The correct answer is the reverse wave I took the test
Answer:
The object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g.
No so sure
Explanation:
Hope it helps
You are given a fixed rate of 15.9 cm³/s. You are also given with the amount of volume in 237 cm³. Through the approach of dimensional analysis, you can manipulate through operations such that the end result of the units must be in seconds. The solution is as follows:
237 cm³ * (1 s/15.9 cm³) = 14.9 seconds
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.
Answer:
D because those are both concerning.
