Rays of light strike a bumpy road and are reflected.

Physics

2 answers:

nekit [7.7K]3 years ago

7 0

<h3><u>Answer;</u></h3>

Each will reflect at an angle equal to its angle of incidence.

<h3><u>Explanation;</u></h3>

<em><u>When light hits a surface or a boundary it bounces of the surface a phenomenon called reflection, which is a property of waves. Reflection occurs such that the incidence angle is equal to the angle of reflection.</u></em>
Reflection of light depends on whether the surface is smooth or rough. When the light hits a smooth surface it bounces off the surface and all the rays of light moves in the same direction, this type of reflection is called specular reflection.
On the other hand , when right rays hits a rough surface the rays bounces off the surface and the rays move in different directions, a type of reflection called diffuse reflection.

Marina CMI [18]3 years ago

5 0

The reflected rays undergo irregular reflection. they are not parallel

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ioda

Answer:

m = 375 [gram]

Explanation:

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For a better understanding, the following image is attached, with values on each beam, which should be read.

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3 years ago

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slamgirl [31]

Answer:

The weight of the body in the new planet is 100 newtons.

Explanation:

From Newton's Law of Gravitation we find that gravitational force is directly proportional to mass of the planet and inversely proportional to the square of its radius. From this fact we can build the following relationship:

$\frac{F_{1}\cdot R_{1}^{2}}{M_{1}} = \frac{F_{2}\cdot R_{2}^{2}}{M_{2}}$ (1)

Where:

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$M_{1}$ , $M_{2}$ - Mass of planet, measured in kilograms.

$R_{1}$ , $R_{2}$ - Radius of the planet, measured in meters.

If we know that $F_{1} = 800\,N$ , $\frac{M_{2}}{M_{1}} = \frac{1}{2}$ and $\frac{R_{2}}{R_{1}} = 2$ , then the expected gravitational force in the new planet is: