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a_sh-v [17]
3 years ago
10

A 125-g coin is placed 8.0 cm from the axis of rotation of a horizontally rotating turntable as shown. The coefficient of static

friction between the coin and the turntable is μs . The turntable makes exactly 1.0 revolution per second and the coin remains in place without slipping. [a] (8 pts) Draw a free-body force diagram for the coin. [b] (10pts)What is the frictional force acting on the coin? [c] (12pts)When the rotational speed is increased to 1.2 revolutions per second, the coin starts to slip. Calculate the coefficient of static friction μs between the coin and the turntable?
Physics
1 answer:
Marianna [84]3 years ago
5 0

Answer:

zhvshshisvdiscdoscd if vdidg

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<h3>How fast was the physicist traveling, according to his own testimony?</h3>

Using the formula for doppler shift for light,

λ' = λ√[(1 + v/c)/(1 - v/c)] where

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Given that the driver is moving away from the stop light, we take the driver as the source. Since, the Doppler shift made the red light of wavelength 650nm appear green to him, with a wavelength of 520nm. we have

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λ' = λ√[(1 + v/c)/(1 - v/c)]

520 nm = 650 nm√[(1 + v/c)/(1 - v/c)]

520/650 = √[(1 + v/c)/(1 - v/c)]

0.8 = √[(1 + v/c)/(1 - v/c)]

Squaring both sides, we have

0.8² = (1 + v/c)/(1 - v/c)

0.64 = (1 + v/c)/(1 - v/c)

0.64(1 - v/c) = (1 + v/c)

0.64 - 0.64v/c = 1 + v/c

0.64 - 1 = v/c + 0.64v/c

-0.36 = 1.64v/c

-0.2195 = v/c

v = -0.22c

v = -0.22 × 3 × 10⁸ m/s

v = -0.66 × 10⁸ m/s

v = -6.6 × 10⁷ m/s

So, the physicist traveling, according to his own testimony at -6.6 × 10⁷ m/s.

Learn more about doppler shift for light here:

brainly.com/question/28499579

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The escape speed from the moon is much smaller than from earth. True or False
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Answer:

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The escape speed from the Moon is much smaller than from Earth.

Explanation:

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Then, replacing those values in equation 1 it is gotten.

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