Zzztayloe, Rex – Homework 10 – Heinz – 81204 – Nov 04, 2004
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
A 1
.
2 m wide by 2
.
0 m long water bed weighs
1084 N.
Find the pressure that the water bed exerts
on the floor.
Assume that the entire lower
surface of the bed makes contact with the
floor.
Correct answer: 451
.
667 Pa.
Explanation:
Basic Concepts:
P
=
F
A
A
=
wl
Given:
F
g
= 1084 N
w
= 1
.
2 m
‘
= 2
.
0 m
Solution:
P
=
F
g
w ‘
=
1084 N
(1
.
2 m)(2 m)
= 451
.
667 Pa
002
(part 1 of 1) 10 points
A piston
A
has a diameter of 0
.
79 cm, as
shown. A second piston
B
has a diameter of
3
.
2 cm.
A
B
440
.
2 N
F
In the absence of friction, determine the
force
~
F
necessary to support the 440
.
2 N
weight.
Correct answer: 26
.
829 N.
Explanation:
Basic Concepts:
P
=
F
A
A
=
πr
2
Given:
r
A
=
0
.
79 cm
2
= 0
.
395 cm
r
B
=
3
.
2 cm
2
= 1
.
6 cm
F
g,B
= 440
.
2 N
Solution:
P
A
=
P
B
F
A
A
A
=
F
g,B
A
B
F
A
π r
2
A
=
F
g,B
π r
2
B
F
A
=
F
g,B
r
2
A
r
2
B
=
(440
.
2 N)(0
.
395 cm)
2
(1
.
6 cm)
2
= 26
.
829 N
downward.
003
(part 1 of 2) 5 points
A submarine is at an ocean depth of 178 m.
Assume that the density of sea water is 1
.
025
×
10
3
kg
/
m
3
and the atmospheric pressure is
1
.
01
×
10
5
Pa.
The acceleration of gravity is 9
.
81 m
/
s
2
.
a) Calculate the absolute pressure at this
depth.
Correct answer: 1
.
89083
×
10
6
Pa.
Explanation:
Basic Concept:
P
=
P
0
+
ρgh
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Zzztayloe, Rex – Homework 10 – Heinz – 81204 – Nov 04, 2004
2
Given:
h
= 178 m
ρ
sea water
= 1
.
025
×
10
3
kg
/
m
3
g
= 9
.
81 m
/
s
2
P
0
= 1
.
01
×
10
5
Pa
Solution:
P
= 101000 Pa
+ (1025 kg
/
m
3
)
·
(9
.
81 m
/
s
2
)(178 m)
= 1
.
89083
×
10
6
Pa
004
(part 2 of 2) 5 points
b) Calculate the magnitude of the force ex
erted by the water at this depth on a cir
cular submarine window with a diameter of
27
.
9 cm.
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 Fall '09
 WOODAHL
 Physics, Work, Kilogram, Standard gravity

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