Physics 262, Final Exam, May 20, 2002, Dr. Baden
Try to do as many problems as you can, and show your work clearly.
There are a lot of problems here,
more than you might have guessed, the reason for this is to try to cover as much material as possible to try
to measure what you’ve learned.
I don’t expect everyone to finish every problem, so please do the
problems you are more sure of first, then go back to the others.
Credit will not be given for answers with
no work shown.
Partial credit will be given.
There are 170 points total you can get on the exam.
Problem 1 (10 points).
Water flows thru a fire hose of diameter
5.4cm
at a rate of
0.105 m
3
/s
.
The fire
hose ends in a nozzle that has a diameter of
2.20 cm
.
Calculate the speed that water exits the nozzle.
The conservation of flow equation is Av = constant=0.105m
3
/s .
The area of a hose with
diameter D is given by
4
2
D
A
π
=
.
The continuity equation is then
v
D
Av
4
2
=
= 0.105m3/s.
Solving for the final velocity gives
s
m
D
v
f
/
2
.
276
4
105
.
0
2
=
=
Problem 2 (10 points).
An astronaut on the Moon wishes to measure the local value of
g
(as in “F=mg”)
on the moon (g
moon
) by timing pulses traveling down a wire that has a large mass suspended from it.
Assume that the wire has a mass of
2.20 g
and a length of
1.10 m
, and that a
7.50 kg
mass is suspended
from it.
A pulse requires
0.015 s
to traverse the length of the wire.
Calculate g
moon
from these data,
neglecting the mass of the wire.
The velocity of waves on a string of mass
m
, length
L
, and under tension
T
is given by
L
m
T
v
=
.
In this problem, the tension comes from gravity due to the weight
M=1.5kg
, so
T=Mg
so we have
L
m
Mg
v
=
.
If the wave takes 6.1s to traverse the string of length 4.1m,
then the velocity is
v=1.1m/0.015s=73.3m/s
.
Solving for
g
gives
2
2
2
2
2
/
43
.
1
1
.
1
5
.
7
0022
.
0
/
3
.
73
s
m
m
kg
kg
s
m
ML
m
v
g
=
⋅
⋅
=
=
Problem 3 (10 points).
A point source emits sound waves with an average power output of
1.3 W
.
a)
find the intensity
and sound level (in dB)
4.50 m
from the source.
b)
find the distance at which the sound level is
95 dB
.
a)
the intensity is defined as power/area, and if the source is a point source then the waves
are emitted as spherical waves, and have an area given by A=4pr
2
at radius r.
The
intensity at r=4.5m is then given by
2
2
/
0051
.
0
5
.
4
4
3
.
1
m
W
W
A
P
I
=
=
=
.
The sound level in dB
is given by
0
log
10
I
I
=
β
where I
0
is the threshold intensity of 10
12
W/m
2
, so at 4.5m the
sound level is
dB
1
.
97
10
0051
.
0
log
10
12
=
=

.
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View Full Documentb)
You use the same formula for sound level, but now you have to solve for the intensity
where the level is 95dB:
2
3
2
5
.
2
5
.
9
12
10
0
/
10
16
.
3
/
10
10
10
10
m
W
x
m
W
I
I



=
=
=
=
β
and
solving for area using the formula
A
P
I
=
gives
2
3
4
5
.
411
10
16
.
3
/
3
.
1
r
x
I
P
A
π
=
=
=
=

.
Solving for the radius gives r =5.7m
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 Spring '10
 mikey
 Physics, Energy, Work, sound level

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