MR. SURRETTE
VAN NUYS HIGH SCHOOL
1

P a g e
PHYSICS
CHAPTER 9
:
FLUID MECHANICS
WORKSHEET
SOLUTIONS
1
.
What is the total pressure at the bottom of a
8
m deep swimming pool? (Note the pressure
contribution from the atmosphere is 1.01 x 10
5
N/m
2
, the density of water is 10
3
kg/m
3
, and
g
= 9.8 m/s
2
)
1
A.
(
1)
p
=
p
a
+
gh
(2)
p
= (1.01 x 10
5
N/m
2
)+(1.0 x 10
3
kg/m
3
)(9.8 m/s
2
)(
8
m)
(3)
p
= 1.79
x 10
5
N/m
2
2
.
A hydraulic lift raises a
1750
kg automobile with
a
6
00 N force is applied to the smaller piston. If
the smaller piston has an area of
8
cm
2
, what is the cross

s
ectional area of the larger piston?
2
A.
(1) F
1
/A
1
= F
2
/A
2
(2) A
2
= (F
2
A
1
)/F
1
(3)
F
2
= w =
mg
(4) A
2
= (mgA
1
)/F
1
(5) A
2
= (
1750
kg
)(
9.8 m/s
2
)(
8
.0 x 10

4
m
2
) /
6
00 N
(6)
A
2
=
2.29
x 10

2
m
3.
An object of mass 0.
61
kg is suspended from a scale and sub
merged in a liquid. If the reading on the
scale is
4
.0
0
N, then the buoyant force that the fluid exerts on the object is:
3A.
(1)
w = mg
(2) w = (0.61
kg)(9.8 m/s
2
)
(3)
w = 5.98
N
(4)
F
B
= w
–
w’
(5) F
B
=
5.98
N
–
4.00
N
(6)
F
B
= 1.9
8
N
4
.
If the
flow rate of a liquid going through a 2.
5
0 cm radius pipe is measured at
1.5
x 10

3
m
3
/s, what is
the average fluid velocity in the pipe?
4
A.
(1)
Flow rate = Av
(2) v = Flow rate / A
(3) v = Flow rate /
r
2
(4) v =
1.5
x 10

3
m
3
/s /
(
2
.5
x 10

2
m
)
2
(
5)
v = 0.76
m/s
MR. SURRETTE
VAN NUYS HIGH SCHOOL
2

P a g e
PHYSICS
5
.
A liquid filled tank has a hole on its vertical surface just above the bottom edge. If
the surface of the
liquid is 0.7
m above the hole, at what speed will the stream of liquid emerge from the hole?
5
A.
(1) v = (2gh)
1/2
(2) v =
[(2)(9.8 m/s
2
)(0.
7
m)]
1/2
(3)
v =
3.7
m/s
6
.
Water (density = 1 x 10
3
kg/m
3
) is flowing thr
ough a pipe whose radius is 0.03
m with a speed of 1
2
m/s. This same pipe goes up to the second floor of the building, 3 m higher, and the pressure remain
s
unch
anged. What is the cross

sectional area
of the pipe on the second floor?
6
A.
(1)
First Equation:
p
1
+ ½
v
1
2
+
杹
1
=
p
2
+ ½
v
2
2
+
杹
2
(2) ½ v
1
2
+
gy
1
= ½ v
2
2
+ gy
2
(3) v
1
2
= v
2
2
+ 2gy
2
(4) v
2
= [(v
1
2
)
–
2gh]
1/2
(5
) v
2
= [(1
2
m/s)
2
–
2(9.8
m/s
2
)(3 m)]
1/2
(6
)
v
2
=
9.23
m/s
(7
)
Second Equation:
A
1
v
1
= A
2
v
2
(8
) A
2
= A
1
v
1
/ v
2
(9
)
Third Equation:
A
1
=
r
2
(10
) A
2
= (
r
2
)v
1
/ v
2
(11
) A
2
=
(0.03
m)
2
(
12 m/s) / (
9.23
m/s
)
(12
)
A
2
=
3.68
x 10

