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.P.hysi9s1.5.0Winter 20111333111 0.219 _ Name;
Section 1.: The following 10 questions are multiple choice problems. There is only one correct
answer for each problem. Please circle your choice. Each multiple choice question is worth 3
points each for a
Chapter 9
2. Our notation is as follows: x1 = 0 and y1 = 0 are the coordinates of the m1 = 3.0 kg particle; x2
= 2.0 m and y2 = 1.0 m are the coordinates of the m2 = 4.0 kg particle; and x3 = 1.0 m and y3 =
2.0 m are the coordinates of the m3 = 8.0 kg par
Chapter 14
1, 2, 5, 8, 9, 10, 28, 31, 34, 81, 82
1. Let the volume of the expanded air sacs be Va and that of the fish with its air sacs collapsed be
V. Then
where w is the density of the water. This implies
fishV = w(V + Va) or (V + Va)/V = 1.08/1.00,
wh
Chapter 2 HW Solutions
2. (a) Using the fact that time = distance/velocity while the velocity is constant, we find
73.2 m 73.2 m
vavg 73.2 m 73.2 m 1.74 m/s.
1.22 m/s 3.05 m
(b) Using the fact that distance = vt while the velocity v is constant, we find
v
Chapter 3 HW Solutions
2. (a) With r = 15 m and = 30, the x component of is given by
r
rx = rcos = (15 m) cos 30 = 13 m.
(b) Similarly, the y component is given by ry = r sin = (15 m) sin 30 = 7.5 m.
3. A vector r can be represented in the magnitude-angle
Chapter 1
1. Various geometric formulas are given in Appendix E.
(a) Expressing the radius of the Earth as
R 6.37 106 m 103 km m 6.37 103 km,
its circumference is
s 2 R 2 (6.37 103 km) 4.00 104 km.
(b) The surface area of Earth is
A 4 R 2 4 6.37 103 km
(c
Chapter 4 HW Solutions
7. Using Eq. 4-3 and Eq. 4-8, we have
m (5.0i 6.0j + 2.0k)
m
r
( 2.0i + 8.0j 2.0k)
m/s.
vavg
(0.70i +1.40j 0.40k)
10 s
8. Our coordinate system has
r
rAB (483 km)i
i
pointed east and
and the second is
j
pointed north. The first
Chapter 15
4. (a) Since the problem gives the frequency f = 3.00 Hz, we have = 2f = 6 rad/s (understood to be valid to three significant figures). Each spring is considered to support one fourth of
the mass mcar so that Eq. 15-12 leads to
(b) If the new m
Chapter 16
5. (a) The motion from maximum displacement to zero is one-fourth of a cycle. One-fourth of a
period is 0.170 s, so the period is T = 4(0.170 s) = 0.680 s.
(b) The frequency is the reciprocal of the period:
(c) A sinusoidal wave travels one wav
Chapter 5 HW Solutions
2. We apply Newtons second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the
chopping block is
, where the vector addition is done using unit-vector notation. The
Fnet F1 F2
acceleration of the block is given b
Chapter 7 HW Solutions
1. (a) From Table 2-1, we have
v v02 2ax
v 2 v02 2ax
2.4 10
7
2
. Thus,
m/s 2 3.6 1015 m/s 2 0.035 m 2.9 107 m/s.
(b) The initial kinetic energy is
2
1
1
Ki mv02 1.67 10 27 kg 2.4 107 m/s 4.8 10 13 J.
2
2
The final kinetic energy
Chapter 10
1. The problem asks us to assume
Thus, with
and are constant. For consistency of units, we write
, the time of flight is
.
During that time, the angular displacement of a point on the balls surface is
11. We assume the sense of initial rotation
Chapter 13
2,4,8,9
2. The gravitational force between you and the moon at its initial position (directly opposite of
Earth from you) is
where
is the mass of the moon,
is the distance between the moon and the Earth, and
is the radius of the Earth. At its f
Chapter 11
11. With
, we solve the problem by applying Eq. 9-14 and Eq. 11-37.
(a) Newtons second law in the x direction leads to
In unit vector notation, we have
, which points leftward.
(b) With R = 0.30 m, we find the magnitude of the angular accelerat
Physics 150 Fall 2011 Exam TWO
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Section I: The following 10 questions are multiple choice probJe:
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Physics 150 Winter 2011 Final Exam
Section 1: The following 26 questions are multiple choice problems. There is only one correct answer for
each problem. Please circle your choice. Each multiple choice question is worth 1.5 points each for a
totai of 39
Physics 150 Winter 2011 Exam Three ' :
Name
Section 1: The following 13 questions are multiple choice problems. There is only one correct
answer for each problem. Please circle your choice. Each multiple choice question is worth 3
points each for a tota
Physics 150 Winter 2011 Exam Two
Name_
Section 1: The following 10 questions are multiple choice problems. There is only one correct
answer for each problem. Please circle your choice. Each multiple choice question is worth 3
points each for a total of 30
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Physics iso Fall 20”. Exam one hams \M t 0?” l
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Section 1: The following l0 questions are multiple choice problems. There is only one correct
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Physics 150 Winter 2011 Exam One
Name_
Section 1: The following 10 questions are multiple choice problems. There is only one correct
answer for each problem. Please circle your choice. Each multiple choice question is worth 3
points each for a total of 30
Physics 150 Winter 2011 Exam Two
Section I: The foliowing 10 questions are multiple choice probierns. There is only one correct
answer for each problem. Please circle your choice. Each multiple choice question is worth 3
points each for a total of 30 po
Physics 150 Fall 2011 Exam Three
Section I: The following 10 questions are multiple choice problems. There is only one correct
answer for each problem. Please circle your choice. Each multiple choice question is worth 3
points each for a total of 30 poi
KM
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Physics 150 Fall 2010 Final Exam Name tit WW5
Section I: The foiiowing 20 questions are multiple choice problems. There is only one correct answer for
each problem. Please circie your choice. Each multiple choice question is worth 1.5 points e
Chapter 8 HW Solutions
3. (a) Noting that the vertical displacement is 10.0 m 1.50 m = 8.50 m downward (same direction as
Fg
), Eq. 7-12 yields
Wg mgd cos (2.00 kg)(9.80 m/s 2 )(8.50 m) cos 0 167 J.
(b) One approach (which is fairly trivial) is to use Eq.
Chapter 6 HW Solutions
1. The greatest deceleration (of magnitude a) is provided by the maximum friction force (Eq. 6-1, with FN
= mg in this case). Using Newtons second law, we find
a = fs,max /m = sg.
Eq. 2-16 then gives the shortest distance to stop: |