,Solution manual for Physics 12th Edition John
D. Cutnell
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,CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the
last vector.
2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to
the shortest distance between the tail of A and the head of B. Thus, R is less than the
magnitude (length) of A plus the magnitude of B.
3. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are
not reversed.
4. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector
A is not reversed.
5. (c) When the two vector components Ax and Ay are added by the tail-to-head method,
the sum equals the vector A. Therefore, these vector components are the correct ones.
6. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar
component along the x axis (Ax = 0 m) and its scalar component along the y axis is
negative.
7. (e) The scalar components are given by Ax′ = −(450 m) sin 35.0° = −258 m and
Ay′ = −(450 m) cos 35.0° = −369 m.
8. (d)
9. Rx = 0 m, Ry = 6.8 m
10. R = 7.9 m, θ = 21 degrees
11. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides
are known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.
⎛ 4.0 km ⎞
12. (b) The angle is found by using the inverse tangent function, θ = tan −1 ⎜ ⎟ = 53° .
⎝ 3.0 km ⎠
,2 INTRODUCTION AND MATHEMATICAL CONCEPTS
13. (e) These vectors form a closed four-sided polygon, with the head of the fourth vector
exactly meeting the tail of the first vector. Thus, the resultant vector is zero.
14. (b) The three vectors form a right triangle, so the magnitude of A is given by the
Pythagorean theorem as A = Ax2 + Ay2 . If Ax and Ay double in size, then the magnitude
of A doubles: ( 2A ) + ( 2A )
2 2
= 4 Ax + 4 Ay = 2 Ax + Ay = 2 A.
2 2 2 2
x y
⎛A ⎞
15. (a) The angle θ is determined by the inverse tangent function, θ = tan −1 ⎜ y ⎟ . If Ax and
⎝ Ax ⎠
Ay both become twice as large, the ratio does not change, and θ remains the same.
16. (d) The distance (magnitude) traveled by each runner is the same, but the directions
are different. Therefore, the two displacement vectors are not equal.
17. (c) Ax and Bx point in opposite directions, and Ay and By point in the same direction.
18. Ay = 3.4 m, By = 3.4 m
, Chapter 1 Problems 3
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
PROBLEMS
1. REASONING We use the fact that 1 m = 3.28 ft to form the following conversion
factor: (1 m)/(3.28 ft) = 1.
SOLUTION To convert ft2 into m2, we apply the conversion factor twice:
⎛ 1 m ⎞⎛ 1 m ⎞
(
Area = 1330 ft 2 ⎜ ) ⎟⎜
⎝ 3.28 ft ⎠ ⎝ 3.28 ft
⎟ = 124 m
⎠
2
2. REASONING
a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we
need to convert miles to kilometers. This conversion is achieved by using the relation
1.609 km = 1 mi (see the page facing the inside of the front cover of the text).
b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must
convert miles to meters and hours to seconds. This is accomplished by using the
conversions 1 mi = 1609 m and 1 h = 3600 s.
SOLUTION
a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi) = 1, we
find the speed of the bicyclists is
⎛ mi ⎞ ⎛ mi ⎞ ⎛ 1.609 km ⎞ km
Speed = ⎜ 34.0 ⎟ (1) = ⎜ 34.0 ⎟⎜ ⎟ = 54.7
⎝ h ⎠ ⎝ h ⎠ ⎝ 1 mi ⎠ h
b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and
(1 h)/(3600 s) = 1, the speed of the bicyclists is
⎛ mi ⎞ ⎛ mi ⎞ ⎛ 1609 m ⎞ ⎛ 1 h ⎞ m
Speed = ⎜ 34.0 ⎟ (1)(1) = ⎜ 34.0 ⎟⎜ ⎟⎜ ⎟ = 15.2
⎝ h ⎠ ⎝ h ⎠ ⎝ 1 mi ⎠ ⎝ 3600 s ⎠ s
3. SSM REASONING We use the facts that 1 mi = 5280 ft, 1 m = 3.281 ft, and
1 yd = 3 ft. With these facts we construct three conversion factors: (5280 ft)/(1 mi) = 1,
(1 m)/(3.281 ft) = 1, and (3 ft)/(1 yd) = 1.
