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Elodia [21]
3 years ago
15

If you shine a light of frequency 375hz on a double slit setup , and you measure the slit separation to be 950 nm and the screen

distance to be 4030 nm away , what is the distance from the zero order fringe to the first order fringe ?
Physics
1 answer:
amm18123 years ago
6 0

Answer:

y = 33.93 10⁵ m

Explanation:

This is an interference exercise, for the contributory interference is described by the expression

           d sin θ = m λ

let's use trigonometry for the angle

           tan θ = y / L

how the angles are small

          tan θ = sin θ / cos tea = sin θ

 

we substitute

         sin θ = y / L

        d y / L = m λ

         y = m λ L / d

the light fulfills the relation of the waves

       c = λ f

       λ = c / f

       λ = 3 10⁸ /375

       λ = 8 10⁵ m

first order m = 1

let's calculate

        y = 1  8 10⁵  4030 10-9 / 950 10-9

        y = 33.93 10⁵ m

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4.91 x 10⁻⁷ m

Explanation:

the applicable formula is

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if we rearrange the equation and substitute the values given above,

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A sound wave travels through water at 1,500 m/s. How far would it travel in 5 s?
hichkok12 [17]
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Which of the following BEST describes all planets in the universe?
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3 years ago
A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of
Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0
3 years ago
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