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An object is constrained by a cord to move in a circular path of radius 1.3-m on a frictionless, horizontal surface. The cord will break if the tension exceeds 15.7-N. The maximum kinetic energy that this object can have is _____ J. Round your answer to the nearest whole number.

Respuesta :

Answer:

Kinetic energy will be equal to  J

Explanation:

We have given radius of the path r = 1.3 m

Tension in the string T = 15.7 N

At breaking point [tex]T=\frac{mv^2}{r}[/tex], here T is tension, m is mass and v is velocity and r is radius

So [tex]mv^2=Tr[/tex]

[tex]mv^2=16\times 0.5=8[/tex]

We have to find the kinetic energy

Kinetic energy is given by [tex]KE=\frac{1}{2}mv^2[/tex]

So [tex]KE=\frac{8}{2}=4J[/tex]

So kinetic energy will be equal to  J

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