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3 years ago

7

A spring with k = 19.5 N/cm is initially stretched 1.39 cm from its equilibrium length. a) How much more energy is needed to fur

ther stretch the spring to 6.93 cm beyond its equilibrium length?

1 answer:

lara31 [8.8K]3 years ago

7 0

Answer:

Energy needed = 54.02 J

Explanation:

the Energy in an elastic spring from hookes law is given as

F= ke , therefore the energy (E) is

E = $\frac{1}{2} Ke^{2}$

K = 19.5 N/cm

e = 1.39cm

E = $\frac{1}{2}$ x 19.5 x 1.39

E = 13.55 J

The energy to stretch the spring for 6.93cm is

E = $\frac{1}{2}$ x 19.5 x 6.93

E = 67.57 J

The more energy needed for the further stretch is

67.57 - 13.55

Energy needed = 54.02 J

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Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

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Put the value in to the formula

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3 years ago

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2 years ago

A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

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Explanation:

The given data is as follows.

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According to the formula of energy conservation,

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2 years ago

Giving brainiest to correct answer.

mixas84 [53]

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3 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

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Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

2 years ago