INSTRUCTOR’S SOLUTIONS MANUAL
PHYSICS
5TH EDITION
CHAPTER NO. 01: INTRODUCTION
Conceptual Questions
1. Knowledge of physics is important for a full understanding of many scientific disciplines, such as chemistry,
biology, and geology. Furthermore, much of our current technology can only be understood with knowledge of
the underlying laws of physics. In the search for more efficient and environmentally safe sources of energy, for
example, physics is essential. Also, many people study physics for the sense of fulfillment that comes with
learning about the world we inhabit.
2. Without precise definitions of words for scientific use, unambiguous communication of findings and ideas would
be impossible.
3. Even when simplified models do not exactly match real conditions, they can still provide insight into the features
of a physical system. Often a problem would become too complicated if one attempted to match the real
conditions exactly, and an approximation can yield a result that is close enough to the exact one to still be useful.
4. After solving a problem, it is a good idea to check that the solution is reasonable and makes intuitive sense. Do the
units work out correctly? In the symbolic version of the answer, before numbers are substituted, would the
expression change in a reasonable way if each parameter were made larger? Smaller? Very much larger or
smaller? It may also be useful to explore other possible methods of solution as a check on the validity of the first.
A good student thinks of a framework of ideas and skills that she is constructing for herself. The problem solution
may extend or strengthen this framework
5. Scientific notation eliminates the need to write many zeros in very large or small numbers, and to count them.
Also, the number of significant digits is indicated unambiguously when a quantity is written this way.
6. In scientific notation the decimal point is often placed after the first (leftmost) numeral. The number of digits
written equals the number of significant figures.
7. Not all of the significant digits are known definitely. The last (rightmost) digit, called the least significant digit, is
an estimate and is less definitely known than the others.
8. It is important to write a quantity with the correct number of significant figures so that we can indicate how
precisely a quantity is known and so that we do not mislead the reader by writing digits that are not at all known
to be correct.
9. The kilogram, meter, and second are three of the base units used in the SI, the international system of units.
10. The international system SI uses a well-defined set of internationally agreed upon standard units and makes
measurements in terms of these units, their combinations, and their powers of ten. The U.S. customary system
contains units that are primarily of historical origin and are not based upon powers of ten. As a result of this
international acceptance and of the ease of manipulation that comes from dealing with powers of ten, scientists
around the world prefer to use SI.
, 11. Fathoms, kilometers, miles, and inches are units with the dimension length. Grams and kilograms are units with
the dimension mass. Years, months, and seconds are units with the dimension time.
12. The first step toward successfully solving almost any physics problem is to thoroughly read the question and
obtain a precise understanding of the scenario. The second step is to visualize the problem, often making a quick
sketch to outline the details of the situation and the known parameters.
13. Trends in a set of data are often the most interesting aspect of the outcome of an experiment. Such trends are more
apparent when data is plotted graphically rather than listed in numerical tables.
14. The statement gives a number for the speed of sound in air, but fails to indicate the units used for the
measurement. Without units, the reader cannot relate the speed to one given in familiar units such as km/s.
Multiple-Choice Questions
1. (b) 2. (b) 3. (a) 4. (c) 5. (d) 6. (d) 7. (b) 8. (d) 9. (b) 10. (c)
Problems
1. Strategy The new fence will be 100% + 37% = 137% of the height of the old fence.
Solution Find the height of the new fence. 1.37 × 1.8 m = 2.5 m
Discussion. Long ago you were told that 37% of 1.8 is 0.37 times 1.8.
2. Strategy. Relate the surface area A to the radius r using A = 4π r 2 .
Solution. Find the ratio of the new radius to the old.
A1 4π r12 and
= = π r22 2.0
A2 4= = A1 2.0(4π r12 ).
4π r22 = 2.0(4π r12 )
r22 = 2.0r12
2
r2
= 2.0
r1
r2
= =2.0 1.4
r1
The radius of the balloon increases by a factor of 1.4.
3. Strategy Relate the surface area A to the radius r using A = 4π r 2 .
Solution Find the ratio of the new radius to the old.
A1 = 4π r12 and A2 = 4π r22 = 1.160 A1 = 1.160(4π r12 ).
