1
Physics and Measurement
CHAPTER OUTLINE
1.1
Standards of Length, Mass, and
Time
1.2
Matter and ModelBuilding
1.3
Dimensional Analysis
1.4
Conversion of Units
1.5
Estimates and Orderof
Magnitude Calculations
1.6
Significant Figures
ANSWERS TO QUESTIONS
* An asterisk indicates an item new to this edition.
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.
Q1.2
(a) 0.3 millimeters (b) 50 microseconds
(c) 7.2 kilograms
*Q1.3
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
Q1.4
No: A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee
is dimensionally correct.
Yes: If an equation is not dimensionally correct, it cannot be correct.
*Q1.5
The meterstick measurement, (a), and (b) can all be 4.31 cm. The meterstick measurement and
(c) can both be 4.24 cm. Only (d) does not overlap. Thus (a) (b) and (c) all agree with the
meterstick measurement.
*Q1.6
41
€
≈
41
€
(1 L 1.3
€
)(1 qt 1 L)(1 gal 4 qt)
≈
(10 1.3) gal
≈
8 gallons, answer (c)
1
SOLUTIONS TO PROBLEMS
Section 1.1
Standards of Length, Mass, and Time
P1.1
Modeling the Earth as a sphere, we fi
nd its volume as
4
3
4
3
6 37
10
1 08
10
3
6
3
21
3
π
π
r
=
×
(
)
=
×
.
.
m
m .
Its density is then
ρ
=
=
×
×
=
×
m
V
5 98
10
1 08
10
5 52
10
24
21
3
3
3
.
.
.
kg
m
kg m
. This value is intermediate
between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000
to 3 000 kg m
3
. The average density of the Earth is significantly higher, so higherdensity material
must be down below the surface.
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P1.2
With
V
=
(
)
(
)
base area
height
V
r
h
=
(
)
π
2
and
ρ
=
m
V
, we have
ρ
π
π
=
=
(
)
(
)
m
r h
2
2
9
1
19 5
39 0
10
1
kg
mm
mm
mm
m
3
3
.
.
⎛
⎝
⎜
⎞
⎠
⎟
=
×
ρ
2 15
10
4
3
.
.
kg m
P1.3
Let
V
represent the volume of the model, the same in
ρ
=
m
V
for both. Then
ρ
iron
kg
=
9 35
.
V
and
ρ
gold
gold
=
m
V
. Next,
ρ
ρ
gold
iron
gold
kg
=
m
9 35
.
and
m
gold
3
3
3
kg
19.3
10
kg/m
kg/m
=
×
×
⎛
9 35
7 86
10
3
.
.
⎝
⎜
⎞
⎠
⎟
=
23 0
.
kg
.
*P1.4
ρ
=
m
V
/
and
V
r
d
d
=
=
=
(
/
)
(
/
) (
/
)
/
4
3
4
3
2
6
3
3
3
π
π
π
where
d
is the diameter.
Then
ρ
π
π
=
=
×
×
=
×
−
−
6
6 1 67
10
2 4
10
2 3
3
27
15
3
m
d
/
( .
)
( .
)
.
kg
m
10
17
3
kg/m
2.3
10
kg/m /(11.3
10
kg/m )=
17
3
3
3
×
×
it is 20
×
10
12
times the density of lead
.
P1.5
For either sphere the volume is
V
r
=
4
3
3
π
and the mass is
m
V
r
=
=
ρ
ρ
π
4
3
3
. We divide
this equation for the larger sphere by the same equation for the smaller:
m
m
r
r
r
r
s
s
s
=
=
=
ρ π
ρ π
4
3
4
3
5
3
3
3
3
.
Then
r
r
s
=
=
(
)
=
5
4 50
1 71
7 69
3
.
.
.
cm
cm
.
Section 1.2
Matter and ModelBuilding
P1.6
From the figure, we may see that the spacing between diagonal planes is half the distance
between diagonally adjacent atoms on a ﬂat plane. This diagonal distance may be obtained from
the Pythagorean theorem,
L
L
L
diag
=
+
2
2
. Thus, since the atoms are separated by a distance
L
=
0 200
.
nm, the diagonal planes are separated by
1
2
0 141
2
2
L
L
+
=
.
nm
.
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 Spring '11
 Houston,C
 Physics, Mass, Orders of magnitude, Physics and Measurement

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