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.

grav potential energy

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.