Physics 40 HW Solutions for Chapter 1 Serway 7th Edition
Q: 1, 3, 5,
P: 4, 5, 9, 11, 15, 17, 20, 30, 31, 32, 33, 49, 60
Questions
Q1.1
Density varies with temperature and pressure. It would be necessary to measure both
mass and volume very accurately in order to use the density of water as a standard.
*Q1.3
In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041 kg (e) 0.27
kg. Then the ranking is c = e > d > a > b
*Q1.5
The answer is yes for (a), (c), and (f). You cannot add or subtract a number of apples
and a number of jokes. The answer is no for (b), (d), and (e). Consider the gauge of a
sausage, 4 kg/2 m, or the volume of a cube, (2 m)3. Thus we have (a) yes (b) no (c)
yes (d) no (e) no (f) yes
Problems
3
*P1.4 ρ = m/V and V = (4/3) ππr3 = (4/3) π (d/2)3 = ππd /6 where d is the diameter.
Then ρ = 6m/πd3 =
6(1.67 × 10−27 kg)
π (2.4 × 10 -15 m)3
= 2.3 × 1017 kg/m3
2.3 × 1017 kg/m3/(11.3 × 103 kg/m3) = it is 20 × 1012 times the density of lead.
P1.5
4
4
For either sphere the volume is V = π r 3 and the mass is m = ρV = ρ π r 3 . We divide
3
3
this equation for the larger sphere by the same equation for the smaller:
ml ρ 4π rl 3 3 rl 3
=
= =5.
ms ρ 4π rs3 3 rs3
P1.9
Then rl = rs 3 5 = 4.50 cm (1.71)= 7.69 cm .
Inserting the proper units for everything except G,
⎡ kg m ⎤ G [kg ]
2
⎢ s2 ⎥ =
⎣
⎦ [m ]
2
.
Multiply both sides by [m] and divide by [kg ] ; the units of G are
2
2
m3
kg ⋅s 2
.
P1.11 Conceptualize: We must calculate the area and convert units. Since a meter is
about 3 feet, we should expect the area to be about A ≈ (30 m )(50 m )= 1 500 m2 .
Categorize: We model the lot as a perfect rectangle to use Area = Length ×
Width. Use the conversion: 1 m = 3.281 ft .
Analyze: A = LW = (100 ft ) ⎛⎜
1m ⎞
1m ⎞
= 1390 m 2 = 1.39 × 103 m 2
(150 ft )⎛⎜⎝
⎝ 3.281 ft ⎟⎠
3.281 ft ⎟⎠
.
Finalize: Our calculated result agrees reasonably well with our initial estimate and
has the proper units of m 2 . Unit conversion is a common technique that is applied to
many problems.
P1.15 From Table 14.1, the density of lead is 1.13 × 10 4 kg m 3 , so we should expect our
calculated value to be close to this number. This density value tells us that lead is about
11 times denser than water, which agrees with our experience that lead sinks.
Density is defined as mass per volume, in ρ =
m
V
. We must convert to SI units in
the calculation.
3
23.94 g ⎛ 1 kg ⎞ ⎛ 100 cm ⎞
23.94 g ⎛ 1 kg ⎞ ⎛ 1 000 000 cm 3 ⎞
4
3
ρ=
=
⎜
⎟
3⎜
⎟⎠ = 1.14 × 10 kg m
⎟
1 m3
2.10 cm ⎝ 1000 g ⎠ ⎝ 1 m ⎠ 2.10 cm 3 ⎜⎝ 1000 g ⎟⎠ ⎜⎝
At one step in the calculation, we note that one million cubic centimeters make
one cubic meter. Our result is indeed close to the expected value. Since the last reported
significant digit is not certain, the difference in the two values is probably due to
measurement uncertainty and should not be a concern. One important common-sense
check on density values is that objects which sink in water must have a density greater
than 1 g cm 3 , and objects that float must be less dense than water.
