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valina [46]
3 years ago
7

We can see our face clearly on mirror why cant we see our face clear on aluminium or irion plate

Physics
1 answer:
Simora [160]3 years ago
8 0

Because specular reflection occurs with a mirror, while diffuse reflection occurs with aluminium or iron plate

Explanation:

Reflection is a phenomenon typical of waves (such as light wave), that occurs when a wave bounces off the surface of a certain material, going back into the original medium at a certain angle, without changing speed, frequency or wavelength.

The direction of the reflected ray is given by the law of reflection, which states that:

  • The incoming ray, the reflected ray and the normal to the surface all lie on the same plane
  • The angle of incidence is equal to the angle of reflection

Depending on the type of surface, two different types of reflection can occur:

  • Specular reflection: this occurs when the surface is perfectly smooth. In such a case, all the incoming rays of light hit the surface with same angle of incidence, therefore they are reflected at the same angle. This means that a perfect and clear image of the original object can be formed: this is the case of a mirror.
  • Diffuse reflection: when the surface is not smooth, the imperfections on the surface are such that the incoming rays of light hit the surface at different angles of incidence. As a result, the rays are reflected at different angles to each other, and therefore no clear image of the object is produced. This is the case of aluminium or iron plate.

Learn more about reflection and other wave phenomena:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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En un barrio un estudiante observa un roedor en medio de dos postes, los cuales están comunicados por un solo cable de fibra ópt
castortr0y [4]

Answer:

7 Newton

Explanation:

Dado

Longitud de la cuerda = 50 m

El cable se dobla en 0,058 m.

Masa de roedor = 350 gramos = 0,35 kg

T = m * a + m × v2 / r

Sustituyendo los valores dados obtenemos

T = 0,35 (10 + 10)

T = 0,35 * 20

T = 7 Newton

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How much heat is released to freeze 47.30 grams of copper at its freezing point of 1,085°C? The latent heat of fusion of copper
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the answer is -9,697

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Could it be hypothesis?
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Describe each of Newton’s Laws of Motion in ice skating. What can you design/develop to improve ice skating?
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Newton's three laws of motion can be used to describe the motion of the ice skating.

<h3>Newton's first law of motion</h3>

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.

  • Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.

<h3>Newton's second law of motion</h3>

Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.

  • Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.

<h3>Newton's third law of motion</h3>

This law states that action and reaction are equal and opposite.

  • Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.

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Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he
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Answer:

E = 2k  \frac{\lambda}{ r}

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

            Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

            λ = q_ {int} /l

            q_ {int} = l λ

we substitute

            E A = l λ /ε₀

is area of ​​cylinder is

           A = 2π r l

we substitute

            E = \frac{ l \ \lambda}{ \epsilon_o \ 2\pi  \ r \ l }

             E = \frac{\lambda}{ 2\pi  \epsilon_o \ r}

the amount

            k = 1 / 4πε₀

            E = 2k  \frac{\lambda}{ r}

5 0
3 years ago
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