141
F
LUID
M
ECHANICS
14.1.
IDENTIFY:
Use Eq.(14.1) to calculate the mass and then use
wm
g
=
to calculate the weight.
SET UP:
/
mV
ρ
=
so
=
From Table 14.1,
33
7.8 10 kg/m .
=×
EXECUTE:
For a cylinder of length
L
and radius
R
,
22
4
3
(
)
(0.01425 m) (0.858 m)
5.474 10 m .
VR
L
ππ
−
==
=
×
Then
4
3
(7.8 10 kg/m )(5.474 10 m )
4.27 kg,
−
== ×
×
=
and
2
(4.27 kg)(9.80 m/s )
41.8 N
g
=
(about
9.4 lbs). A cart is not needed.
EVALUATE:
The rod is less than 1m long and less than 3 cm in diameter, so a weight of around 10 lbs seems
reasonable.
14.2.
IDENTIFY:
Convert gallons to kg. The mass
m
of a volume
V
of gasoline is
=
.
SET UP:
1 gal
3.788 L
3.788 10 m
−
×
.
3
1 m of gasoline has a mass of 737 kg.
EXECUTE:
3
1 gal
1 m
45.0 mi/gal
(45.0 mi/gal)
16.1 mi/kg
737 kg
−
⎛⎞
⎜⎟
×
⎝⎠
EVALUATE:
1 gallon of gasoline has a mass of 2.79 kg. The car goes fewer miles on 1 kg than on 1 gal, since
1 kg of gasoline is less gasoline than 1 gal of gasoline.
14.3.
IDENTIFY:
/
=
SET UP:
The density of gold is
19.3 10 kg/m
×
.
EXECUTE:
333
6
3
(5.0 10 m)(15.0 10 m)(30.0 10 m)
2.25 10 m
V
−−−
−
×
×
=
×
.
63
0.0158 kg
7.02 10 kg/m
m
V
−
= ×
×
. The metal is not pure gold.
EVALUATE:
The average density is only 36% that of gold, so at most 36% of the mass is gold.
14.4.
IDENTIFY:
Find the mass of gold that has a value of
6
$1.00 10
×
. Then use the density of gold to find the volume
of this mass of gold.
SET UP:
For gold,
. The volume
V
of a cube is related to the length
L
of one side by
3
VL
=
.
EXECUTE:
3
6
1 troy ounce
31.1035 10 kg
($1.00 10 )
72.9 kg
$426.60
1 troy ounce
m
−
×
=
.
m
V
=
so
72.9 kg
3.78 10 m
m
V
−
×
.
1/3
0.156 m
15.6 cm
LV
=
.
EVALUATE:
The cube of gold would weigh about 160 lbs.
14.5.
IDENTIFY:
Apply
/
=
to relate the densities and volumes for the two spheres.
SET UP:
For a sphere,
3
4
3
Vr
π
=
. For lead,
l
11.3 10 kg/m
and for aluminum,
a
2.7 10 kg/m
.
EXECUTE:
3
4
3
m
ρ
V
π
r
ρ
. Same mass means
aa
11
r
ρ
r
ρ
=
.
13
3
a1
3
1a
11.3 10
1.6
2.7 10
r
ρ
r
ρ
×
=
×
.
EVALUATE:
The aluminum sphere is larger, since its density is less.
14.6.
IDENTIFY:
Average density is
/
=
.
SET UP:
For a sphere,
3
4
3
=
. The sun has mass
30
sun
1.99 10 kg
M
and radius
8
6.96 10 m
×
.
EXECUTE:
(a)
30
30
sun
83
2
73
4
sun
3
1.409 10 kg/m
(6.96 10 m)
1.412 10 m
M
V
π
××
=
(b)
30
30
16
3
43
1
3
3
4
3
5.94 10 kg/m
(2.00 10 m)
3.351 10 m
π
=
×
14
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View Full Document142
Chapter 14
EVALUATE:
For comparison, the average density of the earth is
33
5.5 10 kg/m .
×
A neutron star is extremely
dense.
14.7.
IDENTIFY:
wm
g
=
and
mV
ρ
=
. Find the volume
V
of the pipe.
SET UP:
For a hollow cylinder with inner radius
1
R
, outer radius
2
R
, and length
L
the volume is
22
21
()
.
VR
R
L
π
=
−
2
1
1.25 10 m
R
−
=×
and
2
2
1.75 10 m
R
−
EXECUTE:
4
3
([0.0175 m]
[0.0125 m] )(1.50 m)
7.07 10 m
V
−
=
−
.
4
3
(8.9 10 kg/m )(7.07 10 m )
6.29 kg
−
== ×
×
=
.
61.6 N
g
==
.
EVALUATE:
The pipe weights about 14 pounds.
14.8.
IDENTIFY:
The gauge pressure
0
pp
−
at depth
h
is
0
g
h
−
=
.
SET UP:
Ocean water is seawater and has a density of
1.03 10 kg/m
×
.
EXECUTE:
2
7
0
(1.03 10 kg/m )(9.80 m/s )(3200 m)
3.23 10 Pa
−
=
×
.
7
0
5
1 atm
(3.23 10 Pa)
319 atm
1.013 10 Pa
⎛⎞
−
=
⎜⎟
×
⎝⎠
.
EVALUATE:
The gauge pressure is about 320 times the atmospheric pressure at the surface.
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 Spring '11
 Shaefer

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