4π r22 = 1.160(4π r12 )
r22 = 1.160r12
2
r2
= 1.160
r1
r2
= 1.160 = 1.077
r1
The radius of the balloon increases by 7.7%.
Discussion. Because the surface area is proportional to the square of the radius, the percentage change in radius is
smaller than the percentage change in area—in fact, a bit less than half as large. The factor of 4π divides out and
, plays no part in determining the answer. The answer just comes from the proportionality of area to the square of
the radius. The circumference also increases by 7.7%.
4. Strategy To find the factor by which Samantha’s height increased, divide her new height by her old height.
Subtract 1 from this value and multiply by 100 % to find the percentage increase.
Solution Find the factor.
1.65 m
= 1.10
1.50 m
Find the percentage.
1.10 − 1 = 0.10, so the percent increase is 10 % .
5. Strategy To find the factor by which the metabolic rate of a 70 kg human exceeds that of a 5.0 kg cat, use a ratio.
3/ 4 3/ 4
m 70
Solution Find the factor: h = = 7.2
mc 5.0
Discussion. The proportionality could be written into an equation as (Metabolic rate) = K (Body mass)3/4 where
K is a proportionality constant (with very odd units). If we write down this equation for a human and again for a
cat, and then divide the two, the K divides out and we obtain the quantity (mh/mc)3/4 that we evaluated. Get used
to using your calculator to follow the order of operations without your having to re-enter any numbers. On my
calculator I type 70 ÷ 5 = ^ 0.75 = .
6. Strategy Let X be the original value of the index. Follow what happens to it.
Solution Find the net percentage change in the index for the two days.
final value = (originalvalue) × (first day change factor) × (second day change factor) =
= X × (1 + 0.0500) × (1 − 0.0500) = 0.9975X
The net percentage change is (0.9975 − 1) × 100% = −0.25%, or down 0.25% .
The index starts higher on day 2 than on day 1, so the decrease on the second day is five percent of a larger
number. This decrease therefore exceeds the increase on the previous day.
7. Strategy Recall that area has dimensions of length squared.
Solution Find the ratio of the area of the park as represented on the map to the area of the actual park.
map length 1 map area
= = 10−4 , so = (10−4 )2 = 10−8 .
actual length 10,000 actual area
8. Strategy We use the given equation to form an equation of ratios comparing heat transferred and thickness.
Solution Represent the first trial, with 86.0 J going through a pad 3.40 cm thick, with the symbols
Q1/∆t1 =κ1A1 ∆T1/d1 . Now the new trial in this problem 8 is represented by Q8/∆t8 =κ8A8 ∆T8/d8 . Dividing the
two equations gives
Q8 ∆t1 κ A ∆T d1
= 8 8 8
∆t8Q1 d8 κ1 A1∆T1
Q8 d
We are given ∆t1 = ∆t8 and ∆T1 = ∆T8 and A1 = A8 and κ1 = κ8 . Then = 1 so
Q1 d8
d1 3.40 cm
Q8 Q=
= 1 86.0 J = 56.2 J
d8 5.20 cm
, 9. Strategy We use the given equation to form an equation of ratios—a proportion—comparing heat transferred,
temperature difference, and thickness
Solution Represent the first trial, with a temperature difference of 37.0°C − 2.0°C = 35.0°C driving 86.0 J to go
through a pad 3.40 cm thick, with the symbols Q1/∆t1 =κ1A1 ∆T1/d1 . Now the new trial in this problem 9 is
represented by Q9/∆t9 =κ9A9 ∆T9/d9 . Dividing the two equations gives
Q9 ∆t1 κ 9 A9 ∆T9 d1
=
∆t9Q1 d9 κ1 A1∆T1
We are given ∆t1 = ∆t9 (same duration), A1 = A9 (same face area), and κ1 = κ9 (same material). Then the full
proportion reduces to
Q9 d1 ∆T9
=
Q1 d9 ∆T1
and
Q1∆T9 86.0 J × 48.0°C
=d9 d=
1 3.40 cm = 8.53 cm
Q9 ∆T1 47.0 J × 35.0°C
Discussion. We could alternatively phrase the solution in terms of ratios (fractions or factors of change) and
proportionalities (patterns of change). The original equation implies that the heat transferred is directly
proportional to the temperature difference and inversely proportional to the thickness of the conductor. The
conductor can equally well be called an insulator. Making the temperature difference 48 degrees instead of 35
degrees would by itself increase the heat flow by a factor of 48/35. To make the actual transfer of heat smaller
instead of larger, the thickness of insulation would have to first be increased by this factor. And then to make the
heat 47 J instead of 86 J, the insulation thickness would need to be further increased by the factor 86/47. Then
the answer 3.40 cm (86/47)(48/35) = 8.53 cm has been assembled.
