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Bezzdna [24]
3 years ago
12

If two coils placed next to one another have a mutual inductance of 3.00 mH, what voltage (in V) is induced in one when the 2.50

A current in the other is switched off in 40.0 ms
Physics
1 answer:
hodyreva [135]3 years ago
3 0

Answer:

-0.1875 V.

Explanation:

Using

E₂ = MdI₁/dt........................ Equation 1

Where E₂ = Voltage induced in the second coil, M = mutual inductance of both coil, dI₁ = change in current in the first coil, dt = change in time.

Given: M = 3.00 mH = 0.003 H, dI₁ = (0-2.50) = -2.5 A, dt = 40 ms = 0.04 s.

Substitute into equation 1

E₂ = 0.003(-2.5)/0.04

E₂ = -0.1875 V.

Hence the induced emf = -0.1875 V.

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Sketch the electric field around these two objects if they have the same sign of charge. Make a separate drawing showing equipot
diamong [38]

Answer:

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

Explanation:

The electric field around two objects with charge of the same sign, for simplicity suppose that the objects have positive point spherical charges,

          E = k q / r2

bold letters indicate vectors, therefore the total electric field is

           E_total = E1 + E2

the module of this field is

           E_total = E1- E2

therefore we can outline this field

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

An outline of these shows in Attachment A

The equipotential surfaces are defined as being perpendicular to the electric field lines since the electric field and the power difference are related

              E = \frac{dV}{dx} i^ + \frac{dV}{dy} j^ + \frac{dV}{dz} k^ = \Delta V

We can schematize some characteristics of these surfaces

* very close to each load are spherical surfaces

* very far from the load is an elliptical surface, which envelops the loads

* between them there is a point of zero potential point C

See attached part B

5 0
3 years ago
Suppose a radar gun sends out radio waves with a frequency of 2,000,000.0 Hz. The waves bounce off a moving car and return with
Igoryamba

Answer:

300,000,030m/s

Explanation:

Using the relationship between frequency and speed of a wave;

F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k

Let F1 be the frequency of the radio wave = 2,000,000.0Hz

V1 be the speed of light = 300,000,000m/s

F2 = frequency produced by the car = 2,000,000.2Hz

V2 be the velocity of the moving car

Substituting this values in the equation above;

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Cross multiplying

2,000,000V2 = 300,000,000×2,000,000.2

2V2 = 300×2,000,000.2

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The velocity of the car is 300,000,030m/s

6 0
3 years ago
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