Student Name_ Student Number_ Lab/Recitation Section Number: 41 42 43 44 (Circle One)
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Answers to PH-131 Fall 2006 Exam 1 This document does not contain the solution to each of the problems. It contains only the final answers. Use the exam as a timed practice test and check your answers. You may check your solutions with any of the TAs. Rem
1. (a) The charge that passes through any cross section is the product of the current and time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is the magnitude of
1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through the battery. Thus,
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is
= E A = EA cos = (1800 N C ) 3.2 10-3 m cos145 = -1.5 10-2 N m 2 C.
2. We use = E A , whe
1. We note that the symbol q2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for q1 = q2 .
The following two sketches are for the cases q1 > q2 (left figure) an
1. (a) With a understood to mean the magnitude of acceleration, Newton's second and third laws lead to m2 a2 = m1a1
c6.3 10 kghc7.0 m s h = 4.9 10 m =
-7 2 2
9.0 m s
(b) The magnitude of the (only) force on particle 1 is
q q q F = m1a1 = k 1 2 2
PH 132 Exam 3
Student Name 5 GK. L) 770 M S
Labeeeitatinn Seetinn Number (1 I ,i . .,3 5)
1.. Fill nut all the inferrnatien abeue. Write yeur nanie en'eaeh page.
2. Clearly indieate year nal answers fer all multi
Student Name_ Student Number_ Lab/Recitation Section Number_(11,34)
1. Fill out all of the information requested above. Write your name on each page. 2. Clearly indicate your final answers for all multiple-choice qu
1. (a) Eq. 28-3 leads to 6.50 10-17 N FB v= = = 4.00 105 m s . -19 -3 eB sin 160 10 C 2.60 10 T sin 23.0 .
(b) The kinetic energy of the proton is
2 1 2 1 mv = 167 10-27 kg 4.00 105 m s = 134 10-16 J. . . 2 2
This is (1.34 10 16 J) / (1.6
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by
With r = 20 ft = 6.10 m, we have
c4 10 B=
hb 2 b6.10 mg
T m A 100 A
g = 3.3 10
T = 3.3 T.
(b) This is about one
Richard Feynman- Game of Chess Theory
Like a game of chess
Learn the moves by watching another player, but you cant ask opposite player the rules
We are the ignorant player, opposite player is the universe.
Classical Mechanics- Newtonian-