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.

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

i would say the answer is food.

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