Answer:
a scientist examines the results and answers the lab question- last choice
the breeding of specimens of a plant or animal by natural processes from the parent stock.
Answer: Outer rings
Explanation:
Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone, ![f_2 = 1227 Hz](https://tex.z-dn.net/?f=f_2%20%3D%201227%20Hz)
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
![f_2=\dfrac{3v}{2l}](https://tex.z-dn.net/?f=f_2%3D%5Cdfrac%7B3v%7D%7B2l%7D)
v is speed of sound
Let f is the fundamental frequency. It is given by :
![f=\dfrac{v}{2l}](https://tex.z-dn.net/?f=f%3D%5Cdfrac%7Bv%7D%7B2l%7D)
The relation between f and f₂ can be written as :
![f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz](https://tex.z-dn.net/?f=f_2%3D3f%5C%5C%5C%5Cf%3D%5Cdfrac%7Bf_2%7D%7B3%7D%5C%5C%5C%5Cf%3D%5Cdfrac%7B1227%7D%7B3%7D%5C%5C%5C%5Cf%3D409%5C%20Hz)
So, the fundamental frequency of the pipe is 409 Hz.