Question #276600

A supersonic bomber which had a momentum of 8,800,000 kg-m/s east, comes to a halt on the ground. What impulse is applied?


1
Expert's answer
2021-12-07T20:12:15-0500

The impulse is the change in momentum:


I=Δp=pfpi=08800000= =8800000 kgm/s.I=\Delta p=p_f-p_i=0-8800000=\\\space\\=-8800000\text{ kg}·\text{m/s}.


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