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julia-pushkina [17]
1 year ago
6

a parallel plate capacitor with circular plates of radius and plate separation is being charged at the rate of .displacementcurr

entdisplacementcurrentoutmagneticfieldwhat is the magnitude of the magnetic field between the capacitor plates at a radius from the axis of the capacitor?
Physics
1 answer:
Kipish [7]1 year ago
6 0

A parallel plate capacitor with circular plates of radius and plate separation is being charged at the rate of Displacement current. Correct answer: letter A.

This is because the plates are parallel and current can flow freely between them. The current flow is perpendicular to the plates, so the charge on the plates is proportional to the area of the plates.

<h3>Magnitude of the magnetic field between the capacitor plates</h3>

The magnitude of the magnetic field between the capacitor plates at a radius from the axis of the capacitor is zero.

That is, it is determined by the amount of charge on the plates and the distance between the plates. Therefore, the closer the plates are to each other, the greater the magnetic field.

Learn more about parallel plate capacitor:

brainly.com/question/15720344

#SPJ4

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Answer:

2

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2

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When riding a roller coaster, you go up and down many hills. When do you have the least amount of kinetic energy?
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at the bottem of the hill

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It has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of f
Anestetic [448]

Answer:

A = 8.34 x 10^(5) m²

Explanation:

The intensity of the solar radiation is the average solar power per unit area. Thus,

I = P/4A = P/(4(πr²))

Where;

P is average solar power.

r is it's distance from centre of sun

A is area

Now, The rate at which the Sun emits energy has a standard value of 3.90 × 10^(26) W

Thus;

I = [3.90 × 10^(26)]/(4πr²)

I = [3 x 10^(25)]/r²

Now, the formula for radiation pressure is;

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Where c is speed of light and has a value of 3 x 10^(8) m/s

Thus,

P_rad = 2([3 x 10^(25)]/r²)/(3 x 10^(8))

P_rad = [2.07 x 10^(17)]/r² N/m²

Also, Radiation pressure on ship; P_rad = F/A

Where Force on ship and A is area.

Thus;

F = P_rad x A

So, F = [2.07 x 10^(17)•A]/r²

Now,

F_grav = GMm/r²

Where;

G is gravitational constant with a value = 6.67 x 10^(-11) Nm²/kg²

M is mass of sun with a value of 1.99 x 10^(30)

m is mass of ship and sail = 1300 kg

Thus, plugging in the relevant values to obtain;

F_grav = (6.67×10^(-11) × 1.99 x 10^(30) × 1300)/r²

F_grav = [17.255 x 10^(22)]/r²

Now, equating F to F_grav, we get;

[2.07 x 10^(17)•A]/r² = [17.255 x 10^(22)]/r²

r² will cancel out to give;

2.07 x 10^(17)•A= [17.255 x 10^(22)]

A = [17.255 x 10^(22)]/2.07 x 10^(17)

A = 8.34 x 10^(5) m²

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4 years ago
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Answer:

An emission-line spectrum.

Explanation:

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I hope you find this information useful and interesting! Good luck!

8 0
3 years ago
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
SVEN [57.7K]

Answer:

the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s

Explanation:

given information

car's mass, m = 1200 kg

h_{A} = 100 m

v_{A} = v_{A}

h_{B} = 150 m

v_{B} = 0

according to conservative energy

the distance from point A to B, h = 150 m - 100 m = 50 m

the initial speed v_{A}

final speed  v_{B} = 0

thus,

v_{B}² = v_{A}² - 2 g h

0 = v_{A}² - 2 g h

v_{A}² = 2 g h

v_{A} = √2 g h

    = √2 (9.8) (50)

    = 31.3 m/s

8 0
3 years ago
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