INSTRUCTOR’S SOLUTIONS MANUAL FOR_SERWAY AND VUILLE’S_COLLEGE PHYSICS NINTH EDITION, VOLUME 1

Using a calculator to multiply the length by the width gives a raw answer of 6783 m2, but this
answer must be rounded to contain the same number of signifi cant fi gures as the least accurate
factor in the product. The least accurate factor is the length, which contains either 2 or 3
signifi cant fi gures, depending on whether the trailing zero is signifi cant or is being used only to
locate the decimal point. Assuming the length contains 3 signifi cant fi gures, answer (c) correctly
expresses the area as 6.78 ×103 m2 . However, if the length contains only 2 signifi cant fi gures,
answer (d) gives the correct result as 6.8 ×103 m2 .
2. Both answers (d) and (e) could be physically meaningful. Answers (a), (b), and (c) must be meaningless
since quantities can be added or subtracted only if they have the same dimensions.
3. According to Newton’s second law, Force = mass×acceleration. Thus, the units of Force must be
the product of the units of mass (kg) and the units of acceleration (m s2 ). This yields kg⋅m s2,
which is answer (a).
4. The calculator gives an answer of 57.573 for the sum of the 4 given numbers. However, this sum
must be rounded to 58 as given in answer (d) so the number of decimal places in the result is the
same (zero) as the number of decimal places in the integer 15 (the term in the sum containing the
smallest number of decimal places).
5. The required conversion is given by:
h =( )⎛
⎝ ⎜
⎞
⎠ ⎟
2 00
1 000 1 00
.
.
m
mm
1.00 m
cubitus
445 mm
⎛
⎝ ⎜
⎞
⎠ ⎟
= 4.49 cubiti
This result corresponds to answer (c).
6. The given area (1 420 ft2 ) contains 3 signifi cant fi gures, assuming that the trailing zero is used
only to locate the decimal point. The conversion of this value to square meters is given by:
A = ( × )⎛
⎝ ⎜
⎞
⎠ ⎟
1 42 103 = 1 32 ×10
2
. ft . 2
1.00 m
3.281 ft
2 m2 = 132 m2
Note that the result contains 3 signifi cant fi gures, the same as the number of signifi cant fi gures in
the least accurate factor used in the calculation. This result matches answer (b).
7. You cannot add, subtract, or equate a number apples and a number of days. Thus, the answer is
yes for (a), (c), and (e). However, you can multiply or divide a number of apples and a number
of days. For example, you might divide the number of apples by a number of days to fi nd the
number of apples you could eat per day. In summary, the answers are (a) yes, (b) no, (c) yes,
(d) no, and (e) yes.
1
2 Chapter 1
8. The given Cartesian coordinates are x = −5.00, and y = 12.00, with the least accurate containing 3
signifi cant fi gures. Note that the specifi ed point (with x < 0 and y > 0) is in the second quadrant.
The conversion to polar coordinates is then given by:
r = x2 + y2 = (− )2 + ( )2 = 5.00 12.00 13.0
tan
.
.
θ = = . θ .
−
y = − = − (−
x
12 00
5 00
2 40 and tan 1 2 40) = −67.3°+180° = 113°
Note that 180° was added in the last step to yield a second quadrant angle. The correct answer is
therefore (b) (13.0, 113°).
9. Doing dimensional analysis on the fi rst 4 given choices yields:
(a) [v]
⎡⎣
⎤⎦
= =
t 2 2 3
L T
T
L
T
(b) [v]
⎡⎣
⎤⎦
= = − −
x2 2
L T 1 1
L
L T
(c)
v2 2 2 2
3
L T
T
L T
T
L
T
⎡⎣
⎤⎦
[ ] =( )= =
t
2
(d)
v2 2 2
2
L T
L
L T
L
L
T
⎡⎣
⎤⎦
[ ] =( )= =
x
2
Since acceleration has units of length divided by time squared, it is seen that the relation given in
answer (d) is consistent with an expression yielding a value for acceleration.
10. The number of gallons of gasoline she can purchase is
# gallons
total expenditure
cost per gallon
E = ≈ 33 uros
Euros
L
L
quart
quarts
1 5
1
1
4
.
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜⎜
⎞
⎠ ⎟⎟
1
5
gal
gal
⎛
⎝ ⎜⎜
⎞
⎠ ⎟⎟
≈
so the correct answer is (b).
11. The situation described is shown in the drawing at the
right.
From this, observe that tan 26
45
° = h
m
, or
h = (45 m) tan 26° = 22 m
Thus, the correct answer is (a).
12. Note that we may write 1.365 248 0 ×107 as 136.524 80 ×105. Thus, the raw answer, including
the uncertainty, is x = (136.524 80 ± 2) ×105. Since the fi nal answer should contain all the
digits we are sure of and one estimated digit, this result should be rounded and displayed as
137 ×105 = 1.37 ×107 (we are sure of the 1 and the 3, but have uncertainty about the 7). We see
that this answer has three signifi cant fi gures and choice (d) is correct.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2. Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly
regular astronomical clocks.
45 m
h
26
Introduction 3
4. (a) ~0.5 lb ≈ 0.25 kg or ~10 −1 kg
(b) ~4 lb ≈ 2 kg or ~100 kg
(c) ~4000 lb ≈ 2000 kg or ~10 3 kg
6. Let us assume the atoms are solid spheres of diameter 10−10 m. Then, the volume of each atom
is of the order of 10−30 m3. (More precisely, volume = 4πr3 3 = πd3 6.) Therefore, since
1 cm3 = 10−6 m3, the number of atoms in the 1 cm3 solid is on the order of 10−6 10−30 = 1024 atoms.