P1.17
(a)
⎛ 8 × 1012
⎜⎝ 1000 $
$ ⎞ ⎛ 1 h ⎞ ⎛ 1 day ⎞ ⎛ 1 yr ⎞
= 250 years
s ⎟⎠ ⎜⎝ 3 600 s ⎟⎠ ⎜⎝ 24 h ⎟⎠ ⎜⎝ 365 days ⎟⎠
(b)
The circumference of the Earth at the equator is 2π 6.378 × 103 m = 4.01× 107 m .
)
(
The length of one dollar bill is 0.155 m so that the length of 8 trillion bills
is 1.24 × 1012 m . Thus, the 8 trillion dollars would encircle the Earth
1.24 × 1012 m
= 3.09 × 10 4 times .
4.01× 107 m
P1.20
(a)
⎛d
⎞
300 ft
⎛
⎞
−3
d nucleus, scale = d nucleus, real ⎜ atom, scale ⎟ = 2.40 × 10 −15 m ⎜
⎟⎠ = 6.79 × 10 ft , or
−10
⎝
d
1.06
×
10
m
⎝ atom, real ⎠
)
(
(
)
d nucleus, scale = 6.79 × 10−3 ft (304.8 mm 1 ft )= 2.07 mm
3
(b)
3
3
/3 ⎛ ratom ⎞ ⎛ d atom ⎞ ⎛ 1.06 × 10−10 m ⎞
4π ratom
Vatom
=
=
=
=
3
Vnucleus 4π rnucleus
/3 ⎜⎝ rnucleus ⎟⎠ ⎜⎝ d nucleus ⎟⎠ ⎜⎝ 2.40 × 10−15 m ⎟⎠
3
= 8.62 × 1013 times as large
P1.30
METHOD ONE
We treat the best value with its uncertainty as a binomial (21.3 ± 0.2 ) cm (9.8 ± 0.1) cm ,
A = ⎡⎣ 21.3(9.8 )± 21.3(0.1)± 0.2 (9.8)± (0.2 )(0.1)⎤⎦ cm2 .
The first term gives the best value of the area. The cross terms add together to
give the uncertainty and the fourth term is negligible.
A = 209 cm 2 ± 4 cm2 .
METHOD TWO We add the fractional uncertainties in the data.
⎛ 0.2 0.1 ⎞
= 209 cm 2 ± 2% = 209 cm 2 ± 4 cm 2
A = (21.3 cm )(9.8 cm )± ⎜
+
⎝ 21.3 9.8 ⎟⎠
3
P1.31 (a)
P1.32
(b)
4
3
(c)
(d)
2
r = (6.50 ± 0.20 ) cm = (6.50 ± 0.20 )× 10−2 m
m = (1.85 ± 0.02 ) kg
ρ=
m
(43 )π r 3
δ ρ δ m 3δ r
.
=
+
ρ
m
r
also,
In other words, the percentages of uncertainty are cumulative. Therefore,
δ ρ 0.02 3(0.20 )
=
+
= 0.103 ,
ρ 1.85
6.50
and
P1.33
ρ=
1.85
( )π (6.5 × 10−2 m)
3
4
3
= 1.61× 103 kg m 3
ρ ± δ ρ = (1.61± 0.17 )× 103 kg m3 = (1.6 ± 0.2 )× 103 kg m3 .
(a)
756.??
37.2?
0.83
+ 2.5?
796. /5/3 = 797
(b)
0.003 2 (2 s.f.)× 356.3(4 s.f.)= 1.140 16 = (2 s.f.) 1.1
(c)
5.620(4 s.f.)× π (>4 s.f.)= 17.656=(4 s.f.) 17.66
P1.49 The scale factor used in the “dinner plate” model is
S=
0.25 m
= 2.5 × 10−6 m lightyears .
1.0 × 10 5 lightyears
The distance to Andromeda in the scale model will be
)(
(
)
Dscale = Dactual S = 2.0 × 106 lightyears 2.5 × 10−6 m lightyears = 5.0 m .
P1.60 The volume of the galaxy is
π r 2t = π (1021 m ) (1019 m )~ 1061 m 3 .
2
If the distance between stars is 4 × 1016 m , then there is one star in a volume on the
order of
3
(4 × 1016 m ) ~ 1050 m3 .
The number of stars is about
1061 m 3
~ 1011 stars .
10 50 m 3 star