hernandez (ejh742) – homework 17 – Turner – (58120)
1
This printout should have 13 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
A 1640 kg car starts From rest and accelerates
uniFormly to 15
.
9 m
/
s in 15
.
5 s
.
Assume that air resistance remains con
stant at 337 N during this time.
±ind the average power developed by the
engine.
Correct answer: 21
.
5196 hp.
Explanation:
m
= 1640 kg
,
v
i
= 0 m
/
s
,
v
f
= 15
.
9 m
/
s
,
and
Δ
t
= 15
.
5 s
.
The acceleration oF the car is
a
=
v
f

v
i
Δ
t
=
v
f
Δ
t
since v
i
= 0, so
a
=
15
.
9 m
/
s
15
.
5 s
= 1
.
02581 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
= 337 N + (1640 kg)
(
1
.
02581 m
/
s
2
)
= 2019
.
32 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
= (2019
.
32 N)
±
15
.
9 m
/
s
2
²±
1 hp
764 W
²
=
21
.
5196 hp
.
002
(part 2 oF 2) 10.0 points
±ind the instantaneous power output oF the
engine at
t
= 15
.
5 s just beFore the car stops
accelerating.
Correct answer: 43
.
0392 hp.
Explanation:
The instantaneous velocity is 15
.
9 m
/
s and
the instantaneous power output oF the engine
is
P
=
F
engine
v
f
= (2019
.
32 N)(15
.
9 m
/
s)
±
1 hp
764 W
²
=
43
.
0392 hp
.
003
(part 1 oF 2) 10.0 points
Three 7 kg masses are located at points in
the
xy
plane as shown.
34 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: 3
.
0425
×
10
−
8
N.
Explanation:
Let :
m
= 7 kg
,
x
= 34 cm = 0
.
34 m
,
y
= 54 cm = 0
.
54 m
,
and
G
= 6
.
6726
×
10
−
11
N
·
m
2
/
kg
2
.
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2
The force from the mass on the right points
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 Spring '08
 Turner
 Physics, Force, Mass, Work, General Relativity, Correct Answer, Celestial mechanics

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