Ed rolls a ball forward across the floor. It hits a wall 12 meters away and rolls back along the same path. It stops 5 meters fr
2 answers:
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Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
- Initial velocity=u=0m/s
- Acceleration=a=10m/s^2
- Time=t=2s
Final velocity=v