wolz (cmw2833) – HW #7 – Antoniewicz – (57420)
1
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printout
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have
19
questions.
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before answering.
001(part1of2)10.0points
A 1560 kg car starts from rest and accelerates
uniformly to 19
.
3 m
/
s in 18
.
1 s
.
Find the average power developed by the
engine.
Assume that air resistance remains
constant at 382 N during this time.
Correct answer: 26
.
4589 hp.
Explanation:
Let :
m
= 1560 kg
,
v
i
= 0 m
/
s
,
v
f
= 19
.
3 m
/
s
,
and
Δ
t
= 18
.
1 s
.
The acceleration of the car is
a
=
v
f

v
i
Δ
t
=
v
f
Δ
t
=
19
.
3 m
/
s
18
.
1 s
= 1
.
0663 m
/
s
2
and the constant forward force due to the
engine is found from
summationdisplay
F
=
F
engine

F
air
=
m a
F
engine
=
F
air
+
m a
= 382 N + (1560 kg)
(
1
.
0663 m
/
s
2
)
= 2045
.
43 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
parenleftBig
v
f
2
parenrightBig
= (2045
.
43 N)
parenleftbigg
19
.
3 m
/
s
2
parenrightbigg parenleftbigg
1 hp
746 W
parenrightbigg
=
26
.
4589 hp
.
002(part2of2)10.0points
Find the instantaneous power output of the
engine at
t
= 18
.
1 s just before the car stops
accelerating.
Correct answer: 52
.
9178 hp.
Explanation:
The instantaneous velocity is 19
.
3 m
/
s and
the instantaneous power output of the engine
is
P
=
F
engine
v
f
= (2045
.
43 N)(19
.
3 m
/
s)
parenleftbigg
1 hp
746 W
parenrightbigg
=
52
.
9178 hp
.
003(part1of2)10.0points
On the way to the moon, the Apollo astro
nauts reach a point where the Moon’s gravi
tational pull is stronger than that of Earth’s.
Find the distance of this point from the
center of the Earth.
The masses of the
Earth and the Moon are 5
.
98
×
10
24
kg and
7
.
36
×
10
22
kg, respectively, and the distance
from the Earth to the Moon is 3
.
84
×
10
8
m.
Correct answer: 3
.
45653
×
10
8
m.
Explanation:
Let :
M
e
= 5
.
98
×
10
24
kg
,
m
m
= 7
.
36
×
10
22
kg
,
and
R
e
= 3
.
84
×
10
8
m
.
Consider the point where the two forces are
equal. If
r
e
is the distance from this point to
the center of the Earth and
r
m
is the distance
from this point to the center of the Moon,
then
G m M
e
r
2
e
=
G m M
m
r
2
m
r
2
m
r
2
e
=
M
m
M
e
r
m
r
e
=
radicalbigg
M
m
M
e
.
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wolz (cmw2833) – HW #7 – Antoniewicz – (57420)
2
It is also true that
R
=
r
e
+
r
m
=
r
e
+
radicalbigg
M
m
M
e
r
e
r
e
=
R
1 +
radicalbigg
M
m
M
e
=
3
.
84
×
10
8
m
1 +
radicalBigg
7
.
36
×
10
22
kg
5
.
98
×
10
24
kg
=
3
.
45653
×
10
8
m
.
004(part2of2)10.0points
What
is
the
acceleration
due
to
the
Earth’s gravity at this point?
The value
of
the
universal
gravitational
constant
is
6
.
672
×
10
−
11
N
·
m
2
/
kg
2
.
Correct answer: 0
.
00333946 m
/
s
2
.
Explanation:
a
=
F
m
=
G M
e
r
2
e
a
= (6
.
672
×
10
−
11
N
·
m
2
/
kg
2
)
×
5
.
98
×
10
24
kg
(3
.
45653
×
10
8
m)
2
=
0
.
00333946 m
/
s
2
.
005(part1of2)10.0points
Given:
G
= 6
.
67259
×
10
−
11
N m
2
/
kg
2
.
A
1050
kg
geosynchronous
satellite
or
bits a planet similar to Earth at a radius
1
.
96
×
10
5
km from the planet’s center.
Its
angular speed at this radius is the same as the
rotational speed of the Earth, and so they ap
pear stationary in the sky. That is, the period
of the satellite is 24 h
.
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 Spring '08
 Turner
 Physics, Energy, Mass, Velocity

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