Instructor’s Manual with
Test Bank
For
The Physical Universe
Eighteenth Edition
Konrad Krauskopf
Arthur Beiser
Elizabeth Shay Carter
Part 1: Instructor Manual (Chapter 1-19) Page 1-33
Part 2: Test Bank (Chapter 1-19) Page 34-250
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
,Part 1: Instructor Manual (Chapter 1-19)
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 important
ptolemaic system, in which the stars are
features of a complex phenomenon, it
fixed at the same distance from the earth in
may permit scientists to determine the
a crystal ball that revolves around a
fundamental origin of the phenomenon
stationary earth.
with out being confused by secondary
b. In the copernican system the explanation
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)/
(0.0431 h ) = 37.4 km/h
that all move in the same direction about
the sun or, in the case of the moon, about
the earth. 34. 42; 7.5 105 ; 3.0 105
2 Answers to Even-Numbered Exercises
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, 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 =
10 cm.
8. a. Directly across the river.
b. t = d /v(boat ) = 0.1875 h = 11 min 15 s . 38. v ( vert ) = 2 gh = 19.8 m/s; v ( horiz ) =
c. d = v ( river ) t = 0.94 km .
30 m/s; v = v(vert) 2 + v(horiz) 2 =
35.9 m/s.
10. F = F12 + F22 = 7.1 tons .
40. The time of fall is t = 2h /g = 10.1 s . In
12. No. An example is the curved path of a ball
this time the pump will have moved
thrown at an angle with the ground.
horizontally d = vt = 606 m.
14. v2 – at = 15 m/s.
42. a. The tensions are the same.
b. The front coupling is under greater
16. a. a = (v f − v0 ) / t = −3.5 m/s2 . tension because it is accelerating a greater
b. t = (v f − v0 ) / a = 5.71 s. mass.
c. t = (v f − v0 ) / a = 2.86 s.
44. a. v = 55.6 m/s, a = v /t = 18.5 m/s 2 =
1.89 g.
18. a. d = v1t + 12 at , a = 2d /t – 2v1 /t =
2 2
b. F = ma = 222 kN.
–0.178 m/s 2 .
b. v2 = v1 + at = 6.65 m/s. 46. m = F /a = 4 kg.
a. F = ma = 4 N.
20. Yes. b. F = ma = 40 N.
22. T = total time of flight = 2 2h /g . Since 48. F = ma = mv f /t , t = mv f /F = 0.015 s.
g is smaller on Venus than on the earth, T
will be greater and the ball will return to 50. v2 − v1 = −4.17 m/s, F = ma = m ( v2 − v1 ) /t =
the ground later.
−667 N.
24. a. The crate appears to move vertically
downward because both the crate and the 52. F = w = mg , a = F /m = g.
observer have the same horizontal speed.
b. The crate appears to move in a curved 54. a = F /m = (m1 – m2 ) g / (m1 + m2 ) = 1.96
path downward, as in Fig. 2-12. m/s 2 .
26. a. The distance remains the same. 56. A force of ma is needed in addition to the
b. The distance increases. force w = mg needed just to lift the box
Answers to Even-Numbered Exercises 3
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, without acceleration. Hence F = mg + ma = 74. Fc = mv2 /r , v = Fc r / m = 20 m/s.
59 N.
76. a. Fc = mv 2 /r = 4.17 105 N.
58. a. a = F /m = 9 m/s , so the elevator is
2
b. w = mg = 490 N ; Fc /w = 850.
falling with a downward acceleration of
c. No.
0.8 m/s 2 .
b. The elevator is either stationary or rising 78. F = mg + mv 2 /r ,
or falling with constant speed.
r = mv 2 /( F − mg ) = 637 m.
c. a = F /m = 10 m/s 2 , so the elevator is
rising with an upward acceleration of
80. The force would be 4 times what it is today.
0.2 m/s 2 .
