Physics 6C HW#5 solution
May 8, 2008
15.6
a)
P
=
P
0
+
ρgh
= 1
.
013
×
10
5
Pa
+ (1024
Kg/m
3
)(9
.
80
m/s
2
)(1000
m
)
=
1
.
01
×
10
7
Pa
b)
The gauge pressure is the diﬀerence in pressure be
tween the water outside and the air inside the sub
marine, which we suppose is at 1.00 atm.
P
gauge
=
P

P
0
=
ρgh
= 1
.
00
×
10
7
Pa
The resultant inward force on the porthole is then.
F
=
P
gauge
A
= 1
.
00
×
10
7
Pa
(
π
×
(0
.
150
m
)
2
)
=
7
.
09
×
10
5
N
15.8
a)
Suppose the “vacuum cleaner” functions as a high
vacuum pump. The air below the brick will exert on
it lifting force.
F
=
PA
= 1
.
013
×
10
5
Pa
(
π
(1
.
43
×
10

2
m
)
2
)
=
65.1 N
b)
The octopus can pull the bottom away from the top
shell with a force that could be no larger than
F
=
PA
= (
P
0
+
ρgh
)
A
=
±
1
.
013
×
10
5
Pa
+ (1030
Kg/m
3
)(9
.
80
m/s
2
)(32
.
3
m
)
²
×
(
π
(1
.
43
×
10

2
m
)
2
)
=
275 N
15.14
a)
Using the deﬁnition of density, we have
h
w
=
m
water
A
2
ρ
water
=
100
g
5
.
00
cm
2
(1
.
00
g/cm
3
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 Spring '01
 staff
 Thermodynamics, Force, Light

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