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:

link

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.

answer: good job it's all correct !

explanation:

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2