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Lostsunrise [7]

4 years ago

5

A magnet attracts a piece of iron. The iron can then attract another piece of iron. On the basis of domain alignment, explain wh

at happens in each piece of iron.

1 answer:

raketka [301]4 years ago

5 0

Answer :

The magnet produces a domain alignment that allows the iron become magnetic such that it is attracted to the original magnet.the magnetic iron can then create the same effect on another piece of iron.

Explanation:

the piece of magnet is brought closer to the iron where the domains of the iron align including the magnetic poles.

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Which statement demonstrates that ultraviolet (UV) rays are electromagnetic waves?

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The answer is B) They do NOT require a medium to travel.

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A 1,000 kg car travels at 15m/s

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A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform ra

sergey [27]

Answer:

$\frac{di}{dt} = 7.31 \ A/s$

Explanation:

From the question we are told that

The number of turns is $N = 450 \ turns$

The radius is $r = 1.17 \ cm = 0.0117 \ m$

The position from the center consider is x = 3.45 cm = 0.0345 m

The induced emf is $e = 8.20 *10^{-6} \ V/m$

Generally according to Gauss law

$\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A$

=> $e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A$

Where A is the cross-sectional area of the solenoid which is mathematically represented as

$A = \pi r ^2$

=> $e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2$

=> $\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}$ ggl;

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} \ N/A^2$

=> $\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}$

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6 0

4 years ago

What is a vector?

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A quantity that has magnitude and direction

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4 years ago

At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu

Ne4ueva [31]

Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$ ..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$ ..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$ .............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$ .............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$ ...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$ ...............................(6)

But $E_{h} = 2.3E_{v}$ ......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\$

$\frac{I_{2} }{I_{1} }= 0.159$

b) When the sunbather lies on his side, the vertical component vanishes, i.e $E_{v} = 0$

$\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{(2.3E_{v} )^{2} }{E_{v} ^{2} +(2.3E_{v} )^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{E_{v} ^{2} +5.29E_{v}^{2} }$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } = \frac{5.29}{6.29} \\\frac{I_{2} }{I_{1} } = 0.84$

8 0

3 years ago