Answer:
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 :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{5\times (2)^2}{0.7}\\\\F=28.57\ N[/tex]
or
F = 29 N
So, the centripetal force on the bucket of water is 29 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
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