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hram777 [196]
3 years ago
12

A coin placed 30.8 cm from the center of a rotating, horizontal turntable slips when its speed is 50.8 cm/s.

Engineering
1 answer:
Anestetic [448]3 years ago
6 0

Answer:

a)(A) static friction

b)\mu _s=0.085

Explanation:

Given that

r = 30.8 cm

V = 50.8 cm/s

a)

(A) static friction

b)

At the condition of motion

Static friction = Centrifugal force

\mu _s\ m\ g=\dfrac{mV^2}{r}

\mu _s\ g=\dfrac{V^2}{r}

\mu _s\times 980=\dfrac{50.8^2}{30.8}

\mu _s=0.085

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Answer:

The correct answer is option (A) 0.060 uF

Note: Kindly find an attached image of the complete question below

Sources: The complete question was well researched from Quizlet.

Explanation:

Solution

Given that:

C₁ = 0.1 μF

C₂ =0.22 μF

C₃ = 0.47 μF

In this case, C₁, C₂ and C₃ are in series

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Their equivalent becomes:

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1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]

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