A150 g egg is dropped from 3.0 meters. the egg is moving at 4.4 m/s right before it hits the ground. the egg comes to a stop in 0.072 seconds.

what is the magnitude of force on the egg?

Magnitude of force is 9.17 m/s²

Explanation:

This question can be solved by momentum formulas.

Momentum (P) is the product of mass and velocity.

                 P = mv

Also from the derivation of Newton's 2nd law of motion, Rate of change of momentum is equal to force  

                F = ΔP/Δt

Given data:

         Mass = m =150 g = 0.15 kg

        Initial Velocity = v₁ = 4.4 m/s

        Final Velocity = v₂ = 0 m/s

        Time = t = 0.072 s

In this question we don't need height

                 P = mv

                  P = 0.15 × 4.4

                  P = 0.66 Ns

Also, Derived from Newton's 2nd law

                  P = F × t

                  F = P/t

                  F = 0.66/0.072

                 F = 9.17 N

9.2 rounded

Explanation: On EDGN 2020

there are no atoms

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm