Solution Manual for University Physics with Modern Physics (3rd Edition) by Wolfgang Bauer and Gary D. Westfall

All 40 Chapters Covered




SOLUTIONS

, UNIVERSITY PHYSICS, 3rd Edition

Table of Contents
PART 1 MECHANICS OF POINT PARTICLES
1 Overview 1
2 Motion in a Straight Line 45
3 Motion in Two and Three Dimensions 108
4 Force 163
5 Kinetic Energy, Work, and Power 223
6 Potential Energy and Energy Conservation 255
7 Momentum and Collisions 308
PART 2 EXTENDED OBJECTS, MATTER, AND CIRCULAR MOTION
8 Systems of Particles and Extended Objects 380
9 Circular Motion 430
10 Rotation 474
11 Static Equilibrium 521
12 Gravitation 574
13 Solids and Fluids 628
PART 3 OSCILLATIONS AND WAVES
14 Oscillations 673
15 Waves 713
16 Sound 747
PART 4 THERMAL PHYSICS
17 Temperature 783
18 Heat and the First Law of Thermodynamics 806
19 Ideal Gases 835
20 The Second Law of Thermodynamics 870
PART 5 ELECTRICITY
21 Electrostatics 898
22 Electric Fields and Gauss’s Law 934
23 Electric Potential 973
24 Capacitors 1007
25 Current and Resistance 1046
26 Direct Current Circuits 1075
PART 6 MAGNETISM
27 Magnetism 1113
28 Magnetic Fields of Moving Charges 1141
29 Electromagnetic Induction 1171
30 Alternating Current Circuits 1197
31 Electromagnetic Waves 1224
PART 7 OPTICS
32 Geometric Optics 1248
33 Lenses and Optical Instruments 1270
34 Wave Optics 1304
PART 8 RELATIVITY AND QUANTUM PHYSICS
35 Relativity 1324
36 Quantum Physics 1354
37 Quantum Mechanics 1382
38 Atomic Physics 1419
39 Elementary Particle Physics 1444
40 Nuclear Physics 1464

, Chapter 1: Overview



Chapter 1: Overview

Concept Checks
1.1. a 1.2. a) 4 b) 3 c) 5 d) 6 e) 2 1.3. a, c and e 1.4. b 1.5. e 1.6. a) 4th b) 2nd c) 3rd d) 1st

Multiple-Choice Questions
1.1. c 1.2. c 1.3. d 1.4. b 1.5. a 1.6. b 1.7. b 1.8. c 1.9. c 1.10. b 1.11. d 1.12. b 1.13. c 1.14
1.15. e 1.16. a

Conceptual Questions
1.17. (a) In Europe, gas consumption is in L/100 km. In the US, fuel efficiency is in
miles/gallon. Let’s relate these two: 1 mile = 1.609 km, 1 gal = 3.785 L.
1 mile 1.609 km 1.609 1  km  1  1
 
  100  0.00425
  

gal 3.785 L 3.785 100  L L/100 235.24 L/100 km
km 
Therefore, 1 mile/gal is the reciprocal of 235.2 L/100 km.
from part (a),
(b) Gas consumption is . 1L 1
12.2 L 
Using
100 km 1  100 km 235.24 miles/gal
12.2 L 12.2
L 1
 1  1
2.2  .
100 km     19.282 miles/gal
100 km  235.24
miles/gal 
Therefore, a car that consumes 12.2 L/100 km of gasoline has a fuel efficiency of 19.3 miles/gal.
(c) If the fuel efficiency of the car is 27.4 miles per gallon, then
27.4 miles 27.4 1
gal 
235.24 L/100 = 8.59 L/100 km
.
km
Therefore, 27.4 miles/gal is equivalent to
8.59 L/100 km. (d)




1.18. A vector is described by a set of components in a given coordinate system, where
the components are the projections of the vector onto each coordinate axis.
Therefore, on a two-dimensional sheet of paper there are two coordinates and
thus, the vector is described by two components. In the real three-dimensional
world, there are three coordinates and a vector is described by three
components. A four-dimensional world would be described by four coordinates, and
a vector would be described by four components.
1.19. A vector contains information about the distance between two points (the
magnitude of the vector). In contrast to a scalar, it also contains information
1

,direction. In many cases knowing a direction can be as important as knowing a
magnitude.




