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netineya [11]
3 years ago
7

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1515

.
Physics
1 answer:
Talja [164]3 years ago
8 0

Complete  Question

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to 1/5.

Answer:

The  angle is  

Explanation:

From the question we are told that

   The light emerging from second Polaroid is 1/5 the  unpolarized

Generally the intensity of light emerging from the first Polaroid is mathematically represented as

             I_1 = \frac{I_o}{ 2 }

Generally from the Malus law the intensity of light emerging from the second Polaroid  is mathematically represented

      I_2  =  I_1 cos^2 (\theta )

=>   cos^2 (\theta ) =  \frac{I_2}{I_1 }

=>   cos (\theta) =  \sqrt{ \frac{I_2}{I_1} }

From the question I_2  =  \frac{I_o}{5}

     cos (\theta) =  \sqrt{ \frac{ \frac{ I_o}{5} }{\frac{I_o}{2} } }

     cos (\theta) =  \sqrt{ \frac{2}{5} }

=>    \theta =   cos ^{-1} [\sqrt{\frac{2}{5}}  ]

=>    \theta  =  50.77^o

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Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

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final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

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2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

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