Final answer:
The velocity of the outer rim of a 12 cm diameter CD, with a centripetal acceleration of 9.0 m/s², is approximately 0.735 m/s.
Step-by-step explanation:
The student is asking about the velocity of the outer rim of a CD with a given centripetal acceleration. To find this, we can use the formula for centripetal acceleration, a = v^2 / r, where a is the centripetal acceleration, v is the tangential velocity, and r is the radius of the circular path.
The question provides us with an acceleration value of 9.0 m/s² and a CD diameter of 12 cm, meaning the radius r is 6 cm or 0.06 m. By rearranging the formula to solve for v (v = √(ar)), we can find the velocity of the CD rim.
Using the given values:
v = √(9.0 m/s² * 0.06 m) = √(0.54 m²/s²) ≈ 0.735 m/s
Therefore, the velocity of the outer rim of a 12 cm diameter CD is approximately 0.735 m/s when the CD player spins it with a centripetal acceleration of 9.0 m/s².