A loaded spring launches a 2.50 kg block, using a force of 450 N. If the change in momentum is 12.0 kg*m/s, how long was the block in contact with the spring?

37.5 seconds

Explanation:

The given parameters are;

The mass of the block on the spring, m = 2.50 kg

The force with which the loaded spring launches the block, F = 450 N

The change in momentum of the block, Δp = 12.0 kg·m/s

We have;

Let the force with which the block was launched = The net force,

By Newton's second law of motion, we have;

F = = Δp × Δt

Where;

Δt = The time the block is in contact with the spring

Therefore;

By plugging in the values for and Δp, we have;

The time duration the block is in contact with the spring, Δt = 37.5 seconds.

the answer is a 3.8cm

explanation:

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr