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.

Respuesta :

Answer:v=8.06 m/s

Explanation:

Given

mass of stone [tex]m=0.7 kg[/tex]

length of string [tex]L=0.7 m[/tex]

Maximum Tension [tex]T=65 N[/tex]

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

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

[tex]65=\frac{0.7\times v^2}{0.7}[/tex]

[tex]v^2=65[/tex]

[tex]v=\sqrt{65}[/tex]

[tex]v=8.06 m/s[/tex]