A(n) 70.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 41.4 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.916 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

4.4 min

Explanation:

To solve this, we use the law of conservation of momentum

0 = -m(a).v(a) + m(c).v(c), where

m(a) = mass of the astronaut

m(c) = mass of the camera

v(a) = velocity of the astronaut

v(c) = velocity of the camera

Making v(a) subject of the formula, we have

v(a) = m(c).v(c)/m(a)

Now, we make one last assumption, that the astronaut is moving at constant speed and time, thus the time needed will be

t = s/v -> since s = vt

Now, we already have our v from above, so we substitute

t = s/[m(c).v(c)/m(a)]

t = s.m(a)/m(c).v(c)

Applying the values, we have

t = (41.4 * 70.1) / (0.916 * 12)

t = 2902.14 / 10.992

t = 264 seconds or 4.4 minutes

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going   so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me