homework 27 – FIERRO, JEFFREY – Due: May 4 2008, 11:00 pm
1
Question 1, chap 18, sect 5.
part 1 of 1
10 points
A small plastic ball (similar to a pingpong
ball) with mass 3
.
5 g is attached by a thread
to the bottom of a beaker. When the beaker is
filled with water so that the ball is totally sub
merged, the tension in the thread is 0
.
033 N
.
The density of water is 1000 kg
/
m
3
and the
acceleration of gravity is 9
.
81 m
/
s
2
.
3
.
5 g
Determine the diameter of the ball.
Correct answer: 2
.
35789
cm (tolerance
±
1
%).
Explanation:
V
sphere
=
4
3
π r
3
=
4
3
π
parenleftbigg
d
2
parenrightbigg
3
=
1
6
π d
3
.
Let :
ρ
w
= 1000 kg
/
m
3
,
g
= 9
.
81 m
/
s
2
,
T
= 0
.
033 N
,
and
m
= 3
.
5 g = 0
.
0035 kg
.
Consider the forces acting on the PingPong
ball
B
m g
T
Applying Archimedes’ principle, we have
B
=
w
f
=
m
f
g
=
ρ
w
V
ball
g
=
1
6
π ρ
w
g d
3
.
Applying
summationdisplay
F
y
= 0 to the ball,
B

m g

T
= 0
1
6
π ρ
w
g d
3

m g

T
= 0
.
Then we have
d
=
3
radicalBigg
6 (
T
+
m g
)
π ρ
w
g
=
3
radicalBigg
6 [0
.
033 N + (0
.
0035 kg) (9
.
81 m
/
s
2
)]
π
(1000 kg
/
m
3
) (9
.
81 m
/
s
2
)
×
100 cm
m
=
2
.
35789 cm
.
Question 2, chap 18, sect 2.
part 1 of 1
10 points
A fountain sends a stream of water 25
.
4 m
up into the air.
The acceleration of gravity is 9
.
8 m
/
s
2
.
If the base of the stream is 6
.
61 cm in
diameter, what power is required to send the
water to this height?
Correct answer:
19058
.
9
W (tolerance
±
1
%).
Explanation:
Let :
h
1
= 25
.
4 m
,
d
= 6
.
61 cm
,
r
=
6
.
61 cm
2
= 0
.
03305 m
,
and
g
= 9
.
8 m
/
s
2
.
Using conservation of energy, we may write
K
max
=
U
max
=
E
total
1
2
m v
2
=
m g h ,
and
v
=
radicalbig
2
g h
=
radicalBig
2 (9
.
8 m
/
s
2
) (25
.
4 m)
= 22
.
3123 m
/
s
.
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homework 27 – FIERRO, JEFFREY – Due: May 4 2008, 11:00 pm
2
is the velocity of the stream at its base. The
flow rate at the base is
π r
2
v
=
π
(0
.
03305 m)
2
(22
.
3123 m
/
s)
= 0
.
0765663 m
3
/
s
.
Using the density of water as 1000 kg
/
m
3
, we
see that 76
.
5663 kg of water emerges from the
fountain each second.
m
t
=
ρ π r
2
v
= (1000 kg
/
m
3
)
π
(0
.
03305 m)
2
×
(22
.
3123 m
/
s)
= 76
.
5663 kg
.
Thus, each second, sufficient work is done to
raise 76
.
5663 kg to a height of 25
.
4 m or
P
=
W
t
=
m g
h
t
=
ρ π r
2
v g
h
t
= (1000 kg
/
m
3
)
π
(0
.
03305 m)
2
×
(22
.
3123 m
/
s) (9
.
8 m
/
s
2
)
(25
.
4 m)
(1 s)
=
19058
.
9 W
.
Question 3, chap 18, sect 5.
part 1 of 1
10 points
An empty rubber balloon has a mass of
0
.
0151 kg.
The balloon is filled with helium
at a density of 0
.
0925 kg
/
m
3
. At this density
the balloon has a radius of 0
.
318 m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
If the filled balloon is fastened to a vertical
line, what is the tension in the line?
Correct answer: 1
.
4328 N (tolerance
±
1 %).
Explanation:
The balloon is in equilibrium under the
action of three forces,
F
b
,
the buoyant force
on the balloon,
w,
its weight, and
T
, the
tension in the string.
We have:
Σ
F
y
= 0
Hence,
T
=
F
b

w .
(1)
F
b
= weightofdisplacedair
=
ρ
air
V g
= (1
.
29 kg
/
m
3
) (9
.
8 m
/
s
2
)
4
3
π
(0
.
318 m)
3
= 1
.
70289 N
.
W
= weight of empty balloon + weight of
enclosed helium, or
W
= (0
.
0151 kg) (9
.
8 m
/
s
2
)
+ (0
.
0925 kg
/
m
3
)
4
3
π
(0
.
318 m)
3
(9
.
8 m
/
s
2
)
= 0
.
270086 N
.
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 Spring '08
 Turner
 Mass, Work, Correct Answer, Orders of magnitude, Bernoulli's principle, JEFFREY

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