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34kurt
3 years ago
10

The angle between the axes of two polarizing filters is 45.0^\circ45.0 ​∘ ​​ . By how much does the second filter reduce the int

ensity of the light coming through the first? Select the correct answer 0.750 0.800 0.250 0.500 Your Answer 0.125 Saved 1 of 3 attempts used Part b (1 points) What now is the direction of polarization of the transmitted light? (You can assume the first polarizer defines the angle \theta = 0θ=0.) Select the correct answer
Physics
1 answer:
suter [353]3 years ago
7 0

Answer

given,                                                                      

angle between two polarizing filters = 45°

filter reduce intensity = ?                          

a) I = I₀ Cos² θ                                

here θ = 45⁰                                

I = \dfrac{I_0}{2}                      

intensity of the light is reduced by 0.500

correct answer from the given option D

b) direction of the polarization                    

                        θ = 45°                  

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Part A.)Six boxes held at rest against identical walls.
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Explanation:

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6 0
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A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of
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8 0
3 years ago
A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is t
vampirchik [111]

The new period is D) √2 T

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy and Period of Simple Pendulum formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

\texttt{ }

\boxed{T = 2\pi \sqrt{ \frac{L}{g} }}

where:

<em>T = period of simple pendulum ( s )</em>

<em>L = length of pendulum ( m )</em>

<em>g = gravitational acceleration ( m/s² )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial length of pendulum = L₁ = L

initial mass = M₁ = M

final length of pendulum = L₂ = 2L

final mass = M₂ = 2M

initial period = T₁ = T

<u>Asked:</u>

final period = T₂ = ?

<u>Solution:</u>

T_1 : T_2 = 2\pi \sqrt{ \frac{L_1}{g} }} : 2\pi \sqrt{ \frac{L_2}{g} }}

T_1 : T_2 = \sqrt{L_1} : \sqrt{L_2}

T : T_2 = \sqrt{L} : \sqrt{2L}

T : T_2 = 1 : \sqrt{2}

\boxed {T_2 = \sqrt{2}\ T}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

3 0
3 years ago
Read 2 more answers
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