A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. The tangential acceleration of a point on its rim is:. a. 12 m/s2.
b. 6.0 m/s2.
c. 5.0 m/s2.
d. 5.0 rad/s2.
e. 3.0 m/s2.

3 m/s²

Explanation:.

To solve this, we use the formula of tangential acceleration. The tangential acceleration is given by the following equation:

a(t) = r * a(a)

where

a(t) is the tangential acceleration, m/s²

r is the radius, m

a(a) is the angular acceleration, rad/s²

Again, we know that the radius can be gotten by saying r = d/2, thus, r = 1.2/2 = 0.6 m

We then multiply this by the angular acceleration, to get our tangential acceleration

a(t) = 0.6 * 5 = 3 m/s²

Therefore, the Tangential acceleration of a point on the flywheel rim has been found to be 3 m/s²

i think it would be 5