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4 years ago

If a particle undergoes shm with amplitude 0.21 m, what is the total distance it travels in one period?

1 answer:

7 0

If a particle undergoes simple harmonic motion with an amplitude of 0.21 meters, this means that the maximum displacement of the particle from its resting position is 0.21. For one period, it traveled from its starting position which is twice the amplitude and then back to its original position which is another distance that is twice the amplitude as well. Therefore, the total distance it traveled is 2*amplitude + 2*amplitude = 2*0.21 + 2*0.21 = 0.42 + 0.42 = 0.84 meters.

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Answer:

Explanation:

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3 years ago

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Answer:

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3 years ago

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Answer:

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3 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

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3 years ago

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<h3>Introduction</h3>

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