The one word you're looking for to fill in the blank
can be "uneven" or "non-uniform".
Answer:
0.5 s
Explanation:
From the question given above, the following data were obtained:
Number of circle (n) = 2
Time (t) = 1 s
Period =?
Period of a wave is simply defined as the time taken to make one complete oscillation. Mathematically, it can be expressed as:
T = t / n
Whereb
T => is the period
t => is the space time
n => is the number of circle or oscillation.
With the above formula, we can obtain the period of the wave as follow:
Number of circle (n) = 2
Time (t) = 1 s
Period =?
T = t / n
T = 1 / 2
T = 0.5 s
Thus, the period of the wave is 0.5 s
I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.
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 think the answer is A because it’s a better explanation