82. The stone’s mass would be the same but its
60. a. F = mg + ma = 648 N. weight would be zero.
b. F = mg – ma = 528 N.
c. F = mg = 588 N. 84. a. d = 12 at 2 = 1.4 mm/s.
d. F = mg = 588 N. b. Since 1 y = (365 days/y)(24 hr/day)
(60 min/hr)(60 s/min ) , this is 44 m/y .
62. The first law is a special case of the second
law, because when F = 0, a = 0. There is no c. The moon never comes closer to the
connection between the second and third earth because its “falling” causes it to move
laws. in an orbit around the earth.
64. a. The upward forces are the reaction forces 86. a. F = Gm1m2 /R 2 = 1.67 10−8 N.
of the road on the car’s tires.
b. 1.67 10−8 N.
b. The downward forces are these reaction
forces plus the force of gravity on the car. c. a2 = F /m2 = 8.3 10−9 m/s 2 ;
a5 = F /m5 = 3.3 10−9 m/s 2 .
66. The second procedure is more likely to
break the string because here the tension in 88. R = m G / F = 5.2 m.
the string is twice as great with the reaction
force exerted by the tree being equal and 90. The earth rotates from west to east. Hence
opposite to the pull of the two children. the satellite sent eastward will have its
launching speed increased because of the
68. The equator; the poles. earth’s rotation, and the one sent westward
will have its launching speed decreased.
70. At the bottom of the circle, since here the The satellite sent eastward will therefore
string must support all of the ball’s weight have the larger orbit.
as well as provide the centripetal force on
the ball.
72. The outward pull is due to the reaction
force of the stone that arises in response to
the inward force applied to it, which is CHAPTER THREE
what causes it to move in a circle. When
the string is released, there is no longer any Exercises:
inward force on the stone, and it proceeds
along in a straight line as predicted by the 2. No work is done by a net force acting on a
first law of motion. moving body when the force is perpendicu-
lar to the direction of the body’s motion.
Answers to Even-Numbered Exercises 4
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
Test Bank
For
The Physical Universe
Eighteenth Edition
Konrad Krauskopf
Arthur Beiser
Elizabeth Shay Carter
Part 1: Instructor Manual (Chapter 1-19) Page 1-33
Part 2: Test Bank (Chapter 1-19) Page 34-250
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
,Part 1: Instructor Manual (Chapter 1-19)
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 important
ptolemaic system, in which the stars are
features of a complex phenomenon, it
fixed at the same distance from the earth in
may permit scientists to determine the
a crystal ball that revolves around a
fundamental origin of the phenomenon
stationary earth.
with out being confused by secondary
b. In the copernican system the explanation
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)/
(0.0431 h ) = 37.4 km/h
that all move in the same direction about
the sun or, in the case of the moon, about
the earth. 34. 42; 7.5 105 ; 3.0 105
2 Answers to Even-Numbered Exercises
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, 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 =
10 cm.
8. a. Directly across the river.
b. t = d /v(boat ) = 0.1875 h = 11 min 15 s . 38. v ( vert ) = 2 gh = 19.8 m/s; v ( horiz ) =
c. d = v ( river ) t = 0.94 km .
30 m/s; v = v(vert) 2 + v(horiz) 2 =
35.9 m/s.
10. F = F12 + F22 = 7.1 tons .
40. The time of fall is t = 2h /g = 10.1 s . In
12. No. An example is the curved path of a ball
this time the pump will have moved
thrown at an angle with the ground.
horizontally d = vt = 606 m.
14. v2 – at = 15 m/s.
42. a. The tensions are the same.
b. The front coupling is under greater
16. a. a = (v f − v0 ) / t = −3.5 m/s2 . tension because it is accelerating a greater
b. t = (v f − v0 ) / a = 5.71 s. mass.
c. t = (v f − v0 ) / a = 2.86 s.
44. a. v = 55.6 m/s, a = v /t = 18.5 m/s 2 =
1.89 g.
