Answer:
Current flows in a resistor-capacitor circuit because of the varying electric field across the plates of a capacitor induced by an AC voltage source <em>(displacement current)</em>
Explanation:
In a capacitor, current does not flow the same way it does in a circuit, that is through conduction. This is because there is a highly resistive material in between the plates of the capacitor. Rather current flows through a phenomenon called displacement current.
Because of change in charge accumulation with time above the plates, the electric field changes causing the displacement current.
Displacement current arises due to the flow of electrons as a result of the varying magnetic fields set up on the plates of the capacitor when supplied with an AC voltage. It is important to note that a DC voltage does not induce any displacement current.
<em>Through this, phenomenon discovered by Maxwell, current is able to flow in a resistor-capacitor circuit despite the absence of an electrically conductive path through the plates.</em>
Answer:
(a) W = 1329.5 J = 1.33 KJ
(b) ΔU = 24.27 KJ
Explanation:
(a)
Work done by the gas can be found by the following formula:
where,
W = Work = ?
P = constant pressure = (0.991 atm)(
) = 100413 Pa
ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)(
) = 0.01324 m³
Therefore,
W = (100413 Pa)(0.01324 m³)
<u>W = 1329.5 J = 1.33 KJ</u>
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(b)
Using the first law of thermodynamics:
ΔU = ΔQ - W (negative W for the work done by the system)
where,
ΔU = change in internal energy of the gas = ?
ΔQ = heat added to the system = 25.6 KJ
Therefore,
ΔU = 25.6 KJ - 1.33 KJ
<u>ΔU = 24.27 KJ</u>
Objects in free fall, disregarding terminal velocity, accelerate at 9.8(m/s)/s. so for every second it was falling, it gained 9.8m/s in speed. 9.8 * 10 = 98m/s
Answer:
The magnitude of the electric field between the plates is half its initial value.
Explanation:
We know the electric field E = V/d where V = voltage applied and d = separation between plates.
Since V is constant and V = Ed,
So, E₁d₁ = E₂d₂ where E₁ = initial electric field at separation d₁, d₁ = initial separation of plates, E₂ = final electric field at separation d₂ and d₂ = final separation of plates.
So, E₂ = E₁d₁/d₂
Now, the distance between the plates is twice their original separation. Thus, d₂ = 2d₁
So, E₂ = E₁d₁/2d₁ = E₁/2
So, E₂ = E₁/2
Thus, the magnitude of the electric field between the plates is half its initial value.
Answer:
If the density of the object is high its molecular arrangement is compact while if the density is lows its molecular arrangement isnt that compact
