Calculate the force constant of a spring which is stretched 5 mm by a force of 12 N.
li

k = 2400 N/m

Explanation:

Given that,

Force acting on the spring, F = 12 N

The spring stretches 5 mm = 0.005 m

We need to find the force constant of the spring.

The force acting on the spring is given by Hooke's law as follows :

F = kx

Where k is force constant

So, the force constant of the spring is 2400 N/m.

wind moving at higher speeds will transfer more energy to the water, resulting in stronger waves.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j