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KengaRu [80]
3 years ago
12

You have a spring with k = 640 N/m connected to a mass with m = 10 kg. You start the system oscillating and measure the velocity

of the mass as it moves through the spring's equilibrium position to be 4 m/s. At what position is the kinetic energy of the system equal to the potential energy?
Physics
1 answer:
sdas [7]3 years ago
4 0

Answer:

X= 0.5 m

Explanation:

Given that

K= 640 N/m

m=10 kg

v= 4 m/s

We know that

kinetic energy of the mass

KE=1/2 m v²

Potential energy

PE=1/2 k X²

Given that kinetic and potential energy are equal

1/2 k X² = 1/2 m v²

k X² =  m v²

Now by putting the values

640  X² = 10 x 4²

X= 0.5 m

So at 0.5 m  kinetic and potential energy will be equal.

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A 4 kg bowling ball moving at 1.4 m/s east impacts a 400 g pin that is stationary. After the impact, the ball is moving at 0.5 m
nignag [31]

The speed of the pin after the elastic collision is 9 m/s east.

<h3>Final speed of the pin</h3>

The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;

m1u1 + mu2 = m1v1 + m2v2

where;

  • m is the mass of the objects
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4(1.4) + 0.4(0) = 4(0.5) + 0.4v2

5.6 = 2 + 0.4v2

5.6 - 2 = 0.4v2

3.6 = 0.4v2

v2 = 3.6/0.4

v2 = 9 m/s

Thus, The speed of the pin after the elastic collision is 9 m/s east.

Learn more about linear momentum here: brainly.com/question/7538238

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Longitudional waves travel through a series of ________ and ___________.
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Explanation:

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28. Ken and Musa shared a cake such that Ken got twice the size
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Required

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Substitute: Ken = 2 * Musa

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Recall that: Ken = 2 * Musa

Ken = 2 * \frac{1}{3}

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