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2 years ago

All of the following are electromagnetic waves except

1 answer:

3 0

B. sound waves
Sound waves need to travel through a medium such as a solid, liquid, or gas. The sound waves move through each of these mediums by vibrating the molecules in the matter. This means it isn't an electromagnetic waves but it is a wave, an example of the difference, all electromagnetic waves can travel in a vacuum at the universal speed limit, this is 299 792 458 m/s.

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Name your favorite color.

Arisa [49]

Answer:

Light blue

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2 years ago

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An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

viktelen [127]

Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

Given:

Current through the wire, passing through the origin, $I_1 = 250\ A.$

Current through the wire, passing through the y axis, $r_y=1.8\ m.$ , $I_2 = 50\ A.$

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to $\mu_o$ times the net current threading the loop.

$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

$\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\$

$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

Therefore,

$B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.$

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

$B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.$

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_2 = 5.310\ m.$

The magnetic field at the given point due to this wire is given by:

$B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.$

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at $r_r=-3.510\ m$ is given by

$B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.$

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2 years ago

How to solve the problem so I can get the right answer

Olin [163]

To solve the problem you think it through do your best i know you have it in you

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2 years ago

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Which of the following is a type of hazard when working with grain? A. Bridging B. Accumulated Grain C. Auger D. All of the answ

Licemer1 [7]

Which of the following is a type of hazard when working with grain?

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2 years ago

A professional soccer player succeeds in scoring a goal on 84% of his penalty kicks. assume that the success of each kick is ind

valina [46]

A. Using the geometric probability: .84^4 * (1-.84) = 0.07966; this means that the probability that he makes his first * the probability he makes his second * the probability he makes his third * the probability he makes his fourth * the probability he misses his fifth.

b. P (X<= 5) = binomcdf (10, 0.84, 5) = 0.0130

c. From the answer in B, if the success rate of the player still continues at 84%, the probability that he will score 5 or few goals in the 10 penalty kicks will still remain at 0.0130. This is quite low, we should examine him about whether he can still hit at 84% of his shots. So given from this data, we could say that his penalty kick success rate has fallen below the given rate of 84%.

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2 years ago

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