Answer:
Explanation:
Givens
vi = 10 m/s
a = 1.5 m/s^2
d = 600 m
vf = ?
Formula
vf^2 = vi^2 + 2*a*d
Solution
vf^2 = 10^2 + 2*1.5 * 600
vf^2 = 100 + 1800
vf^2 = 1900
sqrt(vf^2) = sqrt(1900)
vf = 43.59 m/s
Answer:
The answer is "
".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

Potential energy shifts:


Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.



This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
Answer:
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Explanation:
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Answer:
1,3,5
Explanation:
i think maybe dont come at me