I have honestly never read anything about a car being disseminated,
or any instructions on how to do it, or any of the theory behind it.
Answer:
it probs gos fast then it was it would be easy if u asked us if their was any mass or friton
Explanation:
I'm happy to know that the diagram shows how it's all set up.
If I could see the diagram, then I could probably do a much
better job with an answer. As it is ... 'flying blind' as it were ...
I'm going to wing it and hope it's somewhat helpful.
If the pulley is movable, then I'm picturing one end of the rope
tied to a hook in the ceiling, then the rope passing down through
the pulley, then back up, and you lifting the free end of the rope.
A very useful rule about movable and combination pulleys is:
the force needed to lift the load is
(the weight of the load)
divided by
(the number of strands of rope supporting the load) .
With the setup as I described it, there are 2 strands of rope
supporting the load ... one on each side of the pulley. So the
force needed to lift the load is
(250 N) / 2 = 125 N .
Answer:
<h2>Stunt car A and stunt car B are identical cars with the same mass of 35.9 kg. They are both traveling at 37.1 m/s. Stunt car A crashes into a hard concrete wall, whereas stunt car B crashes into a big soft mattress. They both come to a complete stop after the impact.Stunt Car A experiences a <u>LARGE FORCE</u> over a <u>SHORT PERIOD</u> of time. Stunt Car B experiences a <u>SMALL FORCE</u> over a <u>LONG PERIOD</u> of time. Because of the force experienced by Stunt Car A, it will sustain <u>MORE</u> damage than Stunt Car B.</h2>
Explanation:
As we know that force is given as rate of change in momentum
so it is defined as

so here we know that both the cars are identical and both are moving with same speed so after coming to rest both the cars have same change in momentum
while Car A will experience more force because due to collision with concrete wall the time of collision is very small as compared to Car B
So Force on Car A is more than the force on Car B