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3 years ago

12

Why do we need optics now a days?

1 answer:

AveGali [126]3 years ago

7 0

We need optics to help aid people who have short or long sightedness.
we need optics to be able to watch TV
we need optics to be able to use the internet at high speeds

there are a tonne of reasons why we need optics.

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The energy stored per unit volume of the inductor is called the energy density of the magnetic field. Why?

Marrrta [24]

Answer:

yes because physics want that

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2 years ago

A 7.36 g sample of copper is contaminated with a additional 0.51 g sample of zinc. Suppose an atomic mass measurement was perfor

yawa3891 [41]

The molar mass of the sample is equal to the summation of the molar mass of the elementas multiplied by the abundance of the elements by mole. In this case, copper has an abundance of 93.69 percent while zinc has 6.31 percent. In this case, the average molecular weight is 63.67 g/mol

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3 years ago

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

yan [13]

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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3 years ago

DUE TODAY PLEASE HELP.....WILL GIVE BRANIST!!!!!!!

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3 years ago

Read 2 more answers

One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d

stira [4]

Answer:

a) Diffusion coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

$^{2} = 2Dt$ ..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

$36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr$

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

$^{2} = 2Dt$

$0.1^{2} = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds$

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0

3 years ago