The graph below shows the force required to stretch a spring various distances. review the graph below.

a graph of force versus elongation. elongation (m) is on the x axis and force (n) is on the y axis. a straight line passes through the points 0 comma 0 and 10 comma 15.

what is the best approximate value for the elastic potential energy (epe) of the spring elongated by 7.0 meters?

1.5 j
11 j
37 j
74 j

37 J

Explanation:

Given:

A graph of force versus elongation in which a line passes through the points (0, 0) and (10, 15).

The x-axis represents elongation and y-axis represents force.

Elongation (x) = 7.0 m

Now, we know that the slope of a line of force versus elongation graph is the value of the spring constant.

Therefore, finding the slope of the line passing through (0, 0) and (10, 15), we get:

Therefore, the spring constant (k) = 1.5 N/m.

Now, the elastic potential energy (EPE) of the spring is given as:

Plug in 1.5 for 'k' and 7.0 for 'x'. Solve for EPE.

Therefore, the best approximate value for the elastic potential energy (EPE) of the spring elongated by 7.0 meters is 37 J.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.