,4 INTRODUCTION AND MATHEMATICAL CONCEPTS
SOLUTION By multiplying by the given distance d of the fall by the appropriate
conversion factors we find that
⎛ 5280 ft ⎞ ⎛ 1 m ⎞ ⎛ 3 ft ⎞⎛ 1 m ⎞
(
d = 6 mi ⎜ ) ⎟⎜ ⎟ + (
551 yd ⎜
⎜ 1 yd ) ⎟⎜
⎟ ⎝ 3.281 ft ⎟ = 10 159 m
⎝ 1 mi ⎠ ⎝ 3.281 ft ⎠ ⎝ ⎠ ⎠
4. REASONING The word “per” indicates a ratio, so “0.35 mm per day” means
0.35 mm/d, which is to be expressed as a rate in ft/century. These units differ from the
given units in both length and time dimensions, so both must be converted. For length,
1 m = 103 mm, and 1 ft = 0.3048 m. For time, 1 year = 365.24 days, and
1 century = 100 years. Multiplying the resulting growth rate by one century gives an
estimate of the total length of hair a long-lived adult could grow over his lifetime.
SOLUTION Multiply the given growth rate by the length and time conversion
factors, making sure units cancel properly:
⎛ mm ⎞ ⎛ 1 m ⎞ ⎛ 1 ft ⎞ ⎛ 365.24 d ⎞ ⎛ 100 y ⎞
Growth rate = ⎜ 0.35 ⎟⎜ ⎟⎜ ⎟ ⎜⎜ ⎟⎜ ⎟ = 42 ft/century
⎝ d ⎠ ⎝ 103 mm ⎠ ⎝ 0.3048 m ⎠⎝ 1y ⎟ ⎜ century ⎟
⎠⎝ ⎠
5. REASONING In order to calculate d, the units of a and b must be, respectively, cubed
and squared along with their numerical values, then combined algebraically with each
other and the units of c. Ignoring the values and working first with the units alone, we
have
( m ) = m 3 2 = m2
3
a3
d= 2 →
cb ( m/s )( s )
2
(
m / s ⋅s 21 s )
Therefore, the units of d are m2/s.
SOLUTION With the units known, the numerical value may be calculated:
( 9.7 )
3
d= m 2 /s = 0.75 m 2 /s
( 69 )( 4.2 )
2
6. REASONING The dimensions of the variables v, x, and t are known, and the
numerical factor 3 is dimensionless. Therefore, we can solve the equation for z and
then substitute the known dimensions. The dimensions [ L ] and [ T ] can be treated as
algebraic quantities to determine the dimensions of the variable z.
, Chapter 1 Problems 5
3v
SOLUTION Since v = 13 zxt 2 , it follows that z = . We know the following
xt 2
dimensions: v = [ L ] / [ T ] , x = [ L ] , and t = [ T ] . Since the factor 3 is dimensionless, z
has the dimensions of
v
=
[L] / [T] = 1
[L ][ T] [T]
2 2 3
xt
7. SSM REASONING This problem involves using unit conversions to determine the
number of magnums in one jeroboam. The necessary relationships are
1.0 magnum = 1.5 liters
1.0 jeroboam = 0.792 U.S. gallons
1.00 U.S. gallon = 3.785 ×10 −3 m 3 = 3.785 liters
These relationships may be used to construct the appropriate conversion factors.
