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Sagot :
final angular speed
ω = v / r = 2.20m/s / ½*35.0m = 0.126 rad/s
and so prior to reaching that final speed
the angular acceleration
α = Δω / Δt = 0.126rad/s / 15.0s = 0.00838 rad/s² ◄
and the tangential acceleration
a = Δv / Δt = 2.20m/s / 15.0s = 0.147 m/s² ◄
OR
a = αr = 0.00838rad/s² * ½*35.0m = 0.147 m/s² ◄
ω = v / r = 2.20m/s / ½*35.0m = 0.126 rad/s
and so prior to reaching that final speed
the angular acceleration
α = Δω / Δt = 0.126rad/s / 15.0s = 0.00838 rad/s² ◄
and the tangential acceleration
a = Δv / Δt = 2.20m/s / 15.0s = 0.147 m/s² ◄
OR
a = αr = 0.00838rad/s² * ½*35.0m = 0.147 m/s² ◄
(a) The magnitude of the wheels angular acceleration is 0.0088 rad/s².
(b) The magnitude of the tangential acceleration after the maximum operational speed is reached is 0.153 m/s².
Angular acceleration of the wheel
The angular acceleration of the wheel is calculated as follows;
α = ω/t
ω = v/r
α = v/(rt)
α = (2.3)/(17.5 x 15)
α = 0.0088 rad/s²
Tangential acceleration of the wheel
a = v/t
a = (2.3)/15
a = 0.153 m/s²
Learn more about angular acceleration here: https://brainly.com/question/25129606
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