Answer:
Will be doubled.
Explanation:
For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:
where e₀ is a constant, the electric permittivity.
Now we can isolate V, the potential difference between the plates as:
Now, notice that the separation between the plates is in the numerator.
Thus, if we double the distance we will get a new potential difference V', such that:
So, if we double the distance between the plates, the potential difference will also be doubled.
answer✿࿐
I was not able to write it here
so I did it somewhere else and attached the picture
i hope it helps
have a nice day
#Captainpower
Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>
<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>
<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>
so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
Answer:
The magnitude of the free-fall acceleration at the orbit of the Moon is 
(
, where 
).
Explanation:
According to the Newton's Law of Gravitation, free fall acceleration (
), in meters per square second, is directly proportional to the mass of the Earth (
), in kilograms, and inversely proportional to the distance from the center of the Earth (
), in meters:
(1)
Where:
- Gravitational constant, in cubic meters per kilogram-square second.
- Mass of the Earth, in kilograms.
- Distance from the center of the Earth, in meters.
If we know that 
, 
and 
, then the free-fall acceleration at the orbit of the Moon is:
