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satela [25.4K]
2 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]2 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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(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

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k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

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So, Q=mgε₀πr²/q

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(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

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