Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.