3
m
2
MR. SURRETTE
VAN NUYS HIGH SCHOOL
3

P a g e
PHYSICS
CHAPTERS 8

9
:
TEMPERATURE, HEAT, AND
FLUIDS
QUIZ SOLUTIONS
1.
Three moles of an ideal gas are confined to a 30 liter container at a pressure of 3 atmospheres. What
is the gas temperature? (R = 0.0821 L.atm/mole.K)
1A.
(1)
PV = nRT
(2)
T = (PV) / (nR)
(3)
T = [(3 atm)(30 liters) / [(3
mol)(0.0821 L.atm/mol.K)]
(4)
T = 365.4 K
2.
Suppose the ends of a 35

m

long steel rail are rigidly clamped at 5
o
C to prevent expansion. The rail
has a cross

sectional area of 15 cm
2
. What force does the rail exert when it is heated to 38
o
C?
(
STEEL
= 1.1 x 10

5
/
o
C, Y
STEEL
= 2 x 10
11
N/m
2
)
2A.
(1)
Y = (F/A) / (
L/L)
(2)
(
L/L) = (F/A) / Y
(3)
Y(
L/L) = (F/A)
(4)
(F/A) = Y(
L/L)
(5)
F = (AY
L
) / L
(6)
L = L
T
: F = (AY
L
T
) /
L
(7)
F = AY
T
(8)
F = (1.5 x 10

3
m
2
)(2 x 10
11
N/m
2
)(1.1
x 10

5
/
o
C)(33
o
C)
(9)
F =
1.09 x 10
5
N
3.
An unknown gas condenses into a liquid at approximately 105
o
Kelvin. What temperature, in degrees
Fahrenheit, does this correspond to?
3A.
(1)
T
F
= 9/5
T
C
+ 32
(2)
T
F
= 9/5(
T
–
273
) + 32
(3)
T
F
= 9/5(105
–
27
3) + 32
(4)
T
F
=

270.4
o
4.
An auditorium has dimensions 15 m x 25 m x 20 m. How many molecules of air are needed to fill the
auditorium at standard temperature and pressure?
4A.
(1)
(1 mol / 22.4 liter) (1 liter / 10

3
m
3
) (6.022 x 10
23
molecules / m
ol) (7500 m
3
/ auditorium)
(2)
2.02 x 10
29
molecules
MR. SURRETTE
VAN NUYS HIGH SCHOOL
4

P a g e
PHYSICS
5.
A helium

filled balloon has a volume of 1.5 m
3
. As it rises in the Earth’s atmosphere, its volume
expands. What will be its new volume (in cubic meters) if its original temperature and pressure
are 5
o
C
and 1 atm, and its final temperature and pressure are
–
40
o
C and 0.1 atm?
5A.
(1)
(P
1
V
1
) / T
1
= (P
2
V
2
) / T
2
(2)
(T
2
P
1
V
1
) / T
1
= P
2
V
2
(3)
P
2
V
2
= (T
2
P
1
V
1
) / T
1
(4)
V
2
= (T
2
P
1
V
1
) / (T
1
P
2
)
(5)
V
2
= [(233 K)(1 atm)(1.5 m
3
)] / [(278 K)(0.1 atm)]
(6)
V
2
= 12.57 m
3
6.
If the rms velocity of a helium atom at room temperature is 1350 m/s, what is the rms velocity of an
oxygen (O
2
) molecule (mass of O
2
= 32 grams, mass of He = 4 grams)?
6A.
(1) v
rms
= [(
3RT
)/M]
1/2
(2)
v
rms
O
2
is proportional to (
1
/32
gra
ms
)
1/2
(3) v
rms
He is proportional to (
1
/4
grams
)
1/2
(4) v
rms
(O
2
/ He) is proportional to (32)
1/2
/ (4)
1/2
(5) v
rms
(O
2
/ He) = 2.8
(6) 1350 m/s / 2.8
(7)
v
rms
= 477.3 m/s
7.
A silver bar of length 35 cm and cross

sectional area 1 cm
2
is used to tr
ansfer heat from a 100
o
C
reservoir to a 0
o
C reservoir. How much heat is transferred per second?
(k
SILVER
= 427 W/m
o
C)
7A.
(1) H = kA(
T) / L
(2) H= (427 W/m
o
C)(1 x 10

