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.
Probably 90 j but im not sure I haven’t done any work like this in a while
For the first one 320
second
1200W
Data
R = 12 Ω ∆V = 120V I =? P =?
Solution:
According to Ohm’s law,
∆V = I R
I = ∆V / R
= 120 / 12
= 10 A
Power P = I ∆V
= 10 x 120
= 1200 W
Third
∆V = 120 V P = 60 W I =? R =?
Use the formula, P = I ∆V
I = P / ∆V = 60 / 120 = 0.5 A
∆V = I R
R = ∆V / I = 120 / 0.5 = 240 Ω
Answer:
Answer for the question is given in the attachment.
Explanation:
Answer:
<u>Drag force</u> is the frictional force needed to slow an object in motion
Explanation: