A spring with a spring constant of 150 n/m has an equilibrium length of 8. 0 cm. What is the change in spring potential energy when the spring is stretched from a length of 11. 0 cm to 14. 0 cm?.

Answers

Answer 1

When the spring is stretched to 11.0 cm - 8.0 cm = 3.0 cm = 0.030 m away from equilibrium, it stores

1/2 (150 N/m) (0.030 m)² = 0.0675 J

of potential energy, while stretching it to 14.0 cm - 8.0 cm = 6.0 cm = 0.060 m from equilibrium stores

1/2 (150 N/m) (0.060 m)² = 0.27 J

Then the change in potential energy is 0.27 J - 0.0675 J = 0.2025 J ≈ 0.21 J


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Answer:

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Answer:

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Answers

The spring undergoes a change in potential energy of

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The citation that is formatted correctly using MLA's guidelines is;

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Answer:

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Answer:Hubble’s main mirror being polished before installation

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Answers

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system%28a=b%2B12%2Ca%2Bb=44%29



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what this is brainly

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A on EDGENUNITY

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