Question

1. 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. 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

Explanation:

Step1: Recall centripetal force formula

Centripetal force $F_c = \frac{mv^2}{r}$

Step2: Plug in given values

$m=5.0\,\text{kg}$, $v=2.0\,\text{m/s}$, $r=0.70\,\text{m}$
$F_c = \frac{5.0 \times (2.0)^2}{0.70} = \frac{5.0 \times 4.0}{0.70} = \frac{20}{0.70} \approx 29\,\text{N}$

Answer:

1. 29 N