Question

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.

Answer 1

Answer:v=8.06 m/s

Step-by-step 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=(mv^2)/(r)`

`65=(0.7* v^2)/(0.7)`

`v^2=65`

`v=√(65)`

`v=8.06 m/s`