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Arisa [49]
2 years ago
9

A coin is placed 32 cm from the center of a horizontal turntable, initially at rest. The turntable then begins to rotate. When t

he speed of the coin is 110 cm/s (rotating at a constant rate), the coin just begins to slip. The acceleration of gravity is 980 cm/s^2 . What is the coefficient of static friction between the coin and the turntable?
Physics
1 answer:
pashok25 [27]2 years ago
6 0

Answer:

0.39

Explanation:

distance from the center (r) = 32 cm = 0.32 m

speed of the coin (v) = 110 cm/s = 1.1 m/s

acceleration due to gravity (g) = 980 m/s^{2} = 9.8 m/s^{2}

find the coefficient of static friction (k) between the coin and the turn table

frictional force = kmg

before the table begins to move, the frictional force balances the centripetal force (\frac{mv^{2} }{r})

therefore

frictional force = centripetal force

kmg = \frac{mv^{2} }{r}

kg = \frac{v^{2} }{r}

k = \frac{v^{2} }{r} ÷ g

k = \frac{1.1^{2} }{0.32} ÷ 9.8 = 0.39

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The ejection of electrons form a metal surface when the metal surface is exposed to a monochromatic electromagnetic wave of sufficiently short wavelength or higher frequency (or equivalently, above a threshold frequency),  which leads to the enough energy of the wave to incident and get absorbed to the exposed surface emits electrons. This phenomenon is known as the photoelectric effect or photo-emission.

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