10. Strategy We use the given equation about heat transfer to form an equation of ratios—a proportion—comparing
time and thickness.
Solution Represent the first trial, with a pad 3.40 cm thick, with the symbols Q1/∆t1 =κ1A1 ∆T1/d1 . Now the
new trial in this problem 10 is represented by Q10/∆t10 =κ10A10 ∆T10/d10 . Dividing the two equations gives
Q10 ∆t1 κ10 A10 ∆T10 d1
=
∆t10Q1 d10 κ1 A1∆T1
We are given ∆T1 = ∆T10 and A1 = A10 and κ1 = κ10 and Q1 = Q10. The unknown we can identify not as any
single symbol but as the factor or ratio ∆t10/∆t1 . For it we have
∆t10/∆t1 = d10/d1 = (4.10 cm)/(3.40 cm) = 1.21 .
11. Strategy The area of a rectangular poster is given by A = w. Let the original and final areas be A1 = 1w1 and
A2 = 2 w2 , respectively.
Solution Calculate the percentage reduction of the area.
A2 =
= 2 w2 (0.8001 )(0.800
= w1 ) 0.640=1w1 0.640 A1
A1 − A2 A1 − 0.640 A1
× 100% = × 100% = 36.0%
A1 A1
Discussion. Twenty-percent increases in the two independent factors would contribute to a 44% increase in area,
from 1.20 × 1.20 = 1.44. Twenty-percent decreases in length and width contribute together to a 36% decrease.
Proportional reasoning is so profound that it applies to a triangular, round, star-shaped, or dragon-shaped poster,
as long as the final shape is geometrically similar to the original and length and width are interpreted as two
perpendicular maximum distances across the poster.
PHYSICS
5TH EDITION
CHAPTER NO. 01: INTRODUCTION
Conceptual Questions
1. Knowledge of physics is important for a full understanding of many scientific disciplines, such as chemistry,
biology, and geology. Furthermore, much of our current technology can only be understood with knowledge of
the underlying laws of physics. In the search for more efficient and environmentally safe sources of energy, for
example, physics is essential. Also, many people study physics for the sense of fulfillment that comes with
learning about the world we inhabit.
2. Without precise definitions of words for scientific use, unambiguous communication of findings and ideas would
be impossible.
3. Even when simplified models do not exactly match real conditions, they can still provide insight into the features
of a physical system. Often a problem would become too complicated if one attempted to match the real
conditions exactly, and an approximation can yield a result that is close enough to the exact one to still be useful.
4. After solving a problem, it is a good idea to check that the solution is reasonable and makes intuitive sense. Do the
units work out correctly? In the symbolic version of the answer, before numbers are substituted, would the
expression change in a reasonable way if each parameter were made larger? Smaller? Very much larger or
smaller? It may also be useful to explore other possible methods of solution as a check on the validity of the first.
A good student thinks of a framework of ideas and skills that she is constructing for herself. The problem solution
may extend or strengthen this framework
5. Scientific notation eliminates the need to write many zeros in very large or small numbers, and to count them.
Also, the number of significant digits is indicated unambiguously when a quantity is written this way.
6. In scientific notation the decimal point is often placed after the first (leftmost) numeral. The number of digits
written equals the number of significant figures.
7. Not all of the significant digits are known definitely. The last (rightmost) digit, called the least significant digit, is
an estimate and is less definitely known than the others.
8. It is important to write a quantity with the correct number of significant figures so that we can indicate how
precisely a quantity is known and so that we do not mislead the reader by writing digits that are not at all known
to be correct.
9. The kilogram, meter, and second are three of the base units used in the SI, the international system of units.
10. The international system SI uses a well-defined set of internationally agreed upon standard units and makes
measurements in terms of these units, their combinations, and their powers of ten. The U.S. customary system
contains units that are primarily of historical origin and are not based upon powers of ten. As a result of this
international acceptance and of the ease of manipulation that comes from dealing with powers of ten, scientists
around the world prefer to use SI.
