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Digiron [165]
4 years ago
14

Light that is polarized along the vertical direction is incident on a sheet of polarizing material. Only 92% of the intensity of

the light passes through the sheet and strikes a second sheet of polarizing material. No light passes through the second sheet. What angle does the transmission axis of the second sheet make with the vertical?
Physics
1 answer:
Taya2010 [7]4 years ago
6 0

Answer:

73.5°

Explanation:

I = I₀ Cos²Ф.

.92 = Cos²Ф

Ф = 16.5°

So first sheet makes an angle of 16.5° with the vertical.

Second sheet must be perpendicular to the first one because it blocks the whole light. So second sheet makes an angle of 90 + 16.5 = 106.5° with the vertical.That is 73.5 ° with the vertical. Or it may form an angle of 90-16.5 =

73.5° with the vertical when we consider in counter-clockwise direction.

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In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

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Answer:

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Explanation:

The sun is Bigger than everything else.

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