A more precise calculation would require knowledge of the density of the solid and the mass of
each atom. However, our estimate agrees with the more precise calculation to within a factor of 10.
8. Realistically, the only lengths you might be able to verify are the length of a football fi eld and the
length of a housefl y. The only time intervals subject to verifi cation would be the length of a day
and the time between normal heartbeats.
10. In the metric system, units differ by powers of ten, so it’s very easy and accurate to convert from
one unit to another.
ANSWERS TO EVEN NUMBERED PROBLEMS
2. (a) L T2 (b) L
4. All three equations are dimensionally incorrect.
6. (a) kg⋅m s (b) Ft = p
8. (a) 22.6 (b) 22.7 (c) 22.6 is more reliable
10. (a) 3.00 ×108 m s (b) 2.997 9 ×108 m s (c) 2.997 925×108 m s
12. (a) 346 m2 ±13 m2 (b) 66.0 m ±1.3 m
14. (a) 797 (b) 1.1 (c) 17.66
16. 3.09 cm s
18. (a) 5.60 ×102 km = 5.60 ×105 m = 5.60 ×107 cm
(b) 0.491 2 km = 491.2m = 4.912 ×104 cm
(c) 6.192 km = 6.192 ×103 m = 6.192 ×105 cm
(d) 2.499 km = 2.499 ×103 m = 2.499 ×105 cm
20. 10 6. kmL
22. 9 2. nms
24. 2.9×102 m3 = 2.9×108 cm3
26. 2.57×106 m3
4 Chapter 1
28. ∼ 108 steps
30. ~108 people with colds on any given day
32. (a) 4.2 ×10−18 m3 (b) ~10−1 m3 (c) ~1016 cells
34. (a) ∼ 1029 prokaryotes (b) ~1014 kg
(c) The very large mass of prokaryotes implies they are important to the biosphere. They are
responsible for fi xing carbon, producing oxygen, and breaking up pollutants, among many
other biological roles. Humans depend on them!
36. 2.2 m
38. 8.1 cm
40. Δs = r + r − r r ( − ) 1
2
2
2
1 2 1 2 2 cosθ θ
42. 2.33 m
44. (a) 1.50 m (b) 2.60 m
46. 8.60 m
48. (a) and (b)
(c) y x = tan12.0°, y (x −1.00 km) = tan14.0° (d) 1.44×103 m
50. y
d = ⋅ ⋅
−
tan tan
tan tan
θ φ
φ θ
52. (a) 1.609 km h (b) 88 km h (c) 16 km h
54. Assumes population of 300 million, average of 1 can week per person, and 0.5 oz per can.
(a) ∼ 1010 cans yr (b) ∼ 105 tons yr
56. (a) 7.14×10−2 gal s (b) 2.70×10−4 m3 s (c) 1.03 h
58. (a) A A 2 1 = 4 (b) V V 2 1 = 8
60. (a) ∼ 102 yr (b) ∼ 104 times
62. ∼ 104 balls yr. Assumes 1 lost ball per hitter, 10 hitters per inning, 9 innings per game, and
81 games per year.
Introduction 5
PROBLEM SOLUTIONS
1.1 Substituting dimensions into the given equation T = 2π 􀁃 g , and recognizing that 2π is a dimensionless
constant, we have
T
g
[ ]= [ ]
[ ]
􀁃
or T
L
L T
T T 2
= = 2 =
Thus, the dimensions are consistent .
1.2 (a) From x = Bt2, we fi nd that B
x
t
= 2 . Thus, B has units of
[ ]
[ ]
[ ]
B
x
t
= = 2
L
T2
(b) If x = Asin(2π ft ), then [A] = [x] [sin(2π ft )]
But the sine of an angle is a dimensionless ratio.
Therefore, [A] = [x] = L
1.3 (a) The units of volume, area, and height are:
[V] = L3, [A] = L2, and [h] = L
We then observe that L3 = L2L or [V] = [A][h]
Thus, the equation V = Ah is dimensionally correct .
(b) V Rh R h Ah cylinder=π 2 = (π 2 ) = , where A =π R2
V wh w h Ah rectangular box= 􀁃 = (􀁃 ) = , where A = 􀁃w = length×width
1.4 (a) In the equation 1
2
2 1
2 0
mv = mv2 + mgh, [mv ] [mv ] 2
0
2
2
= = ⎛
⎝ ⎜
⎞
⎠ ⎟
M =
L
T
ML
T
2
2
while mgh ⎡⎣
⎤⎦
= ⎛
⎝ ⎜
⎞
⎠ ⎟
M =
L
T
L
M L
2 T
1
2
. Thus, the equation is dimensionally incorrect .
(b) In v = v + 0
at 2, [v] = [v ] = 0
L
T
but [at 2 ] = [a][t 2 ] = ⎛
⎝ ⎜
⎞
⎠ ⎟
( )= L
T
T L 2
2 . Hence, this equation
is dimensionally incorrect .
(c) In the equation ma = v2, we see that [ma] = [m][a] = ⎛
⎝ ⎜
⎞
⎠ ⎟
M =
L
T
ML
2 T2 , while [v ] 2
2
= ⎛
⎝ ⎜
⎞
⎠ ⎟
= L
T
L
T
2
2 .
Therefore, this equation is also dimensionally incorrect .
1.5 From the universal gravitation law, the constant G is G = Fr2 Mm. Its units are then
G
F r
M m
[ ]=
[ ]⎡⎣
⎤⎦
[ ][ ] = ( ⋅ )( )
⋅
=
⋅
2 kg m s m
kg kg
m
kg s
2 2 3
2
6 Chapter 1
1.6 (a) Solving KE
p
m
=
2
2
for the momentum, p, gives p = 2m(KE) where the numeral 2 is a
dimensionless constant. Dimensional analysis