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Rudiy27
3 years ago
13

An athlete stretches a spring an extra 28.6 cm beyond its initial length. how much energy has he transferred to the spring, if t

he spring constant is 52.9 n/cm?
Physics
1 answer:
g100num [7]3 years ago
6 0
PE = 0.5 × k × x²

PE potential Energy
k spring constant
x stretch/compression of the spring
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Dennis_Churaev [7]

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<h3>Unpolarized light:</h3>

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1 year ago
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Satellite (moon) revolve around the planet
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Answer:

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

Explanation:

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Then the mass of the material required to make such a sphere would be calculated as:

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V_t=\frac{4}{3} \pi.r_2^3

And the volume of the hollow space in the sphere:

V_h=\frac{4}{3} \pi.r_1^3

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<u>Then the mass of the material used is:</u>

m=\rho.V

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

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