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A 5.77 kg object is tied to the end of a cord and whirled in a horizontal circle of radius 355 cm. The object completes 4.4 revolutions every 6.94E-3 hours. What is the centripetal force exerted on the object?

Respuesta :

The centripetal force exerted on the object is 25.24 N.

The given parameters:

  • mass of the object, m = 5.77 kg
  • radius of the circle, r = 355 cm = 3.55 m
  • number of revolutions, N = 4.4 rev for 6.9 x 10⁻³ hours

The angular speed of the object is calculated as follows;

[tex]\omega = \frac{4.4 \ rev}{6.9 \times 10^{-3} \ h} \times \frac{1 \ h}{3600 \ s} \times \frac{2 \ \pi \ rad}{1 \ rev} = 1.11 \ rad/s[/tex]

The centripetal force exerted on the object is calculated as follows;

[tex]F =m \omega ^2 r\\\\F = 5.77 \times (1.11)^2 \times 3.55\\\\F = 25.24 \ N[/tex]

Thus, the centripetal force exerted on the object is 25.24 N.

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