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matrenka [14]
4 years ago
4

Which image illustrates diffraction

Physics
2 answers:
skad [1K]4 years ago
8 0

Answer:

I think it is C

Explanation:

I think C because it is the one. I might be wrong but I helped and your welcome if you thanking me.

lara31 [8.8K]4 years ago
3 0

Answer:

Figure C

Explanation:

Here different figure represents different situations

Figure A

it represents the situation of reflection of light in which light bounce back into the same medium after reflecting by a surface

Figure B

It show refraction of light in which light will bend from its path when light travels from one medium to other medium

Figure C)

here we can see that a plane wavefront incident on the small opening and after that it converts into spherical wavefront. So when light passes through a small opening then it gets diverge at the corners of the opening which is known as diffraction

Figure D

it is just the incident of light ray which get completely absorbed by the medium.

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Force,friction,inertia and momentum

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The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.

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Which material would you take, steel or bricks/stone to design a bridge. Why? ( Question related to poisson's ratio)​
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4 years ago
Which two statements are true about redox reactions?
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Read 2 more answers
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
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