4
m
2
)(100
o
C) / (0.35 m)
(3)
H = 12.20 W
8.
A solar heating system has a 30% conve
rsion efficiency; the solar radiant incident on the panels is
500 W/m
2
. What is the increase in temperature of 30 kg of water in a one hour period by a 5.0 m
2
area
collector? (The specific heat of water is 4186 J.kg.
o
C)
8A.
(1)
Total radiation
: 5.0 m
2
(
500 W / 1.0 m
2
) =
2500 W
(2)
Effective radiation
:
2500 W
x 30% = 750 W
(3) 750 W = 750 Joules / second
(4)
Determine Q
: 3600 s (750 J / s) =
2.70 x 10
6
J
(5) Q = mc
T
(6)
T = Q / mc
(7)
T = 2.70 x 10
6
J / (30 kg)(4186 J/kg.
o
C)
(8)
T = 21.5
o
C
MR. SURRETTE
VAN NUYS HIGH SCHOOL
5

P a g e
PHYSICS
9.
How much heat energy is required to melt a 30 gram block of silver at 10
o
C? The melting point of
silver is 960
o
C, its specific heat is 2.34 x 10
2
J/kg.
o
C, and its h
eat of fusion is 8.83 x 10
4
J/kg.
9A.
(1)
Determine specific heat
:
Q
C
= mc
T
(2) Q
C
= (0.030 kg)(2.34x10
2
J/kg.
o
C)(960
o
C
–
10
o
C)
(3)
Q
C
= 6669 J
(4)
Determine latent heat
:
Q
L
= mL
(5) Q
L
= (0.030 kg)(8.83 x 10
4
J/kg)
(6)
Q
L
= 2649 J
(7)
Determine
total heat
: Q = Q
C
+ Q
L
(8) Q =
6669 J
+
2649 J
(9)
Q = 9318 J
10.
An ideal gas is maintained at an average pressure of 75 N/m
2
during an isobaric process while its
volume decreases by 0.25 m
3
. What work is done by the system on its environment?
10A
.
(1) W = P
V
(2) W = (75 Pa)(

0.25 m
3
)
(3)
W =

18.8 J
11.
A boulder of mass 700 kg tumbles down a mountainside and stops 180 m below. If the temperature
of the boulder, mountain, and surrounding air are all at 300 K, what is the change in entropy of the
Univ
erse?
11A.
(1)
S =
Q / T
(2)
Q = Energy
:
Q = U
g
= mgh
(3)
Q = (700 kg)(9.8 m/s
2
)(180 m)
(4)
Q = 1.24 x 10
6
J
(5)
S =
1.24 x 10
6
J
/ 300 K
(6)
S = 4116 J/K
12.
A bottle containing an ideal gas has a volume of 3 m
3
and a pressure of 1 x 1
0
5
N/m
2
at a
temperature of 300 K. The bottle is placed against a metal block that is maintained at 900 K and the gas
expands as the pressure remains constant until the temperature of the gas reaches 900 K. How much
work is done by the gas?
12A.
(1)
Det
ermine
V
: (
p
1
V
1
) / T
1
= (
p
2
V
2
) / T
2
(2) V
2
= (
p
1
V
1
T
2
) / (T
1
p
2
)
(3) V
2
= [(1 x 10
5
Pa)(3 m
3
)(900 K)] / [(300 K)(1 x 10
5
Pa)]
(4) V
2
= 9 m
3
(5)
V = V
2
–
V
1
(6)
V
= 9 m
3
–
3 m
3
=
6 m
3
(7)
Determine the Work
: W =
p
V
(8) W = (1 x 10
5
Pa)(
6 m
3
)
(9)
W = 6
x 10
5
J
MR. SURRETTE
VAN NUYS HIGH SCHOOL
6

