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kolezko [41]
3 years ago
9

A 92-kg astronaut and a 2000-kg satellite are at rest relative to a space station. the astronaut pushes on the satellite, giving

it a speed of 0.14 m/s directly away from the station. seven and a half seconds later the astronaut comes into contact with the station. part a what was the initial distance from the station to the astronaut?
Physics
2 answers:
just olya [345]3 years ago
8 0
Momentum is conserved. 

<span>Momentum of the satellite is 1600*0.14=224 kg-m/sec </span>
<span>Resulting momentum of the astronaut will be equal -- hence </span>
<span>86*v=224 or the velocity of the astronaut is 2.60 m/sec </span>

<span>In 7.5 seconds the astronaut would travel 19.53 meters</span>
jekas [21]3 years ago
7 0

Answer:

22.8 s

Explanation:

First we calculate the change in momentum of the satellite.

ΔP = 2000 kg (0.14 m/s 0 m/s) = 280 kg·m/s

The change in momentum of the satellite is equal to the change in momentum of the astronaut as a result of Newton's Third law. Therefore:

280 kg·m/s = 92kg * v

                 v = 3.04 m/s

d = speed * time = 3.04 m/s * 7.5 s = 22.8 m

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