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elixir [45]
4 years ago
14

The filament in the bulb is moving back and forth, first pushed one way and then the other. What does this imply about the curre

nt in the filament
Physics
1 answer:
Anestetic [448]4 years ago
5 0

Answer:

energy carried by the current is given by the pointyng vector

Explanation:

The current is defined by

       i = dQ / dt

this is the number of charges per unit area over time.

The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.

But the energy carried by the current is given by the pointyng vector of the electromagnetic wave

            S = 1 / μ₀ EX B

It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement

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Fiesta28 [93]

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3 years ago
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A satellite orbiting Earth has a tangential velocity of 5000 m/s. Earth’s mass is 6 × 1024 kg and its radius is 6.4 × 106 m.
Aloiza [94]
When a satellite is moving around the Earth's orbit, two equal forces are acting on it. The centripetal and the centrifugal force. The centripetal force is the force that attracts the object toward the center of the axis of rotation. The opposite force is the centrifugal force. It draws the object away from the center. When these forces are equal, the satellite uniformly rotates along the orbit.

Centripetal force = Centrifugal force
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Centripetal acceleration = Centrifugal acceleration

\frac{ M_{E}*G }{  ( r_{E} )^{2}  } =ω^2r

where

M_{E} = mass of earth
G = gravitational constant = 6.6742 x 10-11<span> m</span>3<span> s</span>-2<span> kg</span><span>-1
</span>r_{E} = radius of earth
ω = angular velocity
<span>r = radius of orbit

To convert to angular velocity:

</span>Tangential velocity = rω
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Then,

\frac{ (6 \ x \  10^{24}) *(6.6742 \ x \ 10^{-11} ) }{ ( 6.4 \ x \  10^{6})^{2} }= ( \frac{5000}{r} )^{2} r

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4 0
3 years ago
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A 29.0-kg block is initially at rest on a horizontal surface. A horizontal force of 71.0 N is required to set the block in motio
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Answer:

(a) \mu_s=0.25

(b) \mu_k=0.20

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F_a=\mu N=\mu mg\\\mu=\frac{F_a}{mg}

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(b) The  coefficient of kinetic friction is related with the force required to keep the block moving with constant speed:

\mu_k=\frac{56N}{29kg*9.8\frac{m}{s^2}}\\\mu_k=0.20

3 0
3 years ago
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<u>Answer</u>:

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8 0
3 years ago
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