# An object with a mass of #6 kg# is revolving around a point at a distance of #8 m#. If the object is making revolutions at a frequency of #3 Hz#, what is the centripetal force acting on the object?

The centripetal force acting on an object of massm,revolving round a circular path of radius r with frequency n is given by

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The centripetal force acting on the object can be calculated using the formula:

[ F = m \cdot r \cdot (2 \pi f)^2 ]

Where: ( F ) = centripetal force, ( m ) = mass of the object, ( r ) = radius of the circular path, ( f ) = frequency of revolutions.

Substituting the given values:

[ F = 6 , \text{kg} \times 8 , \text{m} \times (2 \pi \times 3 , \text{Hz})^2 ]

[ F ≈ 6 \times 8 \times (2 \times \pi \times 3)^2 , \text{N} ]

[ F ≈ 6 \times 8 \times (2 \times 3.14 \times 3)^2 , \text{N} ]

[ F ≈ 6 \times 8 \times (18.84)^2 , \text{N} ]

[ F ≈ 6 \times 8 \times 354.9 , \text{N} ]

[ F ≈ 17039.2 , \text{N} ]

Therefore, the centripetal force acting on the object is approximately ( 17039.2 , \text{N} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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