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ad-work [718]
3 years ago
14

Unpolarized light of intensity I=10 falls on two successive polarizer wheels with the angle between the polarizer wheels e-60°.

What is the resultant intensity of the polarized light after leaving the 2nd polarizer? A) (0)/2 B) (10)/4 C) (0)/8 D) (o)/16
Physics
1 answer:
yan [13]3 years ago
5 0

Answer:

I=\frac{10}{4}

Explanation:

A polarizer changes the orientation of the oscillations of a light wave.

I₀ = Intensity of unpolarized light = 10

θ = Angle given to the polarizer = 60°

Intensity of light

I = I₀cos²θ

⇒I = 10cos²60

\\\Rightarrow I=10\times \frac{1}{2}\times \frac{1}{2}\\\Rightarrow I=\frac{10}{4}

So, the after passing through the second polarizer is \mathbf{\frac{10}{4}}

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All Physicsts over here plz help in these questions!!!!!!!!!
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Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
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g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

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You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride
marin [14]

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

a)

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

\Delta x = 6.17 cm = 0.0617 m

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

2mg = k'\Delta x (1)

where

m = 76.2 kg is the mass of each children

g=9.8 m/s^2 is the acceleration of gravity

k' is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

k'=k+k=2k

Substituting into (1) and solving for k, we find:

2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m

b)

The period of the oscillating system is given by

T=2\pi \sqrt{\frac{m}{k'}}

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

k'=2k=2(12,103)=24,206 N/m is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg

c)

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

f=\frac{1}{T}

where

f is the frequency

T is the period

In  this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

f=\frac{1}{2.09}=0.478 Hz

d)

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

t=10T=10(2.09)=20.9 s

#LearnwithBrainly

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