Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
I thinks its period because they are on the last column to the right
Hey there!
We are given ,
Acceleration, a = -2m/s^2
Initial velocity , u = 15m/s
Time , t = 5 seconds
We know that ,
V=u+at
Now , final speed ,
V = 15+(-2)(5)
V = 15-10
V = 5 m/s -> final speed
Hope this helps you dear :)
Have a good day <3
Answer:
5070
Explanation:
add them up and then you get <em>your</em><em> </em><em>answers</em><em> </em>