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2 years ago

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A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi

d at its center. The coil's resistance is 9.9 ohms. The mutual inductance of the coil and solenoid is 31 μH. At a given instant, the current in the solenoid is 540 mA, and is decreasing at the rate of 2.5 A/s. At the given instant, what is the magnitude of the induced current in the coil? (μ0 = 4π × 10-7 T ∙ m/A)

1 answer:

Sholpan [36]2 years ago

4 0

Answer:

$i = 7.83 \mu A$

Explanation:

Induced EMF in the coil is given by the equation

$EMF = M\frac{di}{dt}$

so we have

$M = 31 \mu H$

also we know that rate of change in current in solenoid is given as

$\frac{di}{dt} = 2.5 A/s$

so induced EMF of coil is given as

$EMF = (31 \times 10^{-6})(2.5)$

$EMF = 77.5 \times 10^{-6} A/s$

now induced current in the coil will be given as

$i = \frac{EMF}{R}$

$i = \frac{77.5 \times 10^{-6}}{9.9}$

$i = 7.83 \mu A$

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An experiment is conducted in which red light is diffracted through a single slit. Listed below are alterations made, one at a t

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Answer:

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

Explanation:

As we know that the position of first minimum is given as

$a sin\theta = N\lambda$

so we have

$\theta = sin^{-1}(\fracN\lambda}{a})$

so width of minimum is given as

$w = L\times sin^{-1}(\fracN\lambda}{a})$

now if we need to decrease the angular position of minimum

1). so we can decrease the distance of screen from the slit

2). we can decrease the wavelength

3). We can increase the width of the slit

So correct answer will be

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

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2 years ago

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

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Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

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2 years ago

A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum. Its final speed will

Pavel [41]

Answer:

Final Velocity = √(eV/m)

Explanation:

The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference

W = (2e) × V = 2eV

And this work is equal to change in kinetic energy

W = Δ(kinetic energy) = ΔK.E

But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0

ΔK.E = ½ × (mass) × (final velocity)²

(Velocity)² = (2×ΔK.E)/(mass)

Velocity = √[(2×ΔK.E)/(mass)]

ΔK.E = W = 2eV

mass = 4m

Final Velocity = √[(2×W)/(4m)]

Final Velocity = √[(2×2eV)/4m]

Final Velocity = √(4eV/4m)

Final Velocity = √(eV/m)

Hope this Helps!!!

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1 year ago

A truck pulled a car of 2,350 kg a distance of 25 meters. If the car accelerates from 3 m/s to 6 m/s, whats the average force ex

faust18 [17]

Answer:

1,269 N

Explanation:

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2 years ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Juli2301 [7.4K]

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used to solve this question.

From the diagram, the horizontal component of the force is:

$F_X = F_{cos \ \theta}$

Replacing 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

On the other hand, the vertical component is ;

$F_Y = Fsin \ \theta$

Replacing 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

$f = \mu \ N$

Where;

$\mu$ is said to be the coefficient of the friction

N = the normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing $F_Y \ for \ F_{sin \ \theta}$ . The normal reaction can now be:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Making $\mu$ the subject of the formular in the above equation; we have the following:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Replacing the following values: i.e

$F_X \ = \ 64.65 \ N$

m = 73 Kh

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Then:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Thus, the coefficient of friction is = 0.0984

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2 years ago