Acoin is placed 34 cm from the center of a horizontal turntable, initially at rest. the turntable then begins to rotate. when the speed of the coin is 120 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?

0.43

Explanation:

At the time of slipping , centripetal force becomes equal to the limiting static friction whose value is μ R or μ mg

Centripetal force which makes it rotate is given by

m v² / r where v is velocity of rotation and r is radius of circular orbit

So at limiting equilibrium

m v² /r = μ mg

μ =

μ =

= .43

The coefficient of static friction between the coin and the turntable is 0.43

If a coin is placed 34 cm from the center of a horizontal turntable, initially at rest, at the time of slippage, the centripetal force is always equal to the friction force. Mathematically;

Given the following parameters

v = 120 cm/s

g =  980 cm/s²

r = 34 cm

Substitute the given parameters into the formula to have:

Hence the coefficient of static friction between the coin and the turntable is 0.43

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carla slowed down in the beginning of her trip, stopped, and then sped up

i think the answer would be reserve.