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3 years ago

What can we say about the way in which the components of polarized light oscillate with respect to the direction of propagation?

Physics

2 answers:

ICE Princess25 [194]3 years ago

5 0

Answer:

the answer is b. The components oscillate perpendicular to the direction of propagation. for plato users

#platoforlife

Explanation:

Black_prince [1.1K]3 years ago

4 0

We can say that the components of ANY light oscillate perpendicular
to the direction of propagation ... left and right, up and down, upper left
and lower right, upper right and lower left, etc. ... any direction ACROSS
the direction of propagation, just not forward and backward.

If the light is polarized, then the Ɛ-field (electrostatic) component of
the wave can only oscillate in the polarized direction ... say, left and
right across the direction of propagation. (The magnetic component
is always perpendicular to the electrostatic component. So if the
polarizer is left/right, then the magnetic component is up/down.)

Any light that doesn't oscillate in the direction selected by the polarizer
gets absorbed in the polarizer, and doesn't come out the other side.
That's why when you pass light through a polarizer, it comes out dimmer
than it went in.

LASER light is always naturally polarized before it comes out of the diode
or whatever is generating it. That's another advantage of laser light.

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Of the 50 U.S. states, 4 have names that start with the letter W.

professor190 [17]

Answer:

Explanation:

Divide 4 by 50,

4/50=0.08

0.08 as a percent is 8%

8 0

3 years ago

Read 2 more answers

A soccer player attempting to steal the ball from an opponent was extending her knee at 50 deg/s when her foot struck the oppone

denpristay [2]

Answer:

$\alpha = -4.36 rad/s^2$

Explanation:

Initial angular speed of the player is given as

$\omega = 50 deg/s$

here we know that

$1 deg = \frac{\pi}{180} rad$

now we know that

$\omega = 50\frac{\pi}{180}$

$\omega = 0.87 rad/s$

now its speed comes to zero in 0.2 s

so angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

$\alpha = \frac{0 - 0.87 rad/s}{0.2 s}$

$\alpha = -4.36 rad/s^2$

7 0

3 years ago

An electron in the Bohr model of hydrogen revolves around the nucleus to:

RoseWind [281]

An electron in the Bohr model of hydrogen revolves around the nucleus to possess energy.

Answer: Option B

Explanation:

According to Rutherford, it was stated that electrons are the subatomic particles which revolve around the nucleus. But the error of spiral collapse of these electrons is due to absence of electrons in discrete levels. Hence, in Bohr atom model, the atom's negative or electrons are said to rotate in respective energy levels.

So, they keep on revolving around a nucleus in Bohr atom is said to revolve around a nucleus in order to possess energy. Thus, each level will exhibit discrete energy levels which will contain electrons of specific energy only.

5 0

3 years ago

What way would the balls fall? Top Left Right Bottom

stira [4]

Each ball, when released, will fall toward the center of the big ball that everybody is standing on.

Another way to describe it: Each ball, when released, will fall in the direction parallel to a line from the boy's head to his feet.

-- The ball held by the boy on top will fall toward the bottom of the picture.

-- The ball held by the boy on the right will fall toward the left side of the picture.

-- The ball held by the boy on the bottom will fall toward the top of the picture.

-- The ball held by the boy on the left will fall toward the right side of the picture.

In each case, the direction toward the center of the big ball that they're all standing on is the direction that each person calls "down".

3 0

4 years ago

A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe

Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C× $\frac{1mol}{63,546g}$ ×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C× $\frac{1mol}{18,02g}$ ×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK× $\frac{1mol}{63,546g}$ ×25,0g×(X-363K) = -75,2J/molK× $\frac{1mol}{18,02g}$ ×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

X = 295K

Final temperature is 295K

I hope it helps!

6 0

4 years ago