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: temperature of the hot reservoir is 303.52K
Explanation:
Th=temperagture of hot reservoir
Tc=temperature of cold reservoir 263K
Eff=efficiency of the refrigerator
Cop=coefficient of performance
7
Eff=1/cop
Eff=1/7
Eff=0.142
Eff=1-263/th
0.142-1=-263/th
-0.858=-263/th
Th=303.52K
Answer:
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The magnitude of the current in wire 3 is (I₃)= 0.33A
<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>
To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,
I₁ + I₂ + I₃ = 0
Where we are given,
I₁ = current in wire 1
=0.40 A.
I₂ = current in wire 2
= -0.73 A.
We have to calculate the magnitude of the current in wire 3, I₃
Now we put the known values in above equation, we get,
I₁ + I₂ + I₃ = 0
Or, I₃ = -.(I₁ + I₂)
Or, I₃ = -.(0.40 - 0.73)
Or, I₃ = 0.33 A
From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A
Learn more about current:
brainly.com/question/25537936
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