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2 years ago

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A stone with a mass of 0.70 kg is attached to one end of a string 0.70 m long. The string will break if its tension exceeds 65.0

N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed.

1 answer:

mixas84 [53]2 years ago

4 0

Answer:v=8.06 m/s

Explanation:

Given

mass of stone $m=0.7 kg$

length of string $L=0.7 m$

Maximum Tension $T=65 N$

the string will not break if Centripetal Force is less than maximum tension

$T=\frac{mv^2}{r}$

$65=\frac{0.7\times v^2}{0.7}$

$v^2=65$

$v=\sqrt{65}$

$v=8.06 m/s$

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