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Gekata [30.6K]
2 years ago
6

In the shadow of a tree with a dense, leafy canopy, one sees a number of light spots. Surprisingly, they all appear to be circul

ar. Why
Physics
1 answer:
Bad White [126]2 years ago
8 0

The characteristics of the diffraction phenomenon allow to find the result for the shape of the points of light that you pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Diffraction is the phenomenon where the undulatory part of the light becomes evident, it is the interference of the waves that make up each ray of light, for this phenomenon to occur it must be fulfilled that the wavelength is of the order of the space where pass the light.

In the leafy tree it has many leaves, but there are spaces between them, some of these spaces are small and it fulfills the diffraction condition, therefore we see bright spots and not a continuous shadow.

Diffraction can be classified depending on the distance to the observer:

  • Near field or fresnel. In this case the distance from the observer is small and we can see the shape of the object that creates the diffraction.
  • Far field or Fraunhoger. In this case the distance between the obstacle (leaves) and the person is great, here the information on the shape of things is lost and we have two observable forms. Lines for the case of slits and circles for the case of objects with a closed shape.

In this case, the distance from the leaves to the observer is large, therefore we are in the case of far-field diffraction and since the edge of the leaves that forms the diffraction is closed, the observable shape is a circle.

In conclusion using the characteristics of the diffraction phenomenon we can find the result for the shape of the points of light that pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Learn more about diffraction here:  brainly.com/question/20140459

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spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center o
satela [25.4K]

Answer:

the correct result is r = 3.71 10⁸ m

Explanation:

For this exercise we will use the law of universal gravitation

          F = - \frac{m_{1} m_{2} }{r^2}

We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m

rocket force -Earth

          F₁ = - \frac{m' M }{r^2}

rocket force - Moon

          F₂ = - \frac{m' m }{(d-r)^2}

in the problem ask for what point the force has the relation

          2 F₁ = F₂

let's substitute

          2 2 \frac{M}{r^2} = \frac{m}{(d-r)^2}

          (d-r) ² = \frac{m}{2M} r²

           d² - 2rd + r² = \frac{m}{2M} r²

           r² (1 -\frac{m}{2M}) - 2rd + d² = 0

Let's solve this quadratic equation to find the distance r, let's call

           a = 1 - \frac{m}{2M}

           a = 1 - \frac{7.36 10^{22} }{2 \  5398 10^{24}} = 1 - 6.15 10⁻³

           a = 0.99385

         

            a r² - 2d r + d² = 0

           r =  \frac  {2d \frac{+}{-}   \sqrt{4d^2 - 4 a d^2}} {2a}

           r = [2d ± 2d \sqrt{1-a}] / 2a

           r = \frac{d}{a}   (1 ± √ (1.65 10⁻³)) =  \frac{d}{a} (1 ± 0.04)

           r₁ = \frac{d}{a} 1.04

           r₂ = \frac{d}{a} 0.96

let's calculate

           r₁ = \frac{3.84 10^8}{0.99385} 1.04

           r₁ = 401.8 10⁸ m

          r₂ = \frac{3.84 10^8}{0.99385} 0.96

          r₂ = 3.71 10⁸ m

therefore the correct result is r = 3.71 10⁸ m

3 0
2 years ago
You fill two balloons with gas, one with hydrogen and one with carbon dioxide. You hold a match to each balloon. The hydrogen ba
miv72 [106K]

Answer:E. Hydrogen was able to participate in an exergonic reaction and carbon dioxide couldn't

Explanation:

An exergonic reaction releases energy to the environment. The combustion of hydrogen contained in the balloon is a chemical reaction. The reaction can take place because hydrogen combines with oxygen in air, that is, the gas is combustible. CO2 does not support combustion, it does not combine with oxygen in air and it is also denser than air, hence does not participate in the exergonic reaction.

4 0
3 years ago
A 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The
Temka [501]

Answer:

1m/s^2

Explanation:

Mass of block=10 kg

Applied horizontal force =F=20 N

Friction force=f=10 N

We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

ma=F-f

Where F= Horizontal force

f=Friction force

m=Mass of object

a=Acceleration of object

10a=20-10=10

a=\frac{10}{10}=1 m/s^2

Hence, the acceleration of the block=1m/s^2

4 0
3 years ago
4. Identify Problems A student is describing a longitudinal wave in his notebook. He
andrew11 [14]

Answer:

 "The distance between crests is 3 cm." 

Explanation:

If he writes down "The distance between crests is 3 cm." 

That means he is describing the wavelength of a wave and not longitudinal wave. He ought to write something about " direction "

Longitudinal waves are waves in which the displacement of the medium is in the same direction as, or parallel to, the direction of propagation of the wave. While

Wavelength is the distance between the two successfully Crest or trough

3 0
3 years ago
An 8 kg mass moving at 8 m/s collides with a 6 kg mass
steposvetlana [31]

Answer:

10 m/s

Explanation:

Momentum before collision = momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v

64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v

60 kg m/s = (6 kg) v

v = 10 m/s

3 0
3 years ago
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