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a Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a time of 15 s. what is the magnitude of the wheels angular acceleration?
b. what is the magnitude of the tangential acceleration after the maximum operational speed is reached?​

Answer :

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 ◄
Onyebuchinnajigo

(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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