Physics 6C HW#6 solution
May 22, 2008
15.30
Volume flow rate for a steady current is,
A
1
v
1
=
A
2
v
2
a)
20
.
0
l
60
.
0
s
(
1000
cm
3
1
l
) =
π
(1
.
00
cm
)
2
v
hose
v
hose
=
333
cm
3
/s
3
.
14
cm
2
=
106 cm/s
b)
20
.
0
l
60
.
0
s
(
1000
cm
3
1
l
) =
π
(0
.
500
cm
)
2
v
nozzle
v
nozzle
=
333
cm
3
/s
0
.
785
cm
2
=
424 cm/s
15.33
We can use the Bernoulli’s equation,
P
1
+
1
2
ρv
2
1
+
ρgh
1
=
P
2
+
1
2
ρv
2
2
+
ρgh
2
We need two points in order to apply the Bernoulli’s
equation.
One point is where the steam exits the
tank(the point of interest) and the other point can
be any other place in the tank.
In this solution, I
choose the top surface as the second point.
Then,
the equation becomes,
At the first point
P
air
+
1
2
ρv
2
x
+
ρgh
At the second point
P
air
+
1
2
ρ
(0)
2
+
ρgh
0
two equation are equal
P
air
+
1
2
ρv
2
x
+
ρgh
=
P
air
+
1
2
ρ
(0)
2
+
ρgh
0
v
2
x
= 2
g
(
h
0

h
)
v
2
x
=
p
2
g
(
h
0

h
)
Then the horizontal distance, x ,is,
x
=
v
x
t
y
=
h
=
1
2
gt
2
t
=
s
2
h
g
plug t into x
x
=
v
x
s
2
h
g
=
p
2
g
(
h
0

h
)
s
2
h
g
=
2
p
h
(
h
0

h
)
1
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16.12
Δ
V
= (
β
CCl
4

3
α
F e
)
V
i
Δ
T
= (5
.
81
×
10

4

3(11
.
0
×
10

6
))(50
.
0
gal
)(20
.
0
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 Thermodynamics, Current, Energy, Light, mAl CAl, CAl +mc Cw, mAl CAl +mc

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