SOLUTION By multiplying one jeroboam by the appropriate conversion factors we
can determine the number of magnums in a jeroboam as shown below:
⎛ 0.792 gallons ⎞ ⎛ 3.785 liters ⎞ ⎛ 1.0 magnum ⎞
(1.0 jeroboam ) ⎜
⎜ 1.0 jeroboam
⎟⎜
⎟ ⎜ 1.0 gallon
⎟⎜
⎟ ⎝ 1.5 liters ⎟⎠
= 2.0 magnums
⎝ ⎠⎝ ⎠
8. REASONING By multiplying the quantity 1.78 × 10 −3 kg/ ( s ⋅ m ) by the appropriate
conversions factors, we can convert the quantity to units of poise (P). These conversion
factors are obtainable from the following relationships between the various units:
1 kg = 1.00 × 103 g
1 m = 1.00 × 10 2 cm
1 P = 1 g/ ( s ⋅ cm )
SOLUTION The conversion from the unit kg/ ( s ⋅ m ) to the unit P proceeds as follows:
⎛ kg ⎞ ⎛ 1.00 × 103 g ⎞ ⎛ 1 m ⎞⎡ 1P ⎤
⎜1.78 × 10
−3
⎟⎜ ⎟⎜ ⎟ ⎢ ⎥ = 1.78 × 10 −2 P
⎜
⎝ s⋅m ⎟⎜
⎠⎝ 1 kg ⎟ ⎝ 1.00 × 10 cm ⎠ ⎢1 g / s ⋅ cm
⎠
2
⎣ ( ) ⎥
⎦
,6 INTRODUCTION AND MATHEMATICAL CONCEPTS
9. REASONING Since the radius of the sphere is to be found in m, we need to convert
the volume unit of gallons to the volume unit of m3. Once we have the volume of water
in m3, we can then set that equal to the formula for the volume of a sphere from
Appendix E of your text and solve for the radius.
SOLUTION We begin by converting the volume of water in gallons to m3.
3.26 × 1020 gal 3.785 ×10−3 m3
= 1.23 × 1018 m3
1 gal
4
We now set this volume equal to the volume of a sphere: 1.23 ×1018 m3 = π r 3
3
1/3
⎛ (1.23 ×1018 m3 )(3) ⎞
Solve for the radius, r: r =⎜ ⎟ = 6.65 ×105 m
⎝ 4π ⎠
For comparison, the radius of the Moon is over 2½ times larger.
10. REASONING To convert from gallons to cubic meters, use the equivalence
1 U.S. gal = 3.785×10−3 m3. To find the thickness of the painted layer, we use the fact
that the paint’s volume is the same, whether in the can or painted on the wall. The layer
of paint on the wall can be thought of as a very thin “box” with a volume given by the
product of the surface area (the “box top”) and the thickness of the layer. Therefore, its
thickness is the ratio of the volume to the painted surface area: Thickness =
Volume/Area. That is, the larger the area it’s spread over, the thinner the layer of paint.
SOLUTION
a. The conversion is
⎛ 3.785 × 10−3 m3 ⎞
( 0.67 )
U.S. gallons ⎜
⎜ U.S. gallons ⎟⎟
= 2.5 × 10−3 m3
⎝ ⎠
b. The thickness is the volume found in (a) divided by the area,
Volume 2.5 × 10−3 m3
Thickness = = = 1.9 × 10−4 m
Area 13 m2
11. SSM REASONING The dimension of the spring constant k can be determined by
first solving the equation T = 2π m / k for k in terms of the time T and the mass m.
Then, the dimensions of T and m can be substituted into this expression to yield the
dimension of k.
, Chapter 1 Problems 7
SOLUTION Algebraically solving the expression above for k gives k = 4 π 2 m / T 2 . The
term 4π 2 is a numerical factor that does not have a dimension, so it can be ignored in
this analysis. Since the dimension for mass is [M] and that for time is [T], the dimension
of k is
Dimension of k =
[M]
[T ]2
12. REASONING AND SOLUTION The following figure (not drawn to scale) shows the
geometry of the situation, when the observer is a distance r from the base of the arch.
The angle θ is related to r and h by tan θ = h / r .
Solving for r, we find h = 192 m
θ
h 192 m r
r= = = 5.5 × 10 3 m = 5.5 km
tan θ tan 2.0°
13. SSM REASONING The shortest distance between the two towns is along the line
that joins them. This distance, h, is the hypotenuse of a right triangle whose other sides
are ho = 35.0 km and ha = 72.0 km, as shown in the figure below.
SOLUTION The angle θ is given by W
tan θ = ho / ha so that θ
ho
h
−1 ⎛ 35.0km ⎞
θ = tan ⎝ 72.0 km ⎠ = 25.9° S of W θ
h a
We can then use the Pythagorean theorem
to find h.
