Solution Manual For The Physical Universe 2026 Release Konrad B. Krauskopf, Arthur Beiser; Emeritus and Elizabeth Shay Carter

Solution Manual For
The Physical Universe 2026 Release Konrad B. Krauskopf, Arthur Beiser; Emeritus and
Elizabeth Shay Carter
Chapters 1-19

Chapter 1


Notes on Math in
The Physical Universe

Physics and chemistry are quantitative sciences and some math is essential to understand and appreciate
them. How much math is appropriate for a given classroom is for the instructor to decide. The math level
of The Physical Universe is quite low, but it may still be too high in places for some students. Here is a
list of material that can be omitted without loss of continuity. In most cases the instructor may well
choose to outline the material in class without requiring the students to work out relevant problems.

2.2. Vectors. Pythagorean theorem. Can omit.
2.4. Distance, Time, and Acceleration. Although can be omitted, it is useful in providing the origin of
Eq. 2-14 and especially in deriving the KE formula in Sec. 3.3. But calculations need not be
required.
2.12. Circular Motion. Discuss, but need not require calculations.
5.2. Heat. Discuss Eq. 5-3, but need not require calculations.
5.7. The Gas Laws. Discuss, but need not require calculations.
6.3. Coulomb’s Law. Discuss, but need not require calculations.
6.19. Transformers. Discuss, but need not require calculations.
7.12. Refraction. Can omit calculations that involve index of refraction.
7.13. Lenses. Can omit ray tracing, but really quite easy.
7.17. Diffraction. Discuss resolving power, but need not require calculations.
8.7. Binding Energy. Discuss, but need not require calculations.
9.2. Photons. Keep quantum energy calculations, but can omit those that involve the photoelectric
effect.
10.17. Chemical Equations. Discuss, but need not require equation balancing.
12.3. The Mole. 12.4 Formula Units. Could omit entirely, but better to give at least brief discussion since
this is how chemical ideas meet the real world.
Chapter 13. Organic Chemistry. Discuss the various aspects of structural formulas, but need not require
solving problems that involve them.
14.2. Atmospheric Moisture. Discuss relative humidity, but need not require calculations using the graph.

, Answers to Even-Numbered
Exercises

14. The Copernican model, because in it the
CHAPTER ONE distances from the earth, and hence the
apparent brightnesses of the other planets,
Exercises: vary with time.

2. The reliance of the scientific method on 16. Only elliptical orbits agree with observa-
experiment and observation. tional data.

18. a. No explanation is possible in the
4. Because a model isolates the most
ptolemaic system, in which the stars are
important features of a complex
fixed at the same distance from the earth in
phenomenon, it may permit scientists to
a crystal ball that revolves around a
determine the fundamental origin of the
stationary earth.
phenomenon with out being confused by
b. In the copernican system the explanation
secondary details.
follows from the orbital motion of the earth
relative to stars at different distances away.
6. A year is the time the sun takes to complete
a circuit across the sky relative to the stars. 20. The earth would then be more flattened at
the poles and bulge to a greater extent at
the equator.
8. If the moon is seen near a particular star on
one evening, by the next evening it will be 22. Yesterday, because the length of the day
some distance east of that star. has been increasing steadily since the
earth’s formation.
10. a. A year does not correspond to a whole
number of days. In order that the seasons 24. The moon.
do not shift around the calendar, an extra
day must be added to every fourth year 26. (291 km)(0.621 mi/km)  181 mi
with further adjustments at longer intervals.
b. A year does not correspond to a whole 28. 1 mm  103 m so d  (104 )(103 m)  10 m
number of days. In order that the seasons and (10 m)(3.28 ft/m)  32.8 ft
do not shift around the calendar, an extra
day must be added to every fourth year
30. (20.0 m)(7.00 m)(2.00 m)(3.28 ft/m ) 
3
with further adjustments at longer intervals.
9.88  103 ft 3
12. These observations suggest that the mem-
bers of the solar system all lie in or near a 32. 2 m 35 s  (155 s)/(3600 s/h )  0.0431 h;
plane not far from the earth’s equator and 1 mi  1.61 km, speed  (1.61 km)/
that all move in the same direction about (0.0431 h )  37.4 km/h
the sun or, in the case of the moon, about
the earth. 34. 42; 7.5 105 ; 3.0 105

