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REY [17]
3 years ago
5

A beam of visible light is passed through a plexiglass container filled with water. If a beam of light comes in at a slight angl

e, describe the change in the path of the beam of light as it passes through the container, moving in from one side and then out the other.
A) The beam of light will bend twice: entering and leaving the container.
B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
C) The beam of light will refract four times; largest entering container; smaller angles of refraction moving through water, through container wall again, and finally into the air on the outside.
D) The beam of light will bend three times: large angle of refraction through container wall; smaller angle of refraction through water; greater angle passing out of a container.
Physics
2 answers:
pentagon [3]3 years ago
8 0

B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
patriot [66]3 years ago
8 0

Answer:

C) The beam of light will refract four times; largest entering container; smaller angles of refraction moving through water, through container wall again, and finally into the air on the outside.

Explanation:

Just got it right on the test

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I think the answer would be Car C.
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Airida [17]

Answer:

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Explanation:

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3 years ago
Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for
klemol [59]

The value of parameter C for the function in the figure is 2.

<h3>What is amplitude of a wave?</h3>

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(x - C)

where;

  • A is amplitude of the wave
  • C is phase difference of the wave

<h3>What is angular frequency of a wave?</h3>

Angular frequency is  the angular displacement of any element of the wave per unit time.

From the blue colored graph; at y = 1, x = -2 cm

1 = cos(2 - C)

(2 - C) = cos^(1)

(2 - C) = 0

C = 2

Thus, the value of parameter C for the function in the figure is 2.

Learn more about phase angle here: brainly.com/question/16222725

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5 0
1 year ago
Find the value of currents through each branch
Irina-Kira [14]

Answer:

the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

  I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

  (I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

  (I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

  \left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]

These equations have the solution ...

  \left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]

This means the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

  (I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

__

It helps to be familiar with the formulas for resistors in series and parallel.

8 0
3 years ago
PLEASE I NEED HELP ASAP!!!!!!!!!!
max2010maxim [7]

tooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

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