18. a. d = v1t + 12 at , a = 2d /t – 2v1 /t =
2 2
b. F = ma = 222 kN.
–0.178 m/s 2 .
b. v2 = v1 + at = 6.65 m/s. 46. m = F /a = 4 kg.
a. F = ma = 4 N.
20. Yes. b. F = ma = 40 N.
22. T = total time of flight = 2 2h /g . Since 48. F = ma = mv f /t , t = mv f /F = 0.015 s.
g is smaller on Venus than on the earth, T
will be greater and the ball will return to 50. v2 − v1 = −4.17 m/s, F = ma = m ( v2 − v1 ) /t =
the ground later.
−667 N.
24. a. The crate appears to move vertically
downward because both the crate and the 52. F = w = mg , a = F /m = g.
observer have the same horizontal speed.
b. The crate appears to move in a curved 54. a = F /m = (m1 – m2 ) g / (m1 + m2 ) = 1.96
path downward, as in Fig. 2-12. m/s 2 .
26. a. The distance remains the same. 56. A force of ma is needed in addition to the
b. The distance increases. force w = mg needed just to lift the box
Answers to Even-Numbered Exercises 3
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
, without acceleration. Hence F = mg + ma = 74. Fc = mv2 /r , v = Fc r / m = 20 m/s.
59 N.
76. a. Fc = mv 2 /r = 4.17 105 N.
58. a. a = F /m = 9 m/s , so the elevator is
2
b. w = mg = 490 N ; Fc /w = 850.
falling with a downward acceleration of
c. No.
0.8 m/s 2 .
b. The elevator is either stationary or rising 78. F = mg + mv 2 /r ,
or falling with constant speed.
r = mv 2 /( F − mg ) = 637 m.
c. a = F /m = 10 m/s 2 , so the elevator is
rising with an upward acceleration of
80. The force would be 4 times what it is today.
0.2 m/s 2 .
82. The stone’s mass would be the same but its
60. a. F = mg + ma = 648 N. weight would be zero.
b. F = mg – ma = 528 N.
c. F = mg = 588 N. 84. a. d = 12 at 2 = 1.4 mm/s.
d. F = mg = 588 N. b. Since 1 y = (365 days/y)(24 hr/day)
(60 min/hr)(60 s/min ) , this is 44 m/y .
62. The first law is a special case of the second
law, because when F = 0, a = 0. There is no c. The moon never comes closer to the
connection between the second and third earth because its “falling” causes it to move
laws. in an orbit around the earth.
64. a. The upward forces are the reaction forces 86. a. F = Gm1m2 /R 2 = 1.67 10−8 N.
of the road on the car’s tires.
b. 1.67 10−8 N.
b. The downward forces are these reaction
forces plus the force of gravity on the car. c. a2 = F /m2 = 8.3 10−9 m/s 2 ;
a5 = F /m5 = 3.3 10−9 m/s 2 .
66. The second procedure is more likely to
break the string because here the tension in 88. R = m G / F = 5.2 m.
the string is twice as great with the reaction
force exerted by the tree being equal and 90. The earth rotates from west to east. Hence
opposite to the pull of the two children. the satellite sent eastward will have its
launching speed increased because of the
68. The equator; the poles. earth’s rotation, and the one sent westward
will have its launching speed decreased.
70. At the bottom of the circle, since here the The satellite sent eastward will therefore
string must support all of the ball’s weight have the larger orbit.
as well as provide the centripetal force on
the ball.
72. The outward pull is due to the reaction
force of the stone that arises in response to
the inward force applied to it, which is CHAPTER THREE
what causes it to move in a circle. When
the string is released, there is no longer any Exercises:
inward force on the stone, and it proceeds
along in a straight line as predicted by the 2. No work is done by a net force acting on a
first law of motion. moving body when the force is perpendicu-
lar to the direction of the body’s motion.
Answers to Even-Numbered Exercises 4
© McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
