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2 years ago

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A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that t

ime if the angular acceleration is constant?

1 answer:

Akimi4 [234]2 years ago

6 0

Answer:

Explanation:

Given that the grand stone has initial angular velocity of

w(ini)= 6rad/

And it has a final angular velocity of

w(fin)=12.20rad/sec

Time taken is t=16s

Using equation of angular motion

To get angular acceleration (α)

w(fin)=w(ini)+αt

12.20=6+16α

16α=12.20-6

16α=6.2

α=6.2/16

α=0.3875rad/sec²

The angular acceleration is 0.39rad/s²

Angle that he turn using

w(fin)²=w(ini)²+2αθ

12.2²=6²+2×0.3875θ

12.2²-6²=0.775θ

0.775θ=112.84

Then, θ=112.84/0.775

θ=145.6radian

The angular displacement is 145.6rad

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Answer:

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Solution:

As per the question:

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Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

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Now,

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$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

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