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2 years ago

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The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the

bearings and the air brings the wheel to rest in 1.0 $h$. What is the magnitude of the constant angular acceleration of the wheel in $rev/min^2$? Do not enter the units.

1 answer:

Oksana_A [137]2 years ago

6 0

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t

⇒ α = (0 rev/min - 113 rev/min) / (60 min)

⇒ α = - 1.883 rev/min²

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Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

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