1410
Chapter 14
(b)
The density of the liquid is
3
3
3
0.355 kg
1000 kg/m
,
0.355
10
m
−
=
×
and so the volume flow rate is
3
3
3
1.30 kg/s
1.30
10
m /s
1.30 L/s.
1000 kg/m
−
=
×
=
This result may also be obtained from
(
)(
)
220
0.355 L
1.30 L/s.
60.0 s
=
(c)
3
3
1
4
2
1.30
10
m /s
6.50 m/s
2.00
10
m
v
−
−
×
=
=
×
.
2
1
/4
1.63 m/s
v
v
=
=
.
(d)
(
)
(
)
2
2
1
2
2
1
2
1
1
2
p
p
ρ
v
v
ρ
g y
y
=
+
−
+
−
.
(
)
3
2
2
2
1
1
2
152 kPa
(1000 kg/m
)
[(1.63 m/s)
(6.50 m/s)
]
(9.80 m/s
)(
1.35 m)
p
=
+
−
+
−
.
1
119 kPa
p
=
.
E
VALUATE
:
The increase in height and the increase in fluid speed at point 1 both cause the pressure at point 1 to
be less than the pressure at point 2.
14.45.
I
DENTIFY
:
Apply Bernoulli°s equation to the two points.
S
ET
U
P
:
1
2
y
y
=
.
1
1
2
2
v A
v A
=
.
2
1
2
A
A
=
.
E
XECUTE
:
2
2
1
1
1
1
1
2
2
2
2
2
p
gy
v
p
gy
v
ρ
ρ
ρ
ρ
+
+
=
+
+
.
1
1
2
1
2
1
(2.50 m/s)
1.25 m/s
2
A
A
v
v
A
A
⎛
⎞
⎛
⎞
=
=
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
.
2
2
4
3
2
2
4
1
1
2
1
1
2
2
2
(
)
1.80
10
Pa
(1000 kg/m
)([2.50 m/s]
[1.25 m/s]
)
2.03
10
Pa
p
p
v
v
ρ
=
+
−
=
×
+
−
=
×
E
VALUATE
:
The gauge pressure is higher at the second point because the water speed is less there.
14.46.
I
DENTIFY
and
S
ET
U
P
:
Let point 1 be where
1
4.00 cm
r
=
and point 2 be where
2
2.00 cm.
r
=
The volume flow
rate
vA
has the value
3
7200 cm /s at all points in the pipe. Apply Eq.(14.10) to find the fluid speed at points 1 and
2 and then use Bernoulli°s equation for these two points to find
2
.
p
E
XECUTE
:
2
3
1
1
1
1
7200 cm
,
v A
v
r
π
=
=
so
1
1.43 m/s
v
=
2
3
2
2
2
2
7200 cm
,
v A
v
r
π
=
=
so
2
5.73 m/s
v
=
2
2
1
1
1
1
1
2
2
2
2
2
p
gy
v
p
gy
v
ρ
ρ
ρ
ρ
+
+
=
+
+
1
2
y
y
=
and
5
1
2.40
10
Pa,
p
=
×
so
2
2
5
1
2
1
1
2
2
(
)
2.25
10
Pa
p
p
v
v
ρ
=
+
−
=
×
E
VALUATE
:
Where the area decreases the speed increases and the pressure decreases.
14.47.
I
DENTIFY
:
F
pA
=
, where
A
is the crosssectional area presented by a hemisphere. The force
bb
F
that the body
builder must apply must equal in magnitude the net force on each hemisphere due to the air inside and outside the
sphere.
S
ET
U
P
:
2
4
D
A
π
=
.
E
XECUTE
:
(a)
(
)
2
bb
0
.
4
D
F
p
p
π
=
−
(b)
The force on each hemisphere due to the atmosphere is
2
2
5
(5.00
10
m)
(1.013
10
Pa/atm)(0.975 atm)
776
π
−
×
×
=
Ν
. The bodybuilder must exert this force on each
hemisphere to pull them apart.
E
VALUATE
:
The force is about 170 lbs, feasible only for a very strong person. The force required is proportional
to the square of the diameter of the hemispheres.
14.48.
I
DENTIFY
:
Apply
0
p
p
gh
ρ
=
+
and
0
(
)
p V
V
B
Δ
Δ
= −
,where
B
is the bulk modulus.
S
ET
U
P
:
Seawater has density
3
3
1.03
10
kg/m
ρ
=
×
. The bulk modulus of water is
9
2.2
10
Pa
B
=
×
.
5
air
1.01
10
Pa
p
=
×
.
E
XECUTE
:
(a)
5
3
3
2
3
8
0
air
1.01
10
Pa
(1.03
10
kg/m
)(9.80 m/s
)(10.92
10
m)
1.10
10
Pa
p
p
gh
ρ
=
+
=
×
+
×
×
=
×
(b)
At the surface
3
1.00 m
of seawater has mass
3
1.03
10
kg
×
. At a depth of 10.92 km the change in volume is
8
3
3
0
9
(
)
(1.10
10
Pa)(1.00 m
)
0.050 m
2.2
10
Pa
p V
V
B
Δ
×
Δ
= −
= −
= −
×
. The volume of this mass of water at this depth therefore
is
3
0
0.950 m
V
V
V
=
+ Δ
=
.
3
3
3
3
1.03
10
kg
1.08
10
kg/m
0.950 m
m
V
ρ
×
=
=
=
×
. The density is 5% larger than at the surface.
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 Spring '06
 Buchler
 Physics, Force, m/s

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