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MissTica
2 years ago
6

) A 73-mH solenoid inductor is wound on a form that is 0.80 m long and 0.10 m in diameter. A coil having a resistance of is tigh

tly wound around the solenoid at its center. The mutual inductance of the coil and solenoid is At a given instant, the current in the solenoid is and is decreasing at the rate of At the given instant, what is the induced current in the coil
Physics
1 answer:
Aleonysh [2.5K]2 years ago
5 0

Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil

Answer:

6.169 μA

Explanation:

Formula for induced EMF is given by the equation;

EMF = M(di/dt). We are given;

di/dt = 2.5 A/s

M = 19μH = 19 × 10^(-6) H

Thus;

EMF = 19 × 10^(-6) × 2.5.

EMF = 47.5 × 10^(-6) V

Formula for current is;

i = EMF/R. R is resistance given as 7.7 ohms.

Thus; i = 47.5 × 10^(-6)/7.7

i = 6.169 μA

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Answer:

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

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Now, formula for current is; I = V/R

In this case, V which is voltage is denoted by ε.

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3 years ago
What is the difference between a lunar eclipse and a solar eclipse?
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Answer:

A lunar eclipse occurs when Earth blocks the light going to the moon. A solar eclipse occurs when the moon blocks the light coming from the sun.

Explanation:

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3 years ago
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1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin
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Answer:

d = 4 d₀o

Explanation:

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In that case as the car stops v_f = 0

the work is

          W = -fr d

we substitute

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           d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

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using the same expressions we find

           d = ½ m (2v₀)² / fr

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3 years ago
A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan
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Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of a \;\;m/s^2 from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, a=\frac{v-u}{t}=\frac{5-4.2}{t}

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Now, from the equation of motion

s=ut+\frac 12 at^2

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\Rightarrow 115=4.2t+\frac 12 \times 0.8 t [ from equation (i)]

\Rightarrow 115=(4.2+0.4)t

\Rightarrow t= 115/4.6 = 25 seconds.

Hence, she takes 25 seconds to walk the distance.

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