11
U
NITS
,
P
HYSICAL
Q
UANTITIES AND
V
ECTORS
1.1.
IDENTIFY:
Convert units from mi to km and from km to ft.
SET UP:
1 in.
2.54 cm
=
, 1 km = 1000 m , 12 in. 1 ft
=
, 1 mi = 5280 ft .
EXECUTE:
(a)
23
5280 ft
12 in.
2.54 cm
1 m
1 km
1.00 mi
(1.00 mi)
1.61 km
1 mi
1 ft
1 in.
10 cm
10 m
⎛⎞
⎛
⎞
⎛
⎞
⎛
⎞
==
⎜⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝⎠
⎝
⎠
⎝
⎠
⎝
⎠
(b)
32
3
1 in.
1 ft
1.00 km
(1.00 km)
3.28 10 ft
1 km
1 m
2.54 cm
12 in.
⎛
⎞
⎛
⎞
×
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
EVALUATE:
A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km.
1.2.
IDENTIFY:
Convert volume units from L to
3
in.
.
SET UP:
3
1 L
1000 cm
=
. 1 in.
2.54 cm
=
EXECUTE:
3
3
3
1000 cm
1 in.
0.473 L
28.9 in.
.
2.54 cm
××=
EVALUATE:
3
1 in. is greater than
3
1 cm , so the volume in
3
in. is a smaller number than the volume in
3
cm ,
which is
3
473 cm .
1.3.
IDENTIFY:
We know the speed of light in m/s.
/
tdv
=
. Convert 1.00 ft to m and
t
from s to ns.
SET UP:
The speed of light is
8
3.00 10 m/s
v
=×
. 1 ft
0.3048 m
=
.
9
1 s
10 ns
=
.
EXECUTE:
9
8
0.3048 m
1.02 10 s 1.02 ns
t
−
×
=
×
EVALUATE:
In 1.00 s light travels
85
5
3.00 10 m
3.00 10 km 1.86 10 mi
×=
×
=
×
.
1.4.
IDENTIFY:
Convert the units from g to kg and from
3
cm to
3
m .
SET UP:
1 kg
1000 g
=
. 1 m 1000 cm
=
.
EXECUTE:
3
4
33
g
1 kg
100 cm
kg
11.3
1.13 10
cm
1000 g
1 m
m
⎛
⎞
××
=
×
⎜
⎟
⎝
⎠
EVALUATE:
The ratio that converts cm to m is cubed, because we need to convert
3
cm to
3
m .
1.5.
IDENTIFY:
Convert volume units from
3
in. to L.
SET UP:
3
1000 cm
=
. 1 in.
2.54 cm
=
.
EXECUTE:
()
3
327 in.
2.54 cm in.
1 L 1000 cm
5.36 L
=
EVALUATE:
The volume is
3
5360 cm .
3
1 cm is less than
3
1 in. , so the volume in
3
cm is a larger number than the
volume in
3
in.
.
1.6.
IDENTIFY:
Convert
2
ft to
2
m and then to hectares.
SET UP:
42
1.00 hectare 1.00 10 m
. 1 ft
0.3048 m
=
.
EXECUTE:
The area is
2
2
43,600 ft
0.3048 m
1.00 hectare
(12.0 acres)
4.86 hectares
1 acre
1.00 ft
1.00 10 m
⎛
⎞
=
⎜
⎟
×
⎝
⎠
.
EVALUATE:
Since 1 ft
0.3048 m
=
,
22
2
1 ft
(0.3048) m
=
.
1.7.
IDENTIFY:
Convert seconds to years.
SET UP:
9
1 billion seconds
1 10 s
. 1 day
24 h
=
. 1 h
3600 s
=
.
EXECUTE:
9
1 h
1 day
1 y
1.00 billion seconds
1.00 10 s
31.7 y
3600 s
24 h
365 days
⎛
⎞
=
⎜
⎟
⎝
⎠
.
1
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Chapter 1
EVALUATE:
The conversion
7
1 y
3.156 10 s
=×
assumes 1 y
365.24 d
=
, which is the average for one extra day
every four years, in leap years. The problem says instead to assume a 365day year.
1.8.
IDENTIFY:
Apply the given conversion factors.
SET UP:
1 furlong
0.1250 mi and 1 fortnight
14 days.
==
1 day
24 h.
=
EXECUTE:
()
0.125 mi
1 fortnight
1 day
180,000 furlongs fortnight
67 mi/h
1 furlong
14 days
24 h
⎛⎞
⎛
⎞
⎛
⎞
=
⎜⎟
⎜
⎟
⎝⎠
⎝
⎠
EVALUATE:
A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much
smaller number.
1.9.
IDENTIFY:
Convert miles/gallon to km/L.
SET UP:
1 mi 1.609 km
=
. 1 gallon
3.788 L.
=
EXECUTE:
(a)
1.609 km
1 gallon
55.0 miles/gallon
(55.0 miles/gallon)
23.4 km/L
1 mi
3.788 L
⎛
⎞
⎜
⎟
⎝
⎠
.
(b)
The volume of gas required is
1500 km
64.1 L
23.4 km/L
=
.
64.1 L
1.4 tanks
45 L/tank
=
.
EVALUATE:
1 mi/gal
0.425 km/L
=
. A km is very roughly half a mile and there are roughly 4 liters in a gallon,
so
2
4
1 mi/gal
km/L
∼
, which is roughly our result.
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 Fall '10
 KOZHAN,R
 Orders of magnitude, ax

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