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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

Sagot :

onyebuchinnaji

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

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