, 28. v  2 gh  15.3 m/s.
CHAPTER TWO
30. h  12 gt 2  78.4 m.
Exercises:
32. Time of rise  time of fall  t  v /g  1.0 s.
2. t  d /v  0.029 s.
Hence the total time of flight  2t  2.0 s.
4. The snake covers 105 m in t  d /v  70 s.
Therefore your speed must be greater than
34. a. t  v /g  2.04 s.
v  100 m / 70 s   1.43 m/s .
c. h  12 gt 2  20.4 m.
6. 30 lb; 0.
36. t  d /v, h  12 gt 2  12 gd 2 /v 2  0.10 m 
8. a. Directly across the river. 10 cm.
b. t  d /v(boat )  0.1875 h  11 min 15 s .
c. d  v  river  t  0.94 km . 38. v  vert   2 gh  19.8 m/s; v  horiz  
30 m/s; v  v(vert)2  v(horiz) 2 
10. F  F  F  7.1 tons .
1
2
2
2
35.9 m/s.
12. No. An example is the curved path of a ball 40. The time of fall is t  2h/g  10.1 s . In
thrown at an angle with the ground.
this time the pump will have moved
14. v2 – at  15 m/s. horizontally d = vt = 606 m.

42. a. The tensions are the same.
16. a. a  (v f  v0 ) / t  3.5 m/s2 . b. The front coupling is under greater
b. t  (v f  v0 ) / a  5.71 s. tension because it is accelerating a greater
c. t  (v f  v0 ) / a  2.86 s. mass.

44. a. v  55.6 m/s, a  v /t  18.5 m/s 2 
18. a. d  v1t  at , a  2d /t – 2v1 /t 
1 2 2
2 1.89 g.
–0.178 m/s 2 . b. F  ma  222 kN.
b. v2  v1  at  6.65 m/s.
46. m  F /a  4 kg.
20. Yes. a. F  ma  4 N.
b. F  ma  40 N.
22. T  total time of flight  2 2h/g . Since
g is smaller on Venus than on the earth, T 48. F  ma  mv f /t , t  mv f /F  0.015 s.
will be greater and the ball will return to
the ground later. 50. v2  v1  4.17 m/s, F  ma  m  v2  v1  /t 
24. a. The crate appears to move vertically 667 N.
downward because both the crate and the
observer have the same horizontal speed. 52. F  w  mg , a  F /m  g .
b. The crate appears to move in a curved
path downward, as in Fig. 2-12. 54. a  F /m  (m1 – m2 ) g / (m1  m2 )  1.96
m/s2 .
26. a. The distance remains the same.
b. The distance increases.

, 56. A force of ma is needed in addition to the 74. Fc  mv2 /r , v  Fc r / m  20 m/s.
force w  mg needed just to lift the box
without acceleration. Hence F  mg  ma  76. a. Fc  mv 2 /r  4.17  105 N.
59 N. b. w  mg  490 N ; Fc /w  850.
c. No.
58. a. a  F /m  9 m/s , so the elevator is
2

falling with a downward acceleration of
78. F  mg  mv 2 /r ,
0.8 m/s2 .
b. The elevator is either stationary or rising r  mv 2 /( F  mg )  637 m.
or falling with constant speed.
80. The force would be 4 times what it is today.
c. a  F /m  10 m/s 2 , so the elevator is
rising with an upward acceleration of 82. The stone’s mass would be the same but its
0.2 m/s 2 . weight would be zero.

60. a. F  mg  ma  648 N.
84. a. d  12 at 2  1.4 mm/s.
b. F  mg – ma  528 N.
b. Since 1 y  (365 days/y)(24 hr/day)
c. F  mg  588 N.
(60 min/hr )(60 s/min ) , this is 44 m/y .
d. F  mg  588 N.
c. The moon never comes closer to the
earth because its ―falling‖ causes it to move
62. The first law is a special case of the second
in an orbit around the earth.
law, because when F = 0, a = 0. There is no
connection between the second and third
laws. 86. a. F  Gm1m2 /R 2  1.67  108 N.
b. 1.67 108 N.
64. a. The upward forces are the reaction
c. a2  F /m2  8.3  109 m/s 2 ;
forces of the road on the car’s tires.
b. The downward forces are these reaction a5  F /m5  3.3 109 m/s 2 .
forces plus the force of gravity on the car. 88. R  m G / F  5.2 m.
66. The second procedure is more likely to
break the string because here the tension in 90. The earth rotates from west to east. Hence
the string is twice as great with the reaction the satellite sent eastward will have its
force exerted by the tree being equal and launching speed increased because of the
opposite to the pull of the two children. earth’s rotation, and the one sent westward
will have its launching speed decreased.
68. The equator; the poles. The satellite sent eastward will therefore
have the larger orbit.
70. At the bottom of the circle, since here the
string must support all of the ball’s weight
as well as provide the centripetal force on
the ball.

72. The outward pull is due to the reaction
CHAPTER THREE
force of the stone that arises in response to
the inward force applied to it, which is Exercises:
what causes it to move in a circle. When
the string is released, there is no longer 2. No work is done by a net force acting on a
any inward force on the stone, and it moving body when the force is perpendicu-
proceeds along in a straight line as lar to the direction of the body’s motion.
predicted by the first law of motion.