2

,1.20. In order to add vectors in magnitude-direction form, each vector is expressed in
terms of component vectors which lie along the coordinate axes. The
corresponding components of each vector are added to obtain the components of
the resultant vector. The resultant vector can then be expressed in magnitude-
direction form by computing its magnitude and direction.
1.21. The advantage to using scientific notation is two-fold: Scientific notation is more
compact (thus saving space and writing), and it also gives a more intuitive way of
dealing with significant figures since you can only write the necessary significant
figures and extraneous zeroes are kept in the exponent of the base.
1.22. The SI system of units is the preferred system of measurement due to its ease of
use and clarity. The SI system is a metric system generally based on multiples
of 10, and consisting of a set of standard measurement units to describe the
physical world. In science, it is paramount to communicate results in the clearest
and most widely understood manner. Since the SI system is internationally
recognized, and its definitions are unambiguous, it is used by scientists around
the world, including those in the United States.
1.23. It is possible to add three equal-length vectors and obtain a vector sum of zero. The vector
components of
the three vectors must all add to zero. Consider the following T3 :
arrangement with T1 T2




The horizontal components and cancel out, so the T1  is a vertical vector whose
of T1 T2 sum T2
magnitude is T cosT cos 2T cos. The vector sum T1 T2 T3 is zero if
2T cosT 0
1
cos
2
60
Therefore it is possible for three equal-length vectors to sum to zero.
1.24. Mass is not a vector quantity. It is a scalar quantity since it does not make sense
to associate a direction with mass.
1.25. The volume of a sphere is given V  4 / Doubling the volume gives
by
3r .
3


2V 2  r  (2 )r Now, since the distance between the flies is the
3 3/3 3


 (2 r) .
1/3 3


diameter of the d 2r , and doubling the volume increases the 1/3
2 , the
sphere, radius by a factor of
1/3 1/3 1/3
distance between the flies is then increased to 2(2 r) 2 (2r) 2 d.
1/3
Therefore, the distance is increased by a factor of 2 .
1.26. The volume of a cube of side r3 , and the volume of a sphere of  r . The
3

r is Vc radius r is V sp
ratio of the
volumes is:
3

, 3
Vc r 3
  . Vsp 4 3 4
r
3




4

,Bauer/Westfall: University Physics, 2E Chapter 1: Overview
The ratio of the volumes is independent of the value of r.

2
1.27. The surface area of a sphere is given by 4r . A cube of side length s has 6s2 . To
a surface area of determine s set the two surface areas equal:
 s  r
2 2 4r 2 2
6s 4r 6 .
3
30
1.28. The mass of Sun is 2 10 kg, the number of stars in the Milky Way is about 100
9 11 9 11
10 10 , the number of galaxies in the Universe is about 100 10 10 , and
27
the mass of an H-atom is 2 10 kg.
(a) The total mass of the Universe is roughly equal to the number of galaxies in the
Universe multiplied by the number of stars in a galaxy and the mass of the average
star: 30
 11 11  (11113
0)  52

Muniverse
52
(10 )(10 )(2 10 ) 2 10 kg 2 10 kg.
M universe 2 10 kg 79
n
(b) hydrogen  hydrog  10
27
M
en atoms 2  10 kg
3
1.29. The volume of 1 teaspoon 4.93
21
10 L , and the volume of water in the oceans is abou
is about
21
1.35 10 L.
1.35 10 L 2.74 1023 tsp
3
4.93 10 L/tsp
23
There are about 2.74 10 teaspoons of water in the Earth’s oceans.
1.30. The average arm-span of an adult human is d = 2 m. Therefore, with arms fully
2
extended, a person takes up a circular area of r  d / 2 (1 m)  m .
2 2 2


