Asked by Jazmin
a 0.2kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds up at 25 m/s. If the ball is in contact with the sidewalk for 0.025s, what is the magnitude of the average force applied by the sidewalk on the ball?
I need steps on how to do this including equations ASAP!
I need steps on how to do this including equations ASAP!
Answers
Answered by
Henry
a=V-Vo)/t = (-25-30)/0.025=2200m/s^2.
F = m*a = 0.2 * 2200 = 440 N.
Note: The velocity when leaving sidewalk is in opposite direction(neg.).
F = m*a = 0.2 * 2200 = 440 N.
Note: The velocity when leaving sidewalk is in opposite direction(neg.).
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.