P a g e
PHYSICS
13.
A gasoline engine absorbs 2600 J of heat energy and performs 600 J of mechanical work in each
cycle. What is the efficiency of the engine?
13A.
(1)
e
= (Q
H
–
Q
C
) / Q
H
(2)
Q
H
–
Q
C
= Work
(3)
e
=
Work
/ Q
H
(4)
e
= 600 J / 2600 J
(5)
e
= 23.1%
14.
Calculate the entropy change when 2 mol (36 grams) of water at 100
o
C is converted into steam (The
heat of vaporization of water is 540 cal/g).
14A.
(1)
Determine
Q
:
Q = mL
(2)
Q = (36 g)(540 cal/g)
(3)
Q = 19,440 cal
(4)
Convert to Jo
ules
:
19,440 cal
(4.186 J/cal) =
8.138 x 10
4
J
(5)
Determine Entropy
:
S =
Q / T
(6)
S =
8.138 x 10
4
J
/ 373 K
(7)
S = 218.2 J/K
15.
What is the total pressure at the bottom of a
12 m deep swimming pool? (Note the pressure contribution from the
atmosphere is 1.01 x 10
5
N/m
2
, the
density of water is 10
3
kg/m
3
, and
g
= 9.8 m/s
2
)
15A.
(1)
p
=
p
a
+
gh
(2)
p
= (1.01 x 10
5
N/m
2
)+(1.0 x 10
3
kg/m
3
)(9.8 m/s
2
)(12 m)
(3)
p
= 2.19 x 10
5
N/m
2
16.
A hydraulic lift raises a 1325 kg automobile when a 445 N
force is applied to the smaller piston. If
the smaller piston has an area of 6 cm
2
, what is the cross

sectional area of the larger piston?
16A.
(1) F
1
/A
1
= F
2
/A
2
(2) A
2
= (F
2
A
1
)/F
1
(3)
F
2
= w =
mg
(4) A
2
= (mgA
1
)/F
1
(5) A
2
= (
1325 kg
)(
9.8 m/s
2
)(
6 x
10

4
m
2
) /445 N
(6)
A
2
= 2.29 x 10

2
m
MR. SURRETTE
VAN NUYS HIGH SCHOOL
7

P a g e
PHYSICS
17.
An object of mass 1.24 kg is suspended from a scale and submerged in a liquid. If the reading on
the scale is 3.00 N, then the buoyant force that the fluid exerts on the object is:
17A.
(1)
w = mg
(2) w =
(1.24 kg)(9.8 m/s
2
)
(3)
w = 12.15 N
(4)
F
B
= w
–
w’
(5) F
B
=
12.15 N
–
3.00 N
(6)
F
B
= 9.15 N
1
8.
A liquid filled tank has a hole on its vertical surface just above the bottom edge. If the surface of
the liquid is 1.4 m above the hole, at what speed
will the stream of liquid emerge from the hole?
18A.
(1) v = (2gh)
1/2
(2) v = [(2)(9.8 m/s
2
)(1.4 m)]
1/2
(3)
v = 5.24 m/s
19.
Water (density = 1 x 10
3
kg/m
3
) is flowing through a pipe whose radius is 0.02 m with a speed of 11
m/s. This same pipe go
es up to the second floor of the building, 2.5 m higher, and the pressure remains
unchanged. What is the cross

sectional area of the pipe on the second floor?
19A.
(1)
First Equation:
p
1
+ ½
v
1
2
+
杹
1
=
p
2
+ ½
v
2
2
+
杹
2
(2) ½ v
1
2
+
gy
1
= ½ v
2
2
+ gy
2
(3) v
1
2
= v
2
2
+ 2gy
2
(4) v
2
= [(v
1
2
)
–
2gh]
1/2
(5) v
2
= [(11 m/s)
2
–
2(9.8 m/s
2
)(2.5 m)]
1/2
(6)
v
2
= 8.49 m/s
(7)
Second Equation:
A
1
v
1
= A
2
v
2
(8) A
2
= A
1
v
1
/ v
2
(9)
Third Equation:
A
1
=
r
2
(10) A
2
= (
r
2
)v
1
/ v
2
(11) A
2
=
(0.02 m)
2
(1
1 m/s) / (
8.49 m/s
)
(12)
A
2
= 1.63 x 10

3
m
2
Enter the password to open this PDF file:
File name:

File size:

Title:

Author:

Subject:

Keywords:

Creation Date:

Modification Date:

Creator:

PDF Producer:

PDF Version:

Page Count:

Preparing document for printing…
0%
Comments 0
Log in to post a comment