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sweet [91]
4 years ago
13

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

h's surface and a kinetic energy of 4200 MJ . Later in its orbit the satellite's potential energy is 5700 MJ . Part A Use conservation of energy to find its kinetic energy at that point.
Physics
1 answer:
serious [3.7K]4 years ago
5 0

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

E_{initial} = E_{final}

PE_{initial}+KE_{initial} = PE_{final}+KE_{final}

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}

KE_{final} = (5100+4200)-5700

KE_{final} = 3600MJ

Therefore the final kinetic energy is 3600MJ

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Mariulka [41]
Refer to the diagram shown below.

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Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
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t = time, s

To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.

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4 years ago
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<u>Answer:</u>

"The Sun's diameter is about five times that of Earth" is not true about the sun.

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