The percentage of light gets through three successive polarized filters is 24.9 %
From the question,
Given that,
Angle of transmission axis = 32.8°
The intensity of light emerging from the first polarizer is determined by the equation
I₁ = I₀ / 2
where I₀ ⇒ intensity of unpolarized light
The light emerging from the second polarizer can be mathematically represented by,
I₂ = I₁ x cos²θ
Substituting the values,
I₂ =( I₀/2) x cos²θ
= (I₀/2) x cos² (32.8 )
= (I₀/2) x 0.706
= (0.706 / 2 ) x I₀
The light emerging from the third polarizer is represented as,
I₃ = I₂ x cos²θ
Substituting the values in the above equation,
I₃ = ( 0.706 / 2 ) I₀ x cos² (32.8)
= (0.706 / 2 ) I₀ x 0.706
= 0.249 I₀
The percentage of intensity of light that gwts through with respect to the intensity of unpolarized light is given by the equation,
(I₃ / I₀) x 100
Substituting the values
[(0.249 x I₀) / I₀ ] x 100 = 24.9 %
Hence the percentage of light gets through is 24.9%
To learn more about unpolarized light: brainly.com/question/17164167
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The resultant<span> is the vector sum of 2 or more vectors. It is the conclusion of adding 2 or more vectors together. If </span>displacement <span>vectors A, B, and C are added together, the result will be vector R.</span>
The results of the experiments should be similar.
Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:
A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
ΔP=20 kg.m/s
Explanation:
Given data
Mass m=0.2 kg
Initial speed Vi=-44.5m/s
Final speed Vf=55.5 m/s
Required
Change in momentum ΔP
Solution
First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

Now we need to find the initial momentum
So

Substitute the given values

Now for final momentum

So the change in momentum is given as:
ΔP=P₂-P₁
![=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s](https://tex.z-dn.net/?f=%3D%5B%2811.1kg.m%2Fs%29-%28-8.9kg.m%2Fs%29%5D%5C%5C%3D20kg.m%2Fs)
ΔP=20 kg.m/s
Force = 0.20N .F = m ×a .& a = v/t then the f = m×v/t