, 11. Fathoms, kilometers, miles, and inches are units with the dimension length. Grams and kilograms are units with
the dimension mass. Years, months, and seconds are units with the dimension time.
12. The first step toward successfully solving almost any physics problem is to thoroughly read the question and
obtain a precise understanding of the scenario. The second step is to visualize the problem, often making a quick
sketch to outline the details of the situation and the known parameters.
13. Trends in a set of data are often the most interesting aspect of the outcome of an experiment. Such trends are more
apparent when data is plotted graphically rather than listed in numerical tables.
14. The statement gives a number for the speed of sound in air, but fails to indicate the units used for the
measurement. Without units, the reader cannot relate the speed to one given in familiar units such as km/s.
Multiple-Choice Questions
1. (b) 2. (b) 3. (a) 4. (c) 5. (d) 6. (d) 7. (b) 8. (d) 9. (b) 10. (c)
Problems
1. Strategy The new fence will be 100% + 37% = 137% of the height of the old fence.
Solution Find the height of the new fence. 1.37 × 1.8 m = 2.5 m
Discussion. Long ago you were told that 37% of 1.8 is 0.37 times 1.8.
2. Strategy. Relate the surface area A to the radius r using A = 4π r 2 .
Solution. Find the ratio of the new radius to the old.
A1 4π r12 and
= = π r22 2.0
A2 4= = A1 2.0(4π r12 ).
4π r22 = 2.0(4π r12 )
r22 = 2.0r12
2
r2
= 2.0
r1
r2
= =2.0 1.4
r1
The radius of the balloon increases by a factor of 1.4.
3. Strategy Relate the surface area A to the radius r using A = 4π r 2 .
Solution Find the ratio of the new radius to the old.
A1 = 4π r12 and A2 = 4π r22 = 1.160 A1 = 1.160(4π r12 ).
4π r22 = 1.160(4π r12 )
r22 = 1.160r12
2
r2
= 1.160
r1
r2
= 1.160 = 1.077
r1
The radius of the balloon increases by 7.7%.
Discussion. Because the surface area is proportional to the square of the radius, the percentage change in radius is
smaller than the percentage change in area—in fact, a bit less than half as large. The factor of 4π divides out and
, plays no part in determining the answer. The answer just comes from the proportionality of area to the square of
the radius. The circumference also increases by 7.7%.
4. Strategy To find the factor by which Samantha’s height increased, divide her new height by her old height.
Subtract 1 from this value and multiply by 100 % to find the percentage increase.
Solution Find the factor.
1.65 m
= 1.10
1.50 m
Find the percentage.
1.10 − 1 = 0.10, so the percent increase is 10 % .
5. Strategy To find the factor by which the metabolic rate of a 70 kg human exceeds that of a 5.0 kg cat, use a ratio.
3/ 4 3/ 4
m 70
Solution Find the factor: h = = 7.2
mc 5.0
Discussion. The proportionality could be written into an equation as (Metabolic rate) = K (Body mass)3/4 where
K is a proportionality constant (with very odd units). If we write down this equation for a human and again for a
cat, and then divide the two, the K divides out and we obtain the quantity (mh/mc)3/4 that we evaluated. Get used
to using your calculator to follow the order of operations without your having to re-enter any numbers. On my
calculator I type 70 ÷ 5 = ^ 0.75 = .
6. Strategy Let X be the original value of the index. Follow what happens to it.
Solution Find the net percentage change in the index for the two days.
final value = (originalvalue) × (first day change factor) × (second day change factor) =
= X × (1 + 0.0500) × (1 − 0.0500) = 0.9975X
The net percentage change is (0.9975 − 1) × 100% = −0.25%, or down 0.25% .
The index starts higher on day 2 than on day 1, so the decrease on the second day is five percent of a larger
number. This decrease therefore exceeds the increase on the previous day.
