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A force of 5.00 N is applied to a spring whose elastic constant is 0.250 N/cm. Its change in length is

20.0 cm
23.5 cm
25.0 cm
30.5 cm
35.0 cm

Answer :

Hanny25
Hope this answer helps
${teks-lihat-gambar} Hanny25

When the force is applied on the spring, the change in length must be  20 cm. The first option is correct.

What is gravitational potential energy?

If an object is lifted, work is done against gravitational force. The object gains energy.

Given is a force of 5.00 N is applied to a spring whose elastic constant is 0.250 N/cm.

Using Hooke's law, expressed as

Force, F = k (x₂-x₁) = k x

Put the values, we get Its change in length is

5 = 0.250x

x = 20 cm

Thus, the change in length must be  20 cm.

Learn more about gravitational potential energy.

https://brainly.com/question/3884855

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