The diagram below shows a 5.0-kilogram bucket of water being swung in a horizontal circle of 0.70-meter radius at a constant speed of 2.0 meters per second.
5.0-kg Bucket
0.70 m
The magnitude of the centripetal force on the bucket of water is approximately
1) 5.7 N
2) 14 N
3) 29 N
4) 200 N

The magnitude of the centripetal force on the bucket of water is approximately 29 N

Centripetal force is the force that acts to keep an object moving in a circular motion. It is expressed mathematically as:

F = mv² / r

With the above formula, we can obtain the centripetal force acting on the bucket.

•Mass (m) = 5 Kg

•Radius (R) = 0.7 m

•Velocity (v) = 2 m/s

•Centripetal force (F) =?

F = mv² / r

F = (5 × 2²) / 0.7

F = (5 × 4) / 0.7

F = 20 / 0.7

F = 29 N

Learn more about centripetal force:

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F = 29 N

Explanation:

We have,

Mass of the bucket of water is 5 kg

Radius of a horizontal circle is 0.7 m

Speed of the circle is 2 m/s

It is required to find the magnitude of centripetal force on the bucket of water. The formula used to find the magnitude of centripetal force is given by :

or

F = 29 N

So, the centripetal force on the bucket of water is 29 N.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach