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Kazeer [188]
3 years ago
6

The magnetic field in a solenoid is . A circular wire of radius 8 cm is concentric with a solenoid of radius 2 cm and length d =

2 meter, containing 12000 turns. The current increases at a rate of 40 A/s. What is the emf in the wire?
Physics
1 answer:
defon3 years ago
7 0

Answer:

6.03 mV

Explanation:

length of solenoid, L = 2 m, N = 12000, di/dt = 40 A/s,

Magnetic field due to solenoid

B = μ0 n i = μ0 N i / L

dB/dt = μ0 N / L x di / dt

dB /dt = (4 x 3.14 x 10^-7 x 12000 x 40) / 2 = 0.3 T/s

Induced emf, e = rate of change of magnetic flux

e = dΦ / dt = A x dB / dt

e = 3.14 x 0.08 x 0.08 x 0.3 = 6.03 x 10^-3 V = 6.03 mV

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What is the equation for net force
ser-zykov [4K]

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3 0
3 years ago
Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the
aivan3 [116]

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, \theta=135^{\circ}

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

W=Fd\ cos\theta

W=50\times 9\times \ cos(135)

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

7 0
3 years ago
At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead
Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

n_{air}=1\\ n_{water}=1.33=4/3

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3

At critical angle B = 90°

\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)

Therefore

A = 48.6^\circ

With this, we can find the radius of the circle (refer to my diagram)

h* tan (A) = R\\R =11.3 m

And with that we can find the area

A = \pi R^2=404\ m^2

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

v_{water} = \frac{c}{n_{water}} = 0.75c

to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length  which is the apparent depth.

To put it in mathematical term

t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}

So we get apparent depth

h_{apparent}=0.75h = 7.5\ m

4 0
3 years ago
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