Question

A lapidary plate at rest is turned on to cut a gemstone. The plate rotates until it reaches a tangential speed of 12.0 m/s. What is the centripetal acceleration of a point 0.10 m from the center of the plate?

a. 1.2 m/s²
b. 14.4 m/s²
c. 1440 m/s²
d. 120 m/s²

Answer 1

The centripetal acceleration of a point on a lapidary plate is calculated using the formula a_c = v^2/r, where v is the tangential speed and r is the distance from the center of the plate. Plugging in the values, the centripetal acceleration is found to be 1440 m/s^2.

Step-by-step explanation:

The centripetal acceleration of a point on a lapidary plate can be found using the formula:

ac = v2/r

Where v is the tangential speed and r is the distance from the center of the plate. Plugging in the values, we get:

ac = (12.0 m/s)2 / 0.10 m = 1440 m/s2

Therefore, the centripetal acceleration is 1440 m/s2, so the correct answer is (c) 1440 m/s2.