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

by

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

0 thoughts on “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”

  1. 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;

    [tex]\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[/tex]

    The centripetal acceleration is calculated as;

    a = ω²r

    a = (3.93)² x 0.45

    a = 6.95 m/s²

    Reply

Leave a Comment