Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone, 
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
v is speed of sound
Let f is the fundamental frequency. It is given by :
The relation between f and f₂ can be written as :
So, the fundamental frequency of the pipe is 409 Hz.
I think it has 11 but i don't really know for sure.Correct me if i'm wrong.
|Average acceleration| = (change in speed) / (time for the change)
Change in speed = (speed at the end) minus (speed at the beginning)
= (40 m/s) - ( 0 )
= 40 m/s .
|Average acceleration| = (40 m/s) / (60 sec) = 2/3 m/s² .
Answer:
Part A
Time period, designated by the symbol, 'T', is defines as the time it takes for a complete cycle of an oscillation or vibration (of a wave) to transit through a given point.
The longer the time period of a wave, the lower the frequency of the wave
The unit of the time period is seconds, 's'
Part B
Mathematically, the formula for the time period is presented as follows;
f = 1/T
∴ T = 1/f
f = v/λ
∴ T = λ/v
Where;
v = The velocity of the wave;
λ = The wavelength of the wave
Explanation:
Given:
initial angular speed, 
= 21.5 rad/s
final angular speed, 
= 28.0 rad/s
time, t = 3.50 s
Solution:
Angular acceleration can be defined as the time rate of change of angular velocity and is given by:
Now, putting the given values in the above formula:
Therefore, angular acceleration is:
