A1.3 kg ball drops vertically onto a floor, hitting with a speed of 21 m/s. it rebounds with an initial speed of 9.8 m/s. (a) what impulse acts on the ball during the contact? (b) if the ball is in contact with the floor for 0.0156 s, what is the magnitude of the average force on the floor from the ball?

a) Impulse(J)=40.04 kg.m/s

b)F=2566.66 N

Explanation:

Given that

We know that impulse(J) impulse is a vector quantity

We know that P=mV

So

Now by putting the values

J=1.3(9.8-(-21))

So impulse(J)=40.04 kg.m/s

Force is given as

or we can say that

So

F=2566.66 N

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.