physics7B-sp04-final-Smoot-exam

physics7B-sp04-final-Smoot-exam - Department of Physics...

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Unformatted text preview: Department of Physics, University of California, Berkeley Final Examination Physics 7B, Section 2, Prof. Smoot 12:30 AM - 3:30 PM, 14 May 2004 Name: SID No: Discussion Section: Name of TA: Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Score: 1 Answer all eight problems. Write clearly and explain your work. Partial credit will be given for incomplete solutions provided your logic is reasonable and clear. Cross out any parts that you don’t want to be graded. Enclose your answers with boxes. Express all numerical answers in SI units . Answers with no explanation or disconnected comments will not be credited. If you obtain an answer that is questionable, explain why you think it is wrong. Constants and Conversion factors Avogadro number, N A 6 . 022 × 10 23 Permittivity of vacuum, 8 . 85 × 10- 12 F · m- 1 Permeability of vacuum, μ 4 π × 10- 7 T · m · A- 1 Speed of light in vacuum, c 1 / √ μ = 3 × 10 8 m/s Charge of electron, q e- 1 . 602 × 10- 19 C Universal gas constant, R 8.315 J · mol- 1 · K- 1 = 1.99 cal · mol- 1 · K- 1 Boltzmann constant, k 1 . 381 × 10- 23 J · K- 1 Stefan-Boltzmann constant, σ 5 . 67 × 10- 8 W · m- 2 · K- 4 Acceleration due to gravity, g 9.8 m · s- 2 Specific heat of water 1 kcal · kg- 1 · ◦ C- 1 Heat of fusion of water 80 kcal · kg- 1 1 atm 1 . 013 × 10 5 N · m- 2 1 kcal 4 . 18 × 10 3 J x ◦ F = 9 5 y ◦ C + 32 ◦ F 1 hp 746 W 1 ft-lb = 1.356 N-m 1 liter 10 3 cm 3 Equations and formulae: dN v dv = 4 πN m 2 πkT 3 2 v 2 e- mv 2 2 kT v rms = r 3 kT m = r 3 ρ P v = r 8 kT πm T V γ- 1 = constant W = P 1 V 1 γ- 1 " V 1 V 2 γ- 1- 1 # γ = C P C V dQ dt = σ AT 4 dQ dt =- κA dT dx 2 ~ F = q ( ~ E + ~v × ~ B ) p = q d ~μ = NI ~ A ~ τ = p × E ~ τ = ~μ × ~ B U =- p · E U =- μ · ~ B V ab =- Z b a E · d l U ab = qV ab E =-∇ V u E = 1 2 E 2 u B = 1 2 μ B 2 V = IZ J = σ E R = ρ l A E H = v d B d B = μ 4 π Id l × ˆ r r 2 Φ B = Z B · d a ω LRC = r 1 LC- R 2 4 L 2 α = R 2 L φ = tan- 1 R 2 Lω LRC Z LRC = s R 2 + ωL- 1 ωC 2 P = I 2 rms Z cos φ I rms = 1 √ 2 I I d = d Φ E dt S = 1 μ E × B 3 1. [25 points] Short Questions (a) [5 points] Circle T or F for True or False T F (i) The principle of equipartition of energy states that in equilibrium the thermal energy is shared among all active degrees of freedom and is randomly distributed with an average energy of kT/2. T F (ii) If the temperature difference across a conductor triples, the rate at which it transfers heat energy increases by a factor of nine. T F (iii) When a system goes from equilibrium state 1 to state 2, the change in the internal energy is the same for all processes. T F (iv) The internal energy of a given amount of an ideal gas at equilibrium depends only on its absolute temperature....
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This note was uploaded on 12/14/2011 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at Berkeley.

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physics7B-sp04-final-Smoot-exam - Department of Physics...

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