At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

A uniform disk with mass 43.9 kgkg and radius 0.280 mm is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 29.0 NN is applied tangent to the rim of the disk. Part A What is the magnitude vv of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.400 revolution

Sagot :

Answer:

1.36 m/s

Explanation:

I = ½mR²

τ = FR

α = τ/I = FR / (½mR²) = 2F/mR

a = Rα = 2F/m

s = θR

v² = u² + 2as

u = 0

v = √2as = √(2(2F/m)(θR)) = 2√(FθR/m)

v = 2√(29.0(0.400)(2π)(0.280) / 43.9) = 1.3636272...

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.