Version 040 – Final – TSOI – (58160)
1
This
printout
should
have
41
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A 1 meter long hose of 8 cm diameter is con
nected to a faucet.
The end of this hose is
connected to a second hose which is 4 meters
long and 4 cm in diameter. At the open end
of the second hose water flows out at a rate of
8 cm
3
/s.
What is the ratio of the speed of the water
flowing in the second hose to the speed of the
water flowing in the first hose?
1.
v
2
v
1
=
1
3
2.
v
2
v
1
= 8
3.
v
2
v
1
=
1
8
4.
v
2
v
1
= 2
5.
v
2
v
1
=
1
4
6.
v
2
v
1
= 1
7.
Cannot be determined from the informa
tion given.
8.
v
2
v
1
= 4
correct
9.
v
2
v
1
= 3
10.
v
2
v
1
=
1
2
Explanation:
Let :
r
1
= 4 cm
and
r
2
= 2 cm
.
According to the equation of continuity,
v
2
v
1
=
A
1
A
2
=
π r
2
1
π r
2
2
=
(4 cm)
2
(2 cm)
2
=
4
.
0
.
002
10.0 points
A uniform horizontal rod of mass 1
.
1 kg and
length 1
.
7 m is free to pivot about one end
as shown. The moment of inertia of the rod
about an axis perpendicular to the rod and
through the center of mass is given by
I
=
m ℓ
2
12
.
1
.
7 m
F
66
◦
pivot
1
.
1 kg
If a 6 N force at an angle of 66
◦
to the hor
izontal acts on the rod as shown, what is the
magnitude of the resulting angular accelera
tion about the pivot point? The acceleration
of gravity is 9
.
8 m
/
s
2
.
1. 7.14862
2. 0.402982
3. 7.30469
4. 6.58537
5. 12.8831
6. 2.40796
7. 5.32443
8. 4.95164
9. 0.146427
10. 1.87803
Correct answer: 0
.
146427 rad
/
s
2
.
Explanation:
Let :
ℓ
= 1
.
7 m
,
m
= 1
.
1 kg
,
θ
= 66
◦
,
and
F
= 6 N
.
By the parallel axis theorem, the moment
of inertia of a stick pivoted at the end is
I
=
I
cm
+
m d
2
=
1
12
m ℓ
2
+
m
parenleftbigg
ℓ
2
parenrightbigg
2
=
1
3
m ℓ
2
.
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Version 040 – Final – TSOI – (58160)
2
The sum of the torques (counterclockwise ro
tation defined as positive) is
summationdisplay
τ
=
F ℓ
sin
θ
−
m g
ℓ
2
=
I α
F ℓ
sin
θ
−
m g l
ℓ
2
=
parenleftbigg
1
3
m ℓ
2
parenrightbigg
α
6
F
sin
θ
−
3
m g
= 2
m ℓ α
vectorα
=
6
F
sin
θ
−
3
m g
2
m ℓ
Since
6
F
sin
θ
−
3
m g
= 6 (6 N) sin 66
◦
−
3 (1
.
1 kg) (9
.
8 m
/
s
2
)
= 0
.
547636 N
,
then
bardbl
vectorα
bardbl
=
vextendsingle
vextendsingle
vextendsingle
0
.
547636 N
vextendsingle
vextendsingle
vextendsingle
2 (1
.
1 kg) (1
.
7 m)
=
0
.
146427 rad
/
s
2
.
003
10.0 points
Standing at a crosswalk, you hear a frequency
of 555 Hz from the siren on an approaching
police car.
After the police car passes, the
observed frequency of the siren is 475 Hz.
Determine the car’s speed from these obser
vations. The speed of sound in air is 343 m
/
s.
1. 26.1394
2. 32.7622
3. 20.2939
4. 32.0744
5. 38.2577
6. 27.6199
7. 30.3597
8. 28.3635
9. 23.7723
10. 26.6408
Correct answer: 26
.
6408 m
/
s.
Explanation:
Let :
f
′
= 555 Hz
,
f
′′
= 475 Hz
,
and
v
= 343 m
/
s
.
From the Doppler effect,
f
=
v
±
v
0
v
∓
v
s
f
′
.
When the police car is approaching, the ob
served frequency is
f
′
=
v
v
−
v
s
f
and after
the police car passes, the observed frequency
is
f
′′
=
v
v
+
v
s
,
so
f
′
f
′′
=
v
v
−
v
s
f
v
v
+
v
s
f
=
v
+
v
s
v
−
v
s
f
′
(
v
−
v
s
) =
f
′′
(
v
+
v
s
)
v
s
=
f
′
−
f
′′
f
′
+
f
′′
v
=
555 Hz
−
475 Hz
555 Hz + 475 Hz
(343 m
/
s)
=
26
.
6408 m
/
s
.
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 Fall '10
 TSIO
 Acceleration, Force, Mass, Velocity

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