A squash ball hits a wall perpendicularly at 17 m/s and rebounds at 10 m/s along its original path. If it takes 0.8 s for the ball to rebound, what is the magnitude of the average acceleration of the ball?
please show clear steps and answer. Of course it is in english.
If you can, please also give explanasion
physics(motion)
2006-12-06 3:05 am
回答 (2)
2006-12-06 4:12 am
✔ 最佳答案
a = [17-(-10)]/0.8=27/0.8
=33.75 ms^-2
a = [v - u]/t
take the direction of 17 as positive, the 10 m/s would be negative.
2006-12-06 4:14 am
F=(mv-mu)/t
=m(10+17)/0.8 take the rebounding direction be positive
=33.75m
F=ma
33.75m=ma
a=33.75 ms^-2
=m(10+17)/0.8 take the rebounding direction be positive
=33.75m
F=ma
33.75m=ma
a=33.75 ms^-2
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