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.
<h2><u>Question</u><u>:</u><u>-</u></h2>
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?
<h2><u>Answer:</u><u>-</u></h2>
<h3>Given,</h3>
=> Force applied by Ryan = 10N
=> Distance covered by the book after applying force = 30 cm
<h3>And,</h3>
30 cm = 0.3 m (distance)
<h3>So,</h3>
=> Work done = Force × Distance
=> 10 × 0.3
=> 3 Joules
Answer:
the rate of change in volume with time is 280πr² cm³/min
Explanation:
Data provided in the question:
Radius of the sphere as 'r'
= 70 cm/min
Volume of the sphere, V = 
Surface area of the sphere as 4πr²
Now,
Rate of change in volume with time, 
= 
= 
Substituting the value of 
= 
= 280πr² cm³/min
Hence, the rate of change in volume with time is 280πr² cm³/min
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, from the completed question, determine the distance (in meters) the horse covered in the first ten seconds of it's gallop and apply the formula below.
Average velocity = distance (in meters) ÷ time (in seconds as provided in this question)
The unit for velocity (to be used here) is m/s or ms⁻¹
Answer: D
Explanation:
If Rhea is inspired and not too focused on comparing abilities, Biles is a good role model.
