Question

a merry-go-round starts from rest and accelerates at a constant rate of 0.4 rev/s2.note: this is a multi-part question. once an answer is submitted, you will be unable to return to this part.what is its rotational velocity after 6 s?the rotational velocity of the merry-go-round is 7.2 numeric responseedit unavailable. 7.2 incorrect.rev/s.

Answer 1

Answer: The angular acceleration of the merry-go-round is 0.4 rev/s^2. We can use the following equation to find the final angular velocity:

ω_f = ω_i + αt

where ω_f is the final angular velocity, ω_i is the initial angular velocity (which is zero in this case), α is the angular acceleration, and t is the time.

Substituting the given values, we get:

ω_f = 0 + (0.4 rev/s^2)(6 s)

ω_f = 2.4 rev/s

Therefore, the rotational velocity of the merry-go-round after 6 s is 2.4 rev/s.