J1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}
Planck’s constanth=6.63×10−34 J⋅s=4.14×10−15 eV⋅sh = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}
Electron massm=9.109×10−31 kg=0.511 MeV/c2m = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}
Binomial expansion(1+x)n≈1+nx(1 + x)^n \approx 1 + nx for ∣x∣≪1\lvert x\rvert \ll 1
Small angle approximationssin⁡θ≈θ\sin\theta \approx \theta; cos⁡θ≈1−12θ2\cos\theta \approx 1 - \frac{1}{2}\theta^2

Условие

A group of 12 resistors is arranged along the edges of a cube as shown in the diagram below. The vertices of the cube are labeled aa-hh.

A cube drawn in perspective with a resistor on each of its 12 edges; the vertices are labeled a, b, c, d (upper), e, f, g, h (lower): a–b, a–c, a–e, c–d, b–d, c–g, b–f, d–h, e–f, e–g, g–h, f–h.

a. The resistance between each pair of vertices is as follows:

Rab=Rac=Rae=3.0 ΩR_{ab} = R_{ac} = R_{ae} = 3.0\ \Omega

Rcg=Ref=Rbd=8.0 ΩR_{cg} = R_{ef} = R_{bd} = 8.0\ \Omega

Rcd=Rbf=Reg=12.0 ΩR_{cd} = R_{bf} = R_{eg} = 12.0\ \Omega

Rdh=Rfh=Rgh=1.0 ΩR_{dh} = R_{fh} = R_{gh} = 1.0\ \Omega

What is the equivalent resistance between points aa and hh? [13 т.]

b. The three 12.0 Ω12.0\ \Omega resistors are replaced by identical capacitors. Ccd=Cbf=Ceg=15.0 μFC_{cd} = C_{bf} = C_{eg} = 15.0\ \mu\mathrm{F}. A 12.0 V12.0\ \mathrm{V} battery is attached across points aa and hh and the circuit is allowed to operate for a long period of time. What is the charge (QcdQ_{cd}, QbfQ_{bf}, QegQ_{eg}) on each capacitor after this long period of time? [12 т.]

Отговори

Решение

Оригинал в Архива: olympiad_2007_semi-final.pdf · официални решения: olympiad_2007_semi-final_soln.pdf