The current flowing in each resistor of the circuit is 4 A.
<h3>
Equivalent resistance of the series resistors</h3>
The equivalent resistance of the series circuit is calculated as follows;
6 Ω and 4 Ω are in series = 10 Ω
5 Ω and 10Ω are in series = 15 Ω
<h3>Effective resistance of the circuit</h3>
<h3>Current flowing in the circuit</h3>
V = IR
I = V/R
I = 24/6
I = 4 A
Learn more about resistors in parallel here: brainly.com/question/15121871
.012 J is going to be your answer
Answer:
I can't see the picture
Explanation:
Im sorry can you right whats on the picture down in the comments plz.
Answer:
an apple falling off a tree
Explanation:
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.
