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Suppose a 3950-kg space probe expels 3300 kg of its mass at a constant rate with an exhaust speed of 1.85 × 103 m/s. sho...

Suppose a 3950-kg space probe expels 3300 kg of its mass at a constant rate with an exhaust speed of 1.85 × 103 m/s. show answer No Attempt Calculate the increase in speed, in meters per second, of the space probe. You may assume the gravitational force is negligible at the probe’s location.

Δv =

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Answer #1

Given is :-

Exhaust speed  u= 1.85 x 10m/s

Initial mass  mi = 3950kg

Final mass  

mg = 3950kg - 3300kg

mg = 650kg

now,

The increase in velocity for space probe is given by

Av=u In mi

by plugging all the values we get

3950kg Av = (1.85 x 10°m/s) In 650k

which gives us

Av = 3338.32m/s or   Au = 3.33832 x 10 m/s

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