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Volgvan
4 years ago
13

A capacitor is connected to an AC generator. As the generator's frequency is increased, what happens to the current in the capac

itor? A)The current increases B)The current decreases C)The current does not change.
Physics
1 answer:
just olya [345]4 years ago
3 0

Answer:

A) The current increases

Explanation:

In the DC limit (that means, very low frequency of the generator), the capacitor acts as an open circuit. In fact, a capacitor consists of two parallel plates which are separated from each other: this means that the current cannot flow through it, but it can only flow through the rest of the circuit.

In the case of a direct current (DC), therefore, the current in the circuit must be zero. For an AC current with very low frequency, the current is still very low, because the polarity of the generator changes direction not very often, so there is still enough time for the capacitor to "block" the current. However, when the frequency of the generator is increased, the polarity changes so fast that the current in the circuit can flow without having time of "hitting" the capacitor, so it has almost no effect and the current in the circuit is maximum.

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A cat runs 6m in 4seconds what is it speed
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Explanation:

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(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. Th
KonstantinChe [14]

Answer:

Part A:

n=8.85*10^{28}m^{-3}

Part B:

Electron Mobility=3.03*10^{-3} m^2/V

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

N-Au=\frac{Density*Avogadro Number}{atomic weight}

Density = 19.32g/cm^3

Avogadro Number=6.02*10^{23} atoms/mol

Atomic weight=196.97g/mol

So:

n=1.5*\frac{Density*Avogadro Number}{atomic weight}

n=1.5*\frac{19.32*6.02*10^{23}}{196.97}

n=8.85*10^{28}m^{-3}

Part B:

Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}

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4 0
4 years ago
60 kilometers in 4 hours, what is the average speed?​
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8 0
3 years ago
Consider a simple but surprisingly accurate model for the hydrogen molecule: two positive point charges, each having charge e, a
lyudmila [28]

Answer:

a = R/2

Explanation:

Let R be the radius of the sphere and q = -2e be the charge in it. Let q₁ be the charge at radius a where the one of the point positive charges e is located. . Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πa³

q₁ = q(a/R)³ = -2e(a/R)³. The electric force due to q₁ at r is F₁ = kq₁q/a² = kq²(a/R)³/a² = k(-2e)²a/R³ = 4ke²a/R³.

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If the net force on either charge is zero, then

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a = R³/4(2a)² = R³/8a²

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a³ = R³/8

a = ∛(R³/8)

a = R/2

7 0
3 years ago
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