Question

Can you help me can you right the answer normal and also in one significant figure as well

Answer 1

Given data:

* The force applied by Kinley is F = 3000 N.

* The work done on the softball is W = 100 kJ.

Solution:

The work done on the softball in terms of the force and the displacement is,

`W=F* d`

where d is the displacement of the softball,

Substituting the known values,

`\begin{gathered} 100*10^3=3000* d \\ d=(100*10^3)/(3000) \\ d=(100000)/(3000) \\ d=(100)/(3) \end{gathered}`

By simplifying,

`\begin{gathered} d=33.3\text{ m} \\ d\approx0.03*10^3\text{ m} \\ d\approx0.03\text{ km} \end{gathered}`

Thus, the displacement of the ball is 33.3 meters or 0.03 km (in one significant figure).