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4 years ago

Open the Faraday Law simulation and discover what you can about induction. Make a list of ways to cause induction.

Physics

1 answer:

nika2105 [10]4 years ago

5 0

Answer:

Current can be induced in a lot of ways. I'll try and list as much of these methods as I can here.

1. Moving a bar magnet relative to a wire coil

2. Moving a wire coil relative to a magnet.

3. Move a wire coil that has electricity flowing though it relative to a wire coil without electricity flowing through it.

4. Moving a current carrying circuit relative to a non-current carrying circuit

5. Rapidly opening and closing the switch of a current carrying circuit beside a non current carrying circuit.

6. Moving a bar magnet through the middle of a wire coil.

7. Moving a wire coil through a magnet

8. Moving a circuit relative to a magnetic field.

The main idea is to cause a change in the magnetic field, or to change the available area of the wire loop, or to change the angle between the field and the loop.

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A ray of light is moving from a material having a high indexof refraction into a material with a lower index of refraction.

luda_lava [24]

(a) Away from the normal

We can find the direction of bending of the ray of light by using Snell's equation:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where we have:

n1, n2: index of refraction of the first and second medium

$\theta_1, \theta_2$ ; angle that the incident and the refracted ray form with the normal to the surface

Here, the light ray moves from a material with high index of refraction to a material with lower index, so we have

$n_1 > n_2$

Re-arranging Snell's law we find

$sin \theta_2 = \frac{n_1}{n_2} sin \theta_1$

since we have

$\frac{n_1}{n_2}>1$

this implies

$sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1$

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the wavelength of the light in a vacuum

n is the refractive index

The equation can be rewritten as

$\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2$

and again it can be rewritten as

$\lambda_2 = \frac{n_1}{n_2} \lambda_1$

where

$\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1$

Therefore, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency remains the same

Wavelength and speed of a light ray depend on the medium in which the wave is travelling through, however the frequency does not depend on that, so it remains the same in the two mediums.

8 0

3 years ago

Three forces with magnitudes of 95 pounds, 75 pounds, and 146 pounds act on an object at angles 100°, 200°, and 300°, respective

VladimirAG [237]

Answer:

Explanation:

We shall represent all the forces in vector form .

Force of 95 pounds

F₁ = 95cos100 i + 95sin100j

= -16.5 i +93.55 j

force of 75 pounds

F₂ = 75cos200 + 75sin200j

= -70.47 i - 25.65 j

force of 146 pounds

F₃ = 146cos300 i + 146sin300j

= 73i -126.44 j

Resultant force

R = F₁+ F₂ + F₃

= -16.5 i +93.55 j -70.47 i - 25.65 j +73i -126.44 j

= -13.97 i - 58.54 j

Magnitude of R = √ ( 13.97² + 58.54² )

= 60.18

If Ф be angle of resultant with axis

tanФ = - 58.54 / -13.97

Ф = 76.57 + 180 = 256.57 , because both x and y components are negative. So the resultant will be in third quadrant.

7 0

3 years ago

A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel

dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0

4 years ago

Name one material through which continuous flow of charge can be produced from chemical energy

Doss [256]

Answer:

Battery.

Explanation:

A battery is a device that stores chemical energy, and converts it to electricity. This is known as electrochemistry and the system that underpins a battery is called an electrochemical cell. A battery can be made up of one or several (like in Volta's original pile) electrochemical cells

8 0

3 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$