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taurus [48]
3 years ago
10

A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

the turntable surface of μs = 0.820 and μk = 0.440. The turntable very slowly speeds up.1. What is the angular speed in rpm when the coin slides off?
Physics
1 answer:
Komok [63]3 years ago
5 0

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

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Question 15
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Each of 134 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl
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Answer:

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B. T=63N

Explanation:

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Hence using the equation of force we have

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Where m=total mass of blocks,

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T=126m*a

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T=ma

Since the number of blocks before the string is 63, we multiply the mass of each block by 63.

Hence the tension can be computed as

T=63m*a

Since a=1/m then

T=63m*1/m

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