S
h = ho2 + ha2 = (35.0 km)2 + (72.0 km)2 = 80.1 km
, 8 INTRODUCTION AND MATHEMATICAL CONCEPTS
14. REASONING The drawing shows a schematic
representation of the hill. We know that the hill rises
12.0 m vertically for every 100.0 m of distance in
ho
the horizontal direction, so that ho = 12.0 m and
ha = 100.0 m . Moreover, according to Equation 1.3, θ
the tangent function is tan θ = ho / ha . Thus, we can h
a
use the inverse tangent function to determine the
angle θ.
SOLUTION With the aid of the inverse tangent function (see Equation 1.6) we find
that
⎛h ⎞ ⎛ 12.0 m ⎞
θ = tan −1 ⎜⎜ o ⎟⎟ = tan −1 ⎜ ⎟ = 6.84°
⎝ ha ⎠ ⎝ 100.0 m ⎠
15. REASONING Using the Pythagorean theorem (Equation 1.7), we find that the
relation between the length D of the diagonal of the square (which is also the diameter
of the circle) and the length L of one side of the square is D = L2 + L2 = 2 L .
SOLUTION Using the above relation, we have
D 0.35 m
D = 2L or L= = = 0.25 m
2 2
16. REASONING In both parts of the drawing the line of sight, the horizontal dashed
line, and the vertical form a right triangle. The angles θa = 35.0° and θb = 38.0° at
which the person’s line of sight rises above the horizontal are known, as is the
horizontal distance d = 85.0 m from the building. The unknown vertical sides of the
right triangles correspond, respectively, to the heights Ha and Hb of the bottom and
top of the antenna relative to the person’s eyes. The antenna’s height H is the difference
between Hb and Ha: H = H b − H a . The horizontal side d of the triangle is adjacent to
the angles θa and θb, while the vertical sides Ha and Hb are opposite these angles. Thus,
in either triangle, the angle θ is related to the horizontal and vertical sides by Equation
⎛ h ⎞
1.3 ⎜ tan θ = o ⎟ :
⎝ ha ⎠
Ha
tan θ a = (1)
d
H
tan θ b = b (2)
d
D. Cutnell
Hello all ,
We have all what you need with best price
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,CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the
last vector.
2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to
the shortest distance between the tail of A and the head of B. Thus, R is less than the
magnitude (length) of A plus the magnitude of B.
3. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are
not reversed.
4. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector
A is not reversed.
5. (c) When the two vector components Ax and Ay are added by the tail-to-head method,
the sum equals the vector A. Therefore, these vector components are the correct ones.
6. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar
component along the x axis (Ax = 0 m) and its scalar component along the y axis is
negative.
7. (e) The scalar components are given by Ax′ = −(450 m) sin 35.0° = −258 m and
Ay′ = −(450 m) cos 35.0° = −369 m.
8. (d)
9. Rx = 0 m, Ry = 6.8 m
10. R = 7.9 m, θ = 21 degrees
11. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides
are known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.
⎛ 4.0 km ⎞
12. (b) The angle is found by using the inverse tangent function, θ = tan −1 ⎜ ⎟ = 53° .
⎝ 3.0 km ⎠
,2 INTRODUCTION AND MATHEMATICAL CONCEPTS
13. (e) These vectors form a closed four-sided polygon, with the head of the fourth vector
exactly meeting the tail of the first vector. Thus, the resultant vector is zero.
14. (b) The three vectors form a right triangle, so the magnitude of A is given by the
Pythagorean theorem as A = Ax2 + Ay2 . If Ax and Ay double in size, then the magnitude
of A doubles: ( 2A ) + ( 2A )
2 2
= 4 Ax + 4 Ay = 2 Ax + Ay = 2 A.
2 2 2 2
x y
⎛A ⎞
15. (a) The angle θ is determined by the inverse tangent function, θ = tan −1 ⎜ y ⎟ . If Ax and
⎝ Ax ⎠
Ay both become twice as large, the ratio does not change, and θ remains the same.
16. (d) The distance (magnitude) traveled by each runner is the same, but the directions
are different. Therefore, the two displacement vectors are not equal.
17. (c) Ax and Bx point in opposite directions, and Ay and By point in the same direction.
