Physics, asked by kumarbipinbt50441, 1 year ago

A man stands on a platform that is rotating with an angular speed of 1.22 rev/s; his arms are outstretched and he holds a weight in each hand. With his hands in this position the total rotational inertia of the man, the weights, and the platform is 6.13 kg· m2. If by moving the weights the man decreases the rotational inertia to 1.97 kg ∙ m2, (a) what is the resulting angular speed of the platform and (b) what is the ratio of the new kinetic energy to the original kinetic energy? Assume the platform rotates without friction.

Answers

Answered by hiamni46
0

Answer:

sorry

Explanation:

I don't know answer, plz click pic of questions

Similar questions