Answer:
β2= β1+10*f
Explanation:
comparing β2 and β1, it is said that β2 is increased by a factor of f.
for each factor of f, there is a 10*f dB increase.
therefore if the β1 is increases by an intensity of factor f
the new intensity would be β1+ 10*f
Answer:
It remains the same
Explanation:
It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.
You asked the question twice I answered it on the last one
The wavelength of the note is
![\lambda = 39.1 cm = 0.391 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%2039.1%20cm%20%3D%200.391%20m)
. Since the speed of the wave is the speed of sound,
![c=344 m/s](https://tex.z-dn.net/?f=c%3D344%20m%2Fs)
, the frequency of the note is
![f= \frac{c}{\lambda}=879.8 Hz](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7Bc%7D%7B%5Clambda%7D%3D879.8%20Hz%20)
Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
![f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7B1%7D%7B2L%7D%20%5Csqrt%7B%20%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20%20)
where
![\mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m](https://tex.z-dn.net/?f=%5Cmu%3D0.550%20g%2Fm%20%3D%200.550%20%5Ccdot%2010%5E%7B-3%7D%20kg%2Fm)
is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L: