The blade of a windshield wiper moves through an angle of 90.08 in 0.40 s. The tip of the blade moves on the arc of a circle that has a radi

Question

The blade of a windshield wiper moves through an angle of 90.08 in 0.40 s. The tip of the blade moves on the arc of a circle that has a radius of 0.45 m. What is the magnitude of the centripetal acceleration of the tip of the blade

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Nho 6 months 2021-07-28T04:57:31+00:00 1 Answers 71 views 0

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    2021-07-28T04:58:51+00:00

    Answer:

    The centripetal acceleration is 6.95 m/s²

    Explanation:

    Given;

    angular displacement of the blade, θ = 90.08⁰

    duration of motion of the blade, t = 0.4 s

    radius of the circle moved by the blade, r = 0.45 m

    The angular speed of the blade in radian is calculated as;

    \omega  = \frac{\theta}{t} \times \frac{\pi \ radian}{180^0} \\\\\omega  =  \frac{90.08 ^0}{0.4 \ s} \times \frac{\pi \ radian}{180^0} \\\\\omega  = 3.93 \ rad/s

    The centripetal acceleration is calculated as;

    a = ω²r

    a = (3.93)² x 0.45

    a = 6.95 m/s²

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