Asked by chelsea
Reggie Jackson stepped up to the plate and hit a .15 kg fast ball traveling at 36m/s, the impact caused the ball to leave his bat with a velocity of 45 m/s in the opposite direction. If the impact lasted for .002 seconds, with what force did Jackson exert on the baseball?
Answers
Answered by
Jack
F=MA=M*(v2-v1)/T where M=.15kg, v1=-36m/s, v2=45m/s, and T=.002s
F=(.15Kg)*(45m/s-(-36m/s))/(.002s)
F=(.15Kg)*(81m/s)/(.002s)
F=(12.15Kgm/s)/(.002s)
F=(6075Kgm/s^2)
F=6075N
F=(.15Kg)*(45m/s-(-36m/s))/(.002s)
F=(.15Kg)*(81m/s)/(.002s)
F=(12.15Kgm/s)/(.002s)
F=(6075Kgm/s^2)
F=6075N
Answered by
Cock
wiener
Answered by
Alfonso Garcia
I am going to use this problem for my class.
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