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GarryVolchara [31]
2 years ago
6

A parallel plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the p

lates are pulled away from each other, increasing their separation, what happens to the amount of charge on the plates?
a. The amount of the charge decreases, because the capacitance increases.
b. Nothing happens; the amount of charge stays the same.
c. The amount of the charge increases, because the capacitance increases.
d. The amount of the charge increases, because the capacitance decreases.
e. The amount of the charge decreases, because the capacitance decreases.
Physics
1 answer:
Svetllana [295]2 years ago
5 0

Answer:

e. The amount of the charge decreases, because the capacitance decreases.

Explanation:

The capacitance of a parallel plate capacitor is given by the formula:

C=\frac{\epsilon A}{d}

where \epsilon is the permittivity of the medium between the plates, A the area of the plates and d the separation between them, so <em>if we increase the separation d the capacitance C will decrease</em>.

The relationship between the charge on the plate Q and the voltage applied V (the potential difference between the plates) is given by:

Q=CV

This means that if while <em>keeping the potential difference between the plates V constant the capacitance C decreases then the charge Q will decrease as well</em>.

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Complete Question:

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d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that  the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

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The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

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3 years ago
The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10
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The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

  • The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

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\displaystyle Average \ force, \, F_{ave}  \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}

  • The average force exerted on the limestone floor by the droplets of water is F_{ave} ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

brainly.com/question/3999427

brainly.com/question/4197598

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