7. Strategy Recall that area has dimensions of length squared.
Solution Find the ratio of the area of the park as represented on the map to the area of the actual park.
map length 1 map area
= = 10−4 , so = (10−4 )2 = 10−8 .
actual length 10,000 actual area
8. Strategy We use the given equation to form an equation of ratios comparing heat transferred and thickness.
Solution Represent the first trial, with 86.0 J going through a pad 3.40 cm thick, with the symbols
Q1/∆t1 =κ1A1 ∆T1/d1 . Now the new trial in this problem 8 is represented by Q8/∆t8 =κ8A8 ∆T8/d8 . Dividing the
two equations gives
Q8 ∆t1 κ A ∆T d1
= 8 8 8
∆t8Q1 d8 κ1 A1∆T1
Q8 d
We are given ∆t1 = ∆t8 and ∆T1 = ∆T8 and A1 = A8 and κ1 = κ8 . Then = 1 so
Q1 d8
d1 3.40 cm
Q8 Q=
= 1 86.0 J = 56.2 J
d8 5.20 cm
, 9. Strategy We use the given equation to form an equation of ratios—a proportion—comparing heat transferred,
temperature difference, and thickness
Solution Represent the first trial, with a temperature difference of 37.0°C − 2.0°C = 35.0°C driving 86.0 J to go
through a pad 3.40 cm thick, with the symbols Q1/∆t1 =κ1A1 ∆T1/d1 . Now the new trial in this problem 9 is
represented by Q9/∆t9 =κ9A9 ∆T9/d9 . Dividing the two equations gives
Q9 ∆t1 κ 9 A9 ∆T9 d1
=
∆t9Q1 d9 κ1 A1∆T1
We are given ∆t1 = ∆t9 (same duration), A1 = A9 (same face area), and κ1 = κ9 (same material). Then the full
proportion reduces to
Q9 d1 ∆T9
=
Q1 d9 ∆T1
and
Q1∆T9 86.0 J × 48.0°C
=d9 d=
1 3.40 cm = 8.53 cm
Q9 ∆T1 47.0 J × 35.0°C
Discussion. We could alternatively phrase the solution in terms of ratios (fractions or factors of change) and
proportionalities (patterns of change). The original equation implies that the heat transferred is directly
proportional to the temperature difference and inversely proportional to the thickness of the conductor. The
conductor can equally well be called an insulator. Making the temperature difference 48 degrees instead of 35
degrees would by itself increase the heat flow by a factor of 48/35. To make the actual transfer of heat smaller
instead of larger, the thickness of insulation would have to first be increased by this factor. And then to make the
heat 47 J instead of 86 J, the insulation thickness would need to be further increased by the factor 86/47. Then
the answer 3.40 cm (86/47)(48/35) = 8.53 cm has been assembled.
10. Strategy We use the given equation about heat transfer to form an equation of ratios—a proportion—comparing
time and thickness.
Solution Represent the first trial, with a pad 3.40 cm thick, with the symbols Q1/∆t1 =κ1A1 ∆T1/d1 . Now the
new trial in this problem 10 is represented by Q10/∆t10 =κ10A10 ∆T10/d10 . Dividing the two equations gives
Q10 ∆t1 κ10 A10 ∆T10 d1
=
∆t10Q1 d10 κ1 A1∆T1
We are given ∆T1 = ∆T10 and A1 = A10 and κ1 = κ10 and Q1 = Q10. The unknown we can identify not as any
single symbol but as the factor or ratio ∆t10/∆t1 . For it we have
∆t10/∆t1 = d10/d1 = (4.10 cm)/(3.40 cm) = 1.21 .
11. Strategy The area of a rectangular poster is given by A = w. Let the original and final areas be A1 = 1w1 and
A2 = 2 w2 , respectively.
Solution Calculate the percentage reduction of the area.
A2 =
= 2 w2 (0.8001 )(0.800
= w1 ) 0.640=1w1 0.640 A1
A1 − A2 A1 − 0.640 A1
× 100% = × 100% = 36.0%
A1 A1
Discussion. Twenty-percent increases in the two independent factors would contribute to a 44% increase in area,
from 1.20 × 1.20 = 1.44. Twenty-percent decreases in length and width contribute together to a 36% decrease.
Proportional reasoning is so profound that it applies to a triangular, round, star-shaped, or dragon-shaped poster,
as long as the final shape is geometrically similar to the original and length and width are interpreted as two
perpendicular maximum distances across the poster.
