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hinojosa (jlh3938) – homework 19 – Turner – (58185)
1
This printout should have 13 questions.
Multiplechoice questions may continue on
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beFore answering.
001
(part 1 oF 2) 10.0 points
A 1660 kg car starts From rest and accelerates
uniFormly to 16 m
/
s in 16
.
8 s
.
Assume that air resistance remains con
stant at 358 N during this time.
±ind the average power developed by the
engine.
Correct answer: 20
.
7931 hp.
Explanation:
m
= 1660 kg
,
v
i
= 0 m
/
s
,
v
f
= 16 m
/
s
,
and
Δ
t
= 16
.
8 s
.
The acceleration oF the car is
a
=
v
f

v
i
Δ
t
=
v
f
Δ
t
since v
i
= 0, so
a
=
16 m
/
s
16
.
8 s
= 0
.
952381 m
/
s
2
.
Thus the constant Forward Force due to the
engine is Found From
s
F
=
F
engine

F
air
=
ma
F
engine
=
F
air
+
ma
= 358 N + (1660 kg)
(
0
.
952381 m
/
s
2
)
= 1938
.
95 N
.
The average velocity oF the car during this
interval is
v
av
=
v
f
+
v
i
2
,
so the average power output is
P
=
F
engine
v
av
=
F
engine
p
v
f
2
P
= (1938
.
95 N)
±
16 m
/
s
2
²±
1 hp
764 W
²
=
20
.
7931 hp
.
002
(part 2 oF 2) 10.0 points
±ind the instantaneous power output oF the
engine at
t
= 16
.
8 s just beFore the car stops
accelerating.
Correct answer: 41
.
5861 hp.
Explanation:
The instantaneous velocity is 16 m
/
s and
the instantaneous power output oF the engine
is
P
=
F
engine
v
f
= (1938
.
95 N)(16 m
/
s)
±
1 hp
764 W
²
=
41
.
5861 hp
.
003
(part 1 oF 2) 10.0 points
Three 2 kg masses are located at points in
the
xy
plane as shown.
46 cm
54 cm
What is the magnitude oF the resultant
Force (caused by the other two masses) on
the mass at the origin? The universal gravita
tional constant is 6
.
6726
×
10
−
11
N
·
m
2
/
kg
2
.
Correct answer: 1
.
55847
×
10
−
9
N.
Explanation:
Let :
m
= 2 kg
,
x
= 46 cm = 0
.
46 m
,
y
= 54 cm = 0
.
54 m
,
and
G
= 6
.
6726
×
10
−
11
N
·
m
2
/
kg
2
.
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View Full Documenthinojosa (jlh3938) – homework 19 – Turner – (58185)
2
The force from the mass on the right points
in the
x
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 Fall '08
 Turner
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