Question #336417

A 0.5kg egg is falling at a speed of 10 m/s before hitting the ground. What is the force of impact if the impact time is measured to be 3.0 milliseconds?

1
Expert's answer
2022-05-04T12:40:21-0400

The Newton's second law says

F=ma=mΔvΔtF=ma=m\frac{\Delta v}{\Delta t}

F=0.5kg10m/s3.0103s=1.7103NF=\rm 0.5\: kg*\frac{10\: m/s}{3.0*10^{-3}\: s}=1.7*10^3\: N


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