18. Ay = 3.4 m, By = 3.4 m
, Chapter 1 Problems 3
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS
PROBLEMS
1. REASONING We use the fact that 1 m = 3.28 ft to form the following conversion
factor: (1 m)/(3.28 ft) = 1.
SOLUTION To convert ft2 into m2, we apply the conversion factor twice:
⎛ 1 m ⎞⎛ 1 m ⎞
(
Area = 1330 ft 2 ⎜ ) ⎟⎜
⎝ 3.28 ft ⎠ ⎝ 3.28 ft
⎟ = 124 m
⎠
2
2. REASONING
a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we
need to convert miles to kilometers. This conversion is achieved by using the relation
1.609 km = 1 mi (see the page facing the inside of the front cover of the text).
b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must
convert miles to meters and hours to seconds. This is accomplished by using the
conversions 1 mi = 1609 m and 1 h = 3600 s.
SOLUTION
a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi) = 1, we
find the speed of the bicyclists is
⎛ mi ⎞ ⎛ mi ⎞ ⎛ 1.609 km ⎞ km
Speed = ⎜ 34.0 ⎟ (1) = ⎜ 34.0 ⎟⎜ ⎟ = 54.7
⎝ h ⎠ ⎝ h ⎠ ⎝ 1 mi ⎠ h
b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and
(1 h)/(3600 s) = 1, the speed of the bicyclists is
⎛ mi ⎞ ⎛ mi ⎞ ⎛ 1609 m ⎞ ⎛ 1 h ⎞ m
Speed = ⎜ 34.0 ⎟ (1)(1) = ⎜ 34.0 ⎟⎜ ⎟⎜ ⎟ = 15.2
⎝ h ⎠ ⎝ h ⎠ ⎝ 1 mi ⎠ ⎝ 3600 s ⎠ s
3. SSM REASONING We use the facts that 1 mi = 5280 ft, 1 m = 3.281 ft, and
1 yd = 3 ft. With these facts we construct three conversion factors: (5280 ft)/(1 mi) = 1,
(1 m)/(3.281 ft) = 1, and (3 ft)/(1 yd) = 1.
,4 INTRODUCTION AND MATHEMATICAL CONCEPTS
SOLUTION By multiplying by the given distance d of the fall by the appropriate
conversion factors we find that
⎛ 5280 ft ⎞ ⎛ 1 m ⎞ ⎛ 3 ft ⎞⎛ 1 m ⎞
(
d = 6 mi ⎜ ) ⎟⎜ ⎟ + (
551 yd ⎜
⎜ 1 yd ) ⎟⎜
⎟ ⎝ 3.281 ft ⎟ = 10 159 m
⎝ 1 mi ⎠ ⎝ 3.281 ft ⎠ ⎝ ⎠ ⎠
4. REASONING The word “per” indicates a ratio, so “0.35 mm per day” means
0.35 mm/d, which is to be expressed as a rate in ft/century. These units differ from the
given units in both length and time dimensions, so both must be converted. For length,
1 m = 103 mm, and 1 ft = 0.3048 m. For time, 1 year = 365.24 days, and
1 century = 100 years. Multiplying the resulting growth rate by one century gives an
estimate of the total length of hair a long-lived adult could grow over his lifetime.
SOLUTION Multiply the given growth rate by the length and time conversion
factors, making sure units cancel properly:
⎛ mm ⎞ ⎛ 1 m ⎞ ⎛ 1 ft ⎞ ⎛ 365.24 d ⎞ ⎛ 100 y ⎞
Growth rate = ⎜ 0.35 ⎟⎜ ⎟⎜ ⎟ ⎜⎜ ⎟⎜ ⎟ = 42 ft/century
⎝ d ⎠ ⎝ 103 mm ⎠ ⎝ 0.3048 m ⎠⎝ 1y ⎟ ⎜ century ⎟
⎠⎝ ⎠
5. REASONING In order to calculate d, the units of a and b must be, respectively, cubed
and squared along with their numerical values, then combined algebraically with each
other and the units of c. Ignoring the values and working first with the units alone, we
have
( m ) = m 3 2 = m2
3
a3
d= 2 →
cb ( m/s )( s )
2
(
m / s ⋅s 21 s )
Therefore, the units of d are m2/s.