Since there are approximately 6.5 109 humans, the amount of land area required
for all humans to stand without being able to touch each other is
9 2 9 2 10 2
6.5 10 m () 6.5 10 m (3.14) 2.0 10 m . The area of the United States is
about
6 12 2
3.5 10 square miles or 9.110 m . In the United States there is almost five
hundred times the amount of land necessary for all of the population of Earth to
stand without touching each other.
10
1.31. The diameter of a gold atom is about 2.6 10 m. The circumference of the neck of an adult i
roughly
0.40 m. The number of gold atoms necessary to 1 link to make a necklace is given by:
circumference of neck 4.0 10 m 9
n  1.5 110
0
atoms. diameter of atom 2.6 10
m/atom
7
The Earth has a circumference at the equator of about 4.008 10 m . The number of
gold atoms necessary to link to make a chain that encircles the Earth is given by:
circumference of Earth 7
4.008 10 m
N  17 diameter
of a gold atom
1.5 10 atoms. 10
2.6 10 m
Since one mole of substance is 23
6.022 10 atoms , the necklace of gold atoms has
equivalent to about
1.5 10 9
atoms / 6.022 10 23
atoms/mol 2.5 10 15
moles of gold. The gold
chain has

1.5 10 17
atoms /  
6.022 10 atoms/mol
23

3

2.5 10 moles of gold.
7


1.32. The average dairy cow has a mass of about 1.0 10 kg. Estimate the cow’s
3
average density to be that of water, 1000. kg/m .
mass 3 3
volume   1.0 10 kg 1000. kg/m

5

, 3
1.0 m




6

,Bauer/Westfall: University Physics, 2E


Relate this to the volume of a sphere to
1/3
obtain the radius.
4 3

1/3
r  r  3V
  3 1.0
volume   0.62 m

 m
3


3 4  4 
 
A cow can be roughly approximated by a sphere with a radius of 0.62 m.
1.33. The mass of a head can be estimated first approximating its volume. A rough
approximation to the shape of a head is a cylinder. To obtain the volume from
the circumference, recall that the circumference is C 2r , which gives a radius
of r C / 2. The volume is then:
 C 2
V= r2  h 
 
h  C2 h.
  4
2 
The circumference of a head is about 55 cm = 0.55 m, and its height is about 20 cm
= 0.20 m. These values can be used 2
in the volume equation:
0.55 m
V 0.20 m4.8 103 m3 .
4
Assuming that the density of the head is about the same as the density of water,
the mass of a head can then be estimated as follows:
mass = density volume =  3
1.0 10 kg/m
3
 3 3
4.8 10 m  4.8 kg.
1.34. The average adult human head is roughly a cylinder 15 cm in diameter and 20.
cm in height. Assume about 1/3 of the surface area of the head is covered by hair.
1 1 2 2
A   A    2
2r 2rh  

2

r rh  7.5 cm 7.5
2


cm 20. cm 
3  
cylind
hair 3 3
3 er 2 2
4.32 10 cm 2
On average,2
the density of hair on the scalp is  2.3 10
hairs/cm . Therefore, you have A 
hair hair hair
hairs on your head.
A   4.32  1 0
2
cm
2
 2.3  1 0
2
hairs/cm
2



hair hair
 4
9.9 10 hairs.
Exercises
1.35. (a) Three (b) Four (c) One (d) Six (e) One (f) Two (g) Three
1.36. THINK: The known quantities are: F1 an F2 Bot F1 F2 are in the same
2.0031 N d 3.12 h and
N.
direction, and act on the same object. The total force acting on the object is Ftotal .
SKETCH:




 i.
RESEARCH: Forces that act in the same direction are summed, Ftotal F
 i F1 F2
SIMPLIFY: Ftotal F
CALCULATE Ftotal 2.0031 N 3.12 N 5.1231 N
:
ROUND: When adding (or subtracting), the precision of the result is limited by the least prec
value
used in the F1 is precise to four places after the F2 is precise to only two places
calculation. decimal and
after the decimal, so the result should be precise to two places after the decimal: Ftotal 5.12 N
DOUBLE-CHECK: This result is reasonable as it is greater than each of the
7

, individual forces acting on the object.




8