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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The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
Answer:
All materials are superconducting at temperatures near absolute zero kelvin.
Explanation:
All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.
Ruff's image is 50m behind the mirror surface and the image is also 3m tall.
This is because it is a plane mirror.
Answer:
The current through the resistor is 0.5 A
Explanation:
Given;
power of the light bulb = 60 W
voltage in the wall outlet across the plug terminals = 120 V
power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.
power = voltage x current
Therefore, for a 60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A
