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/second in a circle with a radius of 10.0 centimeters? A. 1.82 × 102 meters/second2 B. 3.61 × 103 meters/second2 C. 5.64 × 103 meters/second2 D. 2.49 × 103 meters/second2 E. 1.18 × 103 meters/second2

Answer 1

Answer:$3.61(10)^{3} \frac{m}{s^{2}}$

The centripetal acceleration $a_{c}$ of an object moving in a uniform circular path is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$

Where:

$V=19m/s$ is the velocity

$r=10cm=0.1m$ is the radius of the circle

$a_{c}=\frac{(19m/s)^{2}}{0.1m}$

$a_{c}=3610m/s^{2}=3.61(10)^{3}m/s^{2}$