le (txl65) – Fluid II – mohanty – (81440)
1
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11
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001
10.0points
A fountain sends a stream of water 23
.
8 m up
into the air.
If the base of the stream is 8
.
9 cm in diame
ter, what power is required to send the water
to this height? The acceleration of gravity is
9
.
8 m
/
s
2
.
Correct answer: 31339
.
3 W.
Explanation:
Let :
h
1
= 23
.
8 m
,
d
= 8
.
9 cm = 0
.
089 m
,
r
= 4
.
45 cm = 0
.
0445 m
,
and
g
= 9
.
8 m
/
s
2
.
Using conservation of energy,
1
2
m v
2
=
m g h
v
=
radicalbig
2
g h
=
radicalBig
2 (9
.
8 m
/
s
2
) (23
.
8 m)
= 21
.
5981 m
/
s
is the velocity of the stream at its base. The
flow rate at the base is
π r
2
v
=
π
(0
.
0445 m)
2
(21
.
5981 m
/
s)
= 0
.
134365 m
3
/
s
.
Using the density of water as 1000 kg
/
m
3
, we
see that 134
.
365 kg of water emerges from the
fountain each second.
m
t
=
ρ π r
2
v
= (1000 kg
/
m
3
)
π
(0
.
0445 m)
2
×
(21
.
5981 m
/
s)
= 134
.
365 kg
.
Thus, each second, sufficient work is done to
raise 134
.
365 kg to a height of 23
.
8 m or
P
=
W
t
=
m g
h
t
=
ρ π r
2
v g
h
t
= (1000 kg
/
m
3
)
π
(0
.
0445 m)
2
×
(21
.
5981 m
/
s) (9
.
8 m
/
s
2
)
23
.
8 m
1 s
=
31339
.
3 W
.
002
10.0points
A styrofoam slab has a thickness of 10
.
3 cm
and a density of 319 kg
/
m
3
.
What is the area of the slab if it floats just
awash (top of slab is even with the water sur
face) in fresh water when a 65
.
1 kg swimmer is
aboard? The density of water is 1000 kg
/
m
3
.
Correct answer: 0
.
928104 m
2
.
Explanation:
Let :
d
= 10
.
3 cm
,
ρ
s
= 319 kg
/
m
3
,
and
m
= 65
.
1 kg
.
Since the slab floats awash, the buoyant force
is
B
=
ρ
w
V g
=
ρ
w
d A g ,
and is equal in
magnitude to the weight of the slab and the
swimmer:
ρ
w
d A g
=
ρ
s
d A g
+
m g
A
=
m
d
(
ρ
w

ρ
s
)
=
65
.
1 kg
(10
.
3 cm) (1000 kg
/
m
3

319 kg
/
m
3
)
=
0
.
928104 m
2
.
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 Spring '11
 sdsd
 Correct Answer, kg

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