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Norma-Jean [14]
3 years ago
6

A beam of unpolarized light with intensity I0 falls first upon a polarizer with transmission axis θTA,1 then upon a second polar

izer with transmission axis θTA,2, where θTA,2−θTA,1=90degrees (in other words the two axes are perpendicular to one another). What is the intensity I2 of the light beam emerging from the second polarizer?
Physics
1 answer:
loris [4]3 years ago
5 0

Answer:

The intensity I₂ of the light beam emerging from the second polarizer is zero.

Explanation:

Given:

Intensity of first polarizer = Io/2

For the second polarizer, the intensity is equal:

I_{2} =\frac{I_{o} }{2} (cos\theta )^{2} =\frac{I_{o} }{2} (cos90)^{2} =0

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Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0
3 years ago
HELP ASAP!! WILL TRY TO GIVE BRAINLIEST
VikaD [51]

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Facilitated diffusion and active transport both utilize proteins to transport substances across membranes. Differences between active transport and facilitated diffusion 1. Active transport requires an input of energy, usually ATP, while facilitated transport does not.

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Which "spheres" are interacting when water evaporates from plants
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The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.
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2 years ago
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A student calculates the density of a copper cube to be 4.15 g/cm . If the accepted value is 8.64 g/cm the percentage error in h
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 The percentage error in his experimental value is -51.97%.

<h3>What is percentage error?</h3>

This is the ratio of the error to the actual measurement, expressed in percentage.

To calculate the percentage error of the student, we use the formula below.

Formula:

  • Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1

From the question,

Given:

  • Calculated value = 4.15 g/cm
  • accepted value = 8.64 g/cm

Substitute these values into equation 1

  • Error(%) = (4.15-8.64)100/8.64
  • Error(%) = -4.49(100)/8.64
  • Error(%) = -449/8.64
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Hence, The percentage error in his experimental value is -51.97%.

Learn more percentage error here: brainly.com/question/5493941

8 0
2 years ago
A weather balloon holds 2,850 cubic meters of helium. The density of helium is 0.1685 kilograms per cubic meter. How many kilogr
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To answer the specific problem, the balloon contains 480kg of helium. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

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