Physics: A4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 13.5 rad/s. what is the magnitude of the angular acceleration of the wheel?

A4.50-kg wheel that is 34.5 cm in diameter rotates

aubrey1161

03.01.2022

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

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H7ttaMaan

03.01.2022

Use the formula

${\omega_f}^2-{\omega_i}^2=2\alpha\Delta\theta$

$\implies\left(13.5\dfrac{\rm rad}{\rm s}\right)^2-\left(22.0\dfrac{\rm rad}{\rm s}\right)^2=2\alpha(13.8\,\mathrm{rad})$

$\implies\alpha=-10.9\dfrac{\rm rad}{\mathrm s^2}$

so the magnitude is 10.9 rad/s^2.

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liltonio

03.01.2022

:> CH - HB_HLFC QD

Explanation:

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