SOLUTION With the units known, the numerical value may be calculated:
( 9.7 )
3
d= m 2 /s = 0.75 m 2 /s
( 69 )( 4.2 )
2
6. REASONING The dimensions of the variables v, x, and t are known, and the
numerical factor 3 is dimensionless. Therefore, we can solve the equation for z and
then substitute the known dimensions. The dimensions [ L ] and [ T ] can be treated as
algebraic quantities to determine the dimensions of the variable z.
, Chapter 1 Problems 5
3v
SOLUTION Since v = 13 zxt 2 , it follows that z = . We know the following
xt 2
dimensions: v = [ L ] / [ T ] , x = [ L ] , and t = [ T ] . Since the factor 3 is dimensionless, z
has the dimensions of
v
=
[L] / [T] = 1
[L ][ T] [T]
2 2 3
xt
7. SSM REASONING This problem involves using unit conversions to determine the
number of magnums in one jeroboam. The necessary relationships are
1.0 magnum = 1.5 liters
1.0 jeroboam = 0.792 U.S. gallons
1.00 U.S. gallon = 3.785 ×10 −3 m 3 = 3.785 liters
These relationships may be used to construct the appropriate conversion factors.
SOLUTION By multiplying one jeroboam by the appropriate conversion factors we
can determine the number of magnums in a jeroboam as shown below:
⎛ 0.792 gallons ⎞ ⎛ 3.785 liters ⎞ ⎛ 1.0 magnum ⎞
(1.0 jeroboam ) ⎜
⎜ 1.0 jeroboam
⎟⎜
⎟ ⎜ 1.0 gallon
⎟⎜
⎟ ⎝ 1.5 liters ⎟⎠
= 2.0 magnums
⎝ ⎠⎝ ⎠
8. REASONING By multiplying the quantity 1.78 × 10 −3 kg/ ( s ⋅ m ) by the appropriate
conversions factors, we can convert the quantity to units of poise (P). These conversion
factors are obtainable from the following relationships between the various units:
1 kg = 1.00 × 103 g
1 m = 1.00 × 10 2 cm
1 P = 1 g/ ( s ⋅ cm )
SOLUTION The conversion from the unit kg/ ( s ⋅ m ) to the unit P proceeds as follows:
⎛ kg ⎞ ⎛ 1.00 × 103 g ⎞ ⎛ 1 m ⎞⎡ 1P ⎤
⎜1.78 × 10
−3
⎟⎜ ⎟⎜ ⎟ ⎢ ⎥ = 1.78 × 10 −2 P
⎜
⎝ s⋅m ⎟⎜
⎠⎝ 1 kg ⎟ ⎝ 1.00 × 10 cm ⎠ ⎢1 g / s ⋅ cm
⎠
2
⎣ ( ) ⎥
⎦
,6 INTRODUCTION AND MATHEMATICAL CONCEPTS
9. REASONING Since the radius of the sphere is to be found in m, we need to convert
the volume unit of gallons to the volume unit of m3. Once we have the volume of water
in m3, we can then set that equal to the formula for the volume of a sphere from
Appendix E of your text and solve for the radius.
SOLUTION We begin by converting the volume of water in gallons to m3.
3.26 × 1020 gal 3.785 ×10−3 m3
= 1.23 × 1018 m3
1 gal
4
We now set this volume equal to the volume of a sphere: 1.23 ×1018 m3 = π r 3
3
1/3
⎛ (1.23 ×1018 m3 )(3) ⎞
Solve for the radius, r: r =⎜ ⎟ = 6.65 ×105 m
⎝ 4π ⎠
For comparison, the radius of the Moon is over 2½ times larger.
10. REASONING To convert from gallons to cubic meters, use the equivalence
1 U.S. gal = 3.785×10−3 m3. To find the thickness of the painted layer, we use the fact
that the paint’s volume is the same, whether in the can or painted on the wall. The layer
of paint on the wall can be thought of as a very thin “box” with a volume given by the
product of the surface area (the “box top”) and the thickness of the layer. Therefore, its
thickness is the ratio of the volume to the painted surface area: Thickness =
Volume/Area. That is, the larger the area it’s spread over, the thinner the layer of paint.
