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V125BC [204]
1 year ago
13

a pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25 o from the vertical. if the initial sp

eed of the pendulum is 1.2 m/s, what is its speed at the bottom of the swing?
Physics
1 answer:
stiks02 [169]1 year ago
3 0

Answer:

1 / 2 m v^2 = L m g (1 - cos θ)

This is the KE due to the pendulum falling from a 25 deg displacement

v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2

v = 1.92 m/s      this is the speed due to an initial displacement of 25 deg

Its speed at the bottom would then be

1.92 + 1.2 = 3.12 m/s   since it gains 1.92 m/s from its initial displacement

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Answer:

W = 16.5 Kj

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E = 16471

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t = 5min 30sec = (5×60) + 30 = 330sec

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