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mote1985 [20]
2 years ago
4

If a closed loop lies partially inside and partially outside a region of changing magnetic field, which is true about calculatin

g the induced electric field along the path?
Physics
1 answer:
sweet-ann [11.9K]2 years ago
8 0

If a closed loop lies partially inside and partially outside a region of changing magnetic field, to calculate the induced electric field along the path we use only the area within the flux change to calculate the electric field.

<h3>Induced Electric Fields:</h3>

The electrostatic field is conservative and does no net work over a closed path, whereas the induced electric field is nonconservative and does net work in transferring a charge over a closed channel. As a result, whereas the induced field cannot be connected with electric potential, the electrostatic field can.

Faraday's law can be written in terms of the induced electric field as. ∮→E⋅d→l=−dΦmdt. The electrostatic field created by a fixed charge distribution and the electric field caused by a changing magnetic field is very different from one another.

An electric field is created by a fluctuating magnetic flux. The induced emf from Faraday's law is related to the fluctuating magnetic flux and the induced electric field.

Learn more about induced electric field here:

brainly.com/question/15581205

#SPJ4

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3 years ago
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Two objects are dropped from rest from the same height. Object A falls through a distance <img src="https://tex.z-dn.net/?f=d_A"
Alenkinab [10]

Answer:

The answer to your question is given below

Explanation:

Since both object A and B were dropped from the same height and the air resistance is negligible, both object A and B will get to the ground at the same time.

From the question, we were told that object A falls through a distance to dA at time t and object B falls through a distance of dB at time 2t.

Remember, both objects must get to the ground at the same time..!

Let the time taken for both objects to get to the ground be t.

Time A = Time B = t

But B falls through time 2t

Therefore,

Time A = Time B = 2t

Height = 1/2gt^2

For A:

Time = 2t

dA = 1/2 x g x (2t)^2

dA = 1/2g x 4t^2

For B

Time = t

dB = 1/2 x g x t^2

Equating dA and dB

dA = dB

1/2g x 4t^2 = 1/2 x g x t^2

Cancel out 1/2, g and t^2

4 = 1

4dA = dB

Divide both side by 4

dA = 1/4 dB

8 0
3 years ago
Which of the following changes would double the force between two charged particles? O A. Decreasing the distance between the pa
jarptica [38.1K]

Let's check

\\ \rm\dashrightarrow F=\dfrac{k}{q_1q_2}{r^2}

\\ \rm\dashrightarrow F\propto Q

\\ \rm\dashrightarrow F\propto \dfrac{1}{r^2}

So

Option A and C can be used

6 0
2 years ago
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Plz help!!! Will give brainsiest!!!!
Zanzabum

Answer:

2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)

Explanation:

The reaction is a neutralization reaction since an acid, aqeous H2SO4 reacts completely with an appropriate amount of alkali, aqueous KOH to produce salt, aqueous K2SO4 and liquid water, H2O only.

2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)

Alkali + Acid → Salt + Water.

During this reaction, 2 moles of KOH neutralize 1 mole of H2SO4 to yield 1 mole of K2SO4 and 2 moles of H2O.

7 0
4 years ago
A ski lift carries people along a 220-meter cable up the side of a mountain. Riders are lifted a total of 110 meters in elevatio
jeka94

The ideal mechanical advantage (IMA) can be determined by the following equation:

 IMA= Input distance/Output distance

 The Input distance and Output distance are:

 Input distance=220 meters

 Output distance=110 meters

 When you substitute in the equation of the ideal mechanical advantage (IMA), you obtain:

 IMA= Input distance/Output distance

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3 years ago
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