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laiz [17]
3 years ago
14

A student wearing a frictionless roller skates on a horizontal is pushed by a friend with a constant force of 55N. How far must

the student be pushed, starting from rest, so that her final kinetic energy is 362 J
Physics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

362 = 55d

d = 362/55

d = 6.58m

Hence the student must push at a distance of 6.58m

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VashaNatasha [74]

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F = mg

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so it is given as

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so as if we compare the two forces we can say that since the value of sine is always less than 1 for an angle less than 90 degree

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3 years ago
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Answer:

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Explanation:

6 0
3 years ago
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An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s
DENIUS [597]
A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
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3 years ago
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makkiz [27]

Answer:

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Explanation:

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