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2 years ago

How does light reflect off of mirrors and other types of surfaces?

Physics

2 answers:

Feliz [49]2 years ago

8 0

The light reflect off of mirrors and other types of polished surfaces because of reflection.

Explanation:

The polished surfaces can reflect most of the light falling on them.

They will reflect the light waves falling on them as the polished surface or mirrors like objects will be opaque in nature.

The materials which are opaque in nature does not allow light to pass through them or transmit through them.

When light rays are incident on an object, three kinds of rays will be formed.

They are incident rays which are the source rays hitting the object, the reflected rays which are the rays coming back or get reflected as they are not allowed to penetrate the object and in some cases refracted rays and transmitted rays occur when the rays penetrate partially with different angle or fully with different angle.

Generally, the opaque objects which have a polished front surface and other surface is closed reflect light waves blocking them to penetrate deeper. So the light will get reflected from the surface.

As the mirrors and most of the bulk surface are opaque in nature, light will not be able to penetrate the depth of those surface and they will get reflected.

Marat540 [252]2 years ago

4 0

Light can reflect from mirrors because mirrors are a prism.

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fenix001 [56]

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2 years ago

A luge and its rider, with a total mass of 85 kg, emerge from adownhill track onto a horizontal straight track with an initial s

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<span>(1) </span>Through the Second Law of motion, the equation for Force is:

F = m x a

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D = Vi^2 / 2a

Where Vi is initial velocity

<span>(3) </span>Work is calculated through the equation,

W = F x D

Substituting the known values,

Part A:

<span>(1) </span> F = (85 kg)(2 m/s^2) = 170 N

<span>(2) </span> D = (37 m/s)^2 / (2)(2 m/s^2) = 9.25 m

<span>(3) </span> W = (170 N)(9.25 m) = 1572.5 J

Part B:

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<span>(2) </span>D = (37 m/s)^2 / (2)(4 m/s^2) = 4.625 m

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2 years ago

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tiny-mole [99]

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2 years ago

A series RLC circuit has a resonant frequency = 6.00 kHz. When it is driven at a frequency = 8.00 kHz, it has an

ANEK [815]

The resistance (R) of the circuit is 707.1 ohms and the inductance (L) is 0.032 H.

<h3>Resistance of the circuit</h3>

For the phase constant of 45⁰, impedance is equal to the resistance of the circuit.

$Z= R\sqrt{2} \\\\R = \frac{Z}{\sqrt{2} } \\\\R = \frac{1000}{\sqrt{2} } = 707.1 \ ohms$

<h3>Resonant frequency</h3>

$f = \frac{1}{2\pi \sqrt{LC} } \\\\6000 = \frac{1}{2\pi \sqrt{LC} } \\\\2\pi(6000) = \frac{1}{\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi (6000)} \\\\LC = (\frac{1}{2\pi (6000)})^2\\\\LC = 7.034 \times 10^{-10} \\\\ C = \frac{7.034 \times 10^{-10} }{L} ---(1)$

<h3>At driven frequency</h3>

$X_l- X_c = R\\\\\omega L - \frac{1}{\omega C} = 707.1\\\\2\pi f L - \frac{1}{2\pi fC} = 707.1\\\\2\pi (8000) L - \frac{1}{2\pi (8000) C } = 707.1\ \ --(2)\\\\$

<em>solve 1 and 2 together</em>

$2\pi(8000) L - \frac{L}{2\pi (8000)(7.034 \times 10^{-10})} = 707.1\\\\50272L - 28279.48L = 707.1\\\\L = 0.032 \ H$

Learn more about impedance of RLC circuit here: brainly.com/question/372577

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2 years ago