A(n) 77.5 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 37.3 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.94 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 in units of min

Explanation:

The speed of the astronaut can be found with the help of law of conservation of momentum .

mv = MV , M is mass of astronaut , m is mass of object thrown , v is velocity of object thrown and V is velocity of  astronaut.

Putting the values

77.5 x V = .94 x 12

V = .14554 m /s

This will be the uniform velocity of astronaut.

Distance to be covered = 37.3 m

time taken = distance / velocity

= 37.3 / .14554

= 256.28 s

= 4.27 minutes.

c) up

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards