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3 years ago

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

Physics

2 answers:

Feliz [49]3 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]3 years ago

4 0

Light can reflect from mirrors because mirrors are a prism.

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3 years ago

A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the c

DedPeter [7]

Answer:

The induced current in the coil at the time 2 s is 0.00263 A

Explanation:

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θ = angle

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3 years ago

Two stationary positive point charges, charge 1 of magnitude 3.05 nC and charge 2 of magnitude 1.85 nC, are separated by a dista

luda_lava [24]

To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

$V = \frac{kq}{r}$

Here,

k = Coulomb's constant

q = Charge

r = Distance to the center point between the charge

From each object the potential will be

$V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}$

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Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

$V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}$