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satela [25.4K]
3 years ago
9

The amplitude of a lightly damped oscillator decreases by 4.2% during each cycle. What percentage of the mechanical energy of th

e oscillator is lost in each cycle?
Physics
1 answer:
ra1l [238]3 years ago
6 0

Answer:

V= A ω      maximum KE of object in SHM

V2 / V1 = .958     ratio of amplitudes since ω is constant

KE2 / KE1 = 1/2 m V2^2 / (1/2 m V1^2) = (V2 / V1)^2

KE2 / KE1 = .958^2 = .918

So KE2 = .918 KE1 and .082 = 8.2% of the energy is lost in one cycle

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       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

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in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

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        x = 124.8 m

Impact speed when it hits the ground

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        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

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