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jarptica [38.1K]

3 years ago

5

If an object is rolling without slipping, how does its linear speed compare to its rotational speed? If an object is rolling wit

hout slipping, how does its linear speed compare to its rotational speed? v = Rω v = ω/R They are unrelated. v = ω

1 answer:

salantis [7]3 years ago

6 0

Answer:

$v = r\omega$

Explanation:

If the object is rolling without slipping, every unit of rotated angle equals to a distance perimeter rotated.

Suppose the object complete 1 revolution within time t. The angular distance is 2π rad. Its angular velocity is 2π/t

The distance it covered is its circumference, which is 2πr, and so the speed is 2πr/t

So the linear speed compared to angular speed is

$\frac{v}{\omega} = \frac{2\pi r/t}{2\pi /t} = r$

$v = r\omega$

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Bob is threatening Tom’s life with a giant laser with wavelength (650 nm), a distance (D = 10 m) from the wall James is shackled

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Option c

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Read 2 more answers

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3 years ago

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IRISSAK [1]

A. $4.64\cdot 10^{11}m$

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$v=\frac{2\pi r}{T}$

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$v=30,000 km/s = 3\cdot 10^7 m/s$ is the orbital speed

r is the orbital radius

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Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

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and considering the value of the solar mass

$M_s = 2\cdot 10^{30}kg$

the mass of the black hole as a multiple of our sun's mass is

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C. $9.28\cdot 10^9 m$

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Substituting numbers into the formula, we find

$R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m$

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