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vovangra [49]
2 years ago
11

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 w

hen the spring is stretched from a length of 11. 0 cm to 14. 0 cm?.
SAT
1 answer:
natima [27]2 years ago
8 0

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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C=59(F−32)The equation above shows how temperature F, measured in degrees Fahrenheit, relates to a temperature C, measured in de
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Our answer choices are:

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