Pretty sure it’s A. Hope this helps.
Answer:
The two possible frequencies of the other player's note are 437.3 Hz and 442.7 Hz
Explanation:
Given Data
f₁=440 Hz
fₐ=2.7 beats per seconds
The beat frequency is equal to the difference in the frequencies of the two original waves:
So
f
=|(f₁±fₐ)|
First to Solve for (-) Sign we get
f=|(440-2.7)|
f=437.3 Hz
Now for Solve for (+) Sign we get
f=|(440+2.7)|
f=442.7 Hz
The two possible frequencies of the other player's note are 437.3 Hz and 442.7 Hz
Answer:
Explanation:
give data:
inside diameter = 5.0 cm
charge q = 0.25 nC
Outside diameter = 15 cm
potential V at inside sphere is = 
potential V at outside sphere is = 
k is constant whose value is = 
then potential difference between two point is
![\Delta V = kq \left [\frac{1}{R}-\frac{1}{r} \right ]](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20kq%20%5Cleft%20%5B%5Cfrac%7B1%7D%7BR%7D-%5Cfrac%7B1%7D%7Br%7D%20%20%5Cright%20%5D)
![\Delta V = 9*10^{9}*0.25*10^{-9} \left [\frac{1}{0.05}-\frac{1}{0.15} \right ]](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%209%2A10%5E%7B9%7D%2A0.25%2A10%5E%7B-9%7D%20%5Cleft%20%5B%5Cfrac%7B1%7D%7B0.05%7D-%5Cfrac%7B1%7D%7B0.15%7D%20%5Cright%20%5D)
