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Furkat [3]
4 years ago
14

The way light reflects off a smooth surface is described by the law of reflection. This law states that the angle of reflection

equals the ___________________.
Physics
2 answers:
bearhunter [10]4 years ago
6 0

Answer:

Angle of incidence

Explanation:

The angle of reflection is equal to the angle of incidence.

Reflection is defined as the repropagation of light rays incidented on a plane or smooth surface. When a light ray strikes a plane surface we say it is incidented on the surface and all incident light tends to reflect out at the same angle that they strike the surface.

There is a third ray acting on the smooth surface and perpendicular to the surface, this ray is called the normal ray. The angle that the incident ray make with this normal ray is angle of incidence while the angle that the reflected ray makes with the normal ray is the angle of reflection. When light ray strikes the surface, this two angles are always equal according to the second law of reflection.

matrenka [14]4 years ago
5 0

Answer:

Angle of incidence ray

Explanation:

The way light reflects off a smooth surface is described by the law of reflection. This law states that the angle of reflection equals the ANGLE OF INCIDENCE RAY.

This is because, when light is reflected off a smooth surface, the light is reflected by all the points that is present on that smooth surface.

Therefore the speed at which the light rays falls on the smooth surface is the same as the speed at which the light ray is reflected back. Due to their speed been the same, the distance of the incidence ray would also be the same as the distance of the reflected ray.

Hence the angle of incidence = angle of reflection.

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pishuonlain [190]

Explanation:

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It is also capable of forming hydrogen bonds with polar molecules. Each water molecule can form two hydrogen bonds involving their hydrogen atoms and two further hydrogen bonds using the hydrogen atoms attached to neighboring water molecules.

5 0
3 years ago
sebuah cermin cembung memiliki jari-jari kelengkungan 30 cm. sebuah benda diletakkan pada jarak 30 cm di depan cermin. hitunglah
stiks02 [169]
A convex mirror has a radius of curvature of 30 cm. an object is placed at a distance of 30 cm in front of the mirror. compute the location of the shadows and shadow enlargement!
5 0
4 years ago
A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.500 m2 . At
Anastaziya [24]

Answer:

31.8 × 10⁻⁴ J = 3.18 mJ

Explanation:

We know the intensity I of a wave is I = P/A where P = power and A = area = 0.500 m²

The intensity of an electromagnetic wave is also equal to I = E₀²/μ₀c

where E₀ = maximum electric field strength = √2E where E = rms value of electric field = 0.0200 N/C, μ₀ = 4π × 10⁻⁷ H/m ,c = 3 × 10⁸ m/s

P/A = E₀²/μ₀c = 2E²/μ₀c

P = 2E²A/μ₀c = 2 × (0.02 N/C)² × 0.5 m²/(4π × 10⁻⁷ H/m × 3 × 10⁸ m/s)

  = 1.06 × 10⁻⁴ W = 0.106 mW

Since P = E/t where E = Energy and t = time

E = Pt with t = 30 s

E = 1.06 × 10⁻⁴ W × 30 s =  31.8 × 10⁻⁴ J = 3.18 mJ

So the wave carries 3.18 mJ of energy through the window in 30 s

5 0
3 years ago
When Maggie applies the brakes of her car, the car slows uniformly from 15.3 m/s to 0 m/s in 2.23 s. How far ahead of a stop sig
nirvana33 [79]

Answer:

The answer is 34.119m

Explanation:

You need to pay attention to the word "uniformly". It means there is no acceleration thus the physics of this problem respond to the uniform rectilinear motion equations:

  • Xf = Xo + vt
  • v = constant
  • a = 0

where:

  • Xf = Final position
  • Xo = Origin position
  • v = velocity (speed)
  • t = time
  • a = acceleration

Xf = 0 + (15.3)(2.23)

Xf = 34.119m

3 0
4 years ago
A 0.22 kg mass at the end of a spring oscillates 2.9 times per second with an amplitude of 0.13 m .
andre [41]

Answer:

2.3687599 m/s

0.91106 m/s

0.617213012 J

Explanation:

f = Frequency = 2.9\ Hz

A = Amplitude = 0.13 m

k = Spring constant

m = Mass of object = 0.22 kg

Angular velocity is given by

\omega=2\pi f\\\Rightarrow \omega=2\pi 2.9\\\Rightarrow \omega=18.22123\ rad/s

Velocity is given by

V=A\omega\\\Rightarrow V=0.13\times 18.22123\\\Rightarrow V=2.3687599\ m/s

Speed when it passes the equilibrium point is 2.3687599 m/s

Frequency is given by

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow k=m4\pi^2f^2 \\\Rightarrow k=0.22\times 4\pi^2\times 2.9^2\\\Rightarrow k=73.04296\ N/m

x = Displacement = 0.12 m

In this system the energies are conserved

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{73.04296(0.13^2-0.12^2)}{0.22}}\\\Rightarrow v=0.91106\ m/s

The speed when it is 0.12 m from equilibrium is 0.91106 m/s

The energy in the system is given by

E=\dfrac{1}{2}kA^2\\\Rightarrow E=\dfrac{1}{2}\times 73.04296\times 0.13^2\\\Rightarrow E=0.617213012\ J

The total energy of the system is 0.617213012 J

4 0
4 years ago
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