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MrMuchimi
2 years ago
6

How is the polarization of an electromagnetic wave defined? Group of answer choices The polarization is defined by the plane in

which the magnetic field is oscillating. The polarization is defined by the direction the wave is traveling. None of these choices. The polarization is defined by the plane in which the electric field is oscillating.
Physics
1 answer:
Anarel [89]2 years ago
7 0

Answer:

The polarization is defined by the plane in which the electric field is oscillating.

Explanation:

The polarization of an electromagnetic wave depends on the direction of the electric field. If the electric component is oscillating along the x-axis, the polarization will also be along the x-axis.

The polarization of the electromagnetic wave describes the magnitude and the direction of the electric field of the wave.

<u>The polarization is defined by the plane in which the electric field is oscillating.</u>

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5016.2446

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For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

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no idea

Explanation:

7 0
3 years ago
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
Aleksandr-060686 [28]
B4 the tackle: 

<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>

<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>


<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>

<span>The vector triangle is right angled: </span>

<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>

<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>

<span>v(f) = 5.6 m/s (to 2 sig figs) </span>


<span>direction of v(f) is the same as the direction of the final momentum </span>

<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>


<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>




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4 0
3 years ago
Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

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It's the angle made by the incident ray when it's perpendicular to the surface. (Perpendicular lines are the lines that form a graph or like a 90-degree angle)
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