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if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? * 0.013 n 4 n 0. 32 n 20.1 n

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KyanLin

Answer:

0.013N

Explanation:

[tex]F=\frac{mv^{2} }{r}[/tex]

m=0.04, v=4.4m/s, r=60

F=0.013 N

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