Urgent ! a spring has a spring constant of 105 n/m. if you compress the spring 0.1 m past its natural length, what force does the spring apply? a. 500n, b. 10.5n, c. 0.0005n, d. 1050n

Force, F = -10.5 N

Explanation:

It is given that,

Spring constant of a spring, k = 105 N/m

The compression in the spring, x = 0.1 m

We need to find the force applied by the spring. We know that the Hooke's law gives the relationship between the force and the compression in the spring. It is given by :

F = -kx

F = -10.5 N

So, the force applied by the spring is 10.5 N. Hence, this is the required solution.

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

trig 15.3 cos