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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

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The Answer Is...............:

What is the centripetal acceleration of a small laboratory centrifuge in which the-example-1
User New Thrall
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Answer:
3.61(10)^(3) (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)=(V^(2))/(r)

Where:


V=19m/s is the velocity


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


a_(c)=((19m/s)^(2))/(0.1m)


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

User Helfer
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