SOLUTION
a. The conversion is
⎛ 3.785 × 10−3 m3 ⎞
( 0.67 )
U.S. gallons ⎜
⎜ U.S. gallons ⎟⎟
= 2.5 × 10−3 m3
⎝ ⎠
b. The thickness is the volume found in (a) divided by the area,
Volume 2.5 × 10−3 m3
Thickness = = = 1.9 × 10−4 m
Area 13 m2
11. SSM REASONING The dimension of the spring constant k can be determined by
first solving the equation T = 2π m / k for k in terms of the time T and the mass m.
Then, the dimensions of T and m can be substituted into this expression to yield the
dimension of k.
, Chapter 1 Problems 7
SOLUTION Algebraically solving the expression above for k gives k = 4 π 2 m / T 2 . The
term 4π 2 is a numerical factor that does not have a dimension, so it can be ignored in
this analysis. Since the dimension for mass is [M] and that for time is [T], the dimension
of k is
Dimension of k =
[M]
[T ]2
12. REASONING AND SOLUTION The following figure (not drawn to scale) shows the
geometry of the situation, when the observer is a distance r from the base of the arch.
The angle θ is related to r and h by tan θ = h / r .
Solving for r, we find h = 192 m
θ
h 192 m r
r= = = 5.5 × 10 3 m = 5.5 km
tan θ tan 2.0°
13. SSM REASONING The shortest distance between the two towns is along the line
that joins them. This distance, h, is the hypotenuse of a right triangle whose other sides
are ho = 35.0 km and ha = 72.0 km, as shown in the figure below.
SOLUTION The angle θ is given by W
tan θ = ho / ha so that θ
ho
h
−1 ⎛ 35.0km ⎞
θ = tan ⎝ 72.0 km ⎠ = 25.9° S of W θ
h a
We can then use the Pythagorean theorem
to find h.
S
h = ho2 + ha2 = (35.0 km)2 + (72.0 km)2 = 80.1 km
, 8 INTRODUCTION AND MATHEMATICAL CONCEPTS
14. REASONING The drawing shows a schematic
representation of the hill. We know that the hill rises
12.0 m vertically for every 100.0 m of distance in
ho
the horizontal direction, so that ho = 12.0 m and
ha = 100.0 m . Moreover, according to Equation 1.3, θ
the tangent function is tan θ = ho / ha . Thus, we can h
a
use the inverse tangent function to determine the
angle θ.
SOLUTION With the aid of the inverse tangent function (see Equation 1.6) we find
that
⎛h ⎞ ⎛ 12.0 m ⎞
θ = tan −1 ⎜⎜ o ⎟⎟ = tan −1 ⎜ ⎟ = 6.84°
⎝ ha ⎠ ⎝ 100.0 m ⎠
15. REASONING Using the Pythagorean theorem (Equation 1.7), we find that the
relation between the length D of the diagonal of the square (which is also the diameter
of the circle) and the length L of one side of the square is D = L2 + L2 = 2 L .
SOLUTION Using the above relation, we have
D 0.35 m
D = 2L or L= = = 0.25 m
2 2
16. REASONING In both parts of the drawing the line of sight, the horizontal dashed
line, and the vertical form a right triangle. The angles θa = 35.0° and θb = 38.0° at
which the person’s line of sight rises above the horizontal are known, as is the
horizontal distance d = 85.0 m from the building. The unknown vertical sides of the
right triangles correspond, respectively, to the heights Ha and Hb of the bottom and
top of the antenna relative to the person’s eyes. The antenna’s height H is the difference
between Hb and Ha: H = H b − H a . The horizontal side d of the triangle is adjacent to
the angles θa and θb, while the vertical sides Ha and Hb are opposite these angles. Thus,
in either triangle, the angle θ is related to the horizontal and vertical sides by Equation
⎛ h ⎞
1.3 ⎜ tan θ = o ⎟ :
⎝ ha ⎠
Ha
tan θ a = (1)
d
H
tan θ b = b (2)
d
