Question

What is the centripetal acceleration of a small Laboratory Centrifuge in which the tip of the test tube is moving at 19.0 meters per second in a circle with a radius of 10.0 cm

Answer 1

Answer:

3610 m/s^2

Explanation:

The centripetal acceleration of the tip of the test tube is given by:

$a=\frac{v^2}{r}$

where

v = 19.0 m/s is the velocity of the tip

r = 10.0 cm = 0.1 m is the radius of the circular trajectory of the tip

Substituting the numbers into the formula, we find

$a=\frac{(19.0 m/s)^2}{0.1 m}=3610 m/s^2$