Question

A(n) 74.2 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 39.8 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.872 kg camera in her hand and decides to get back to the shuttle by throwing the camera at a speed of 12 m/s in the direction away from the shuttle.

How long will it take for her to reach the shuttle? Answer in minutes.

Answer 1

4.6Min

Step-by-step explanation:

This is a very simple momentum problem

`\left \{ {{M_(Ast)*V_(1) = M_(Cam) *V_(2) \\} \atop {t=(x)/(V_(1) ) }} \right.`

`(74.2KG-0.872KG)V_(1)=0.872KG*12M/S`

`V_(1)=0.1427M/S`

`t=(39.8M)/(0.1427M/S) =279s=4.6Min`

about use 4.6Min back shuttle.