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What is the precision (relative error) of the centripetal force divided by the mass if the velocity and the radius are each determined with a precision of 1%

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The precision (relative error) of the centripetal force is 1%.

Relative error

This is the error in measurement of a variable obtained in comparison with other variables.

F = mv²/r

where;

  • F is centripetal force
  • m is mass
  • v is velocity
  • r is radius

F/m = v²/r

F/m = (0.01v)²/(0.01r)

F/m = 0.01v²/r

F/m = 1%(v²/r)

Thus, the precision (relative error) of the centripetal force is 1%.

Learn more about relative error here: https://brainly.com/question/13370015

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