I’m gonna have to say “Ocean waves” as the answer
Charge will decreases.
A parallel plate capacitor when it is fully charged to voltage V is given as:
C = Q/V
The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is
C = ε₀ A /d
since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.
So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.
Thus, Charge will decrease.
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Answer:
The answer is explained below.
Explanation:
The energy emitted during the de-excitation of an electron from a higher energy level to a lower energy level is directly proportional to the frequency of the emitted light.
Here, the total sum of the energies of 2 frequencies of light emitted in different stages is equal to the energy of a single frequency of light during the de-excitation of fourth level to ground level directly.
Hence the total sum of of the frequencies of 2 lights emitted in different stages is equal to the frequency of single frequency of light emitted during the de-excitation from fourth level to ground level directly.
The some of the energies of 2 frequencies emitted by one electron is equal to the energy of a single frequency when electron jumps directly.
They are alike because they both have vibrations to make sound
Any point in the wire has 1.12 x 
. electrons flow per second.
<h3>What causes a current in a wire?</h3>
- Electric current in a wire, where electrons serve as the charge carriers, is a measurement of the amount of charge that moves through any point of the wire in a given amount of time.
- A free electron is drawn to a proton to become neutral if an electron is added to the wire.
- Lack of electrons can result from pushing electrons out of their orbits.
- Electric current is the name given to the constantly moving electrons in wire.
The current is the quantity of charge Q flowing through a certain point of the wire in a time interval of 
.
I = 
.
by using this relationship
I=1.80 A, we can find the charge passing any point in the wire in 1 second:
Electric Charge, Q = 1.80 C.
To find how many electrons corresponds to this charge, we should divide this value by the charge of a single electron
charge of the electron = 1.6 x 
C.
No. of Electrons = Q/q = 
= 1.12 x .
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