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.
You use more significant figures. 5 sigfigs (1.0985) is more accurate than 2 sigfigs (1.0)
A. Acceleration.
acceleration is m/s^2. speed is m/s
Answer:
Ratio of magnetic field will be 
Explanation:
We have given radius of the loop r = 30 mm = 0.03 m
We know that magnetic field at the center of the loop is given by
---------eqn 1
Number of turns in the solenoid is given as n = 3 turn per mm = 3000 turn per meter
We know that magnetic field due to solenoid is given by
-------------eqn 2
Now dividing eqn 1 by eqn 2

C.<span>a stable internal attribution</span>