Ask question

Ask question

All categories

3 years ago

9

What is the bouncing of light when it changes media

1 answer:

matrenka [14]3 years ago

5 0

When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.

The part of the light that continues on into the new medium is <em>transmitted</em>
light. Its forward progress at any point in its journey is <em>transmission</em>.

Its direction usually changes as it crosses the boundary. The bending is <em>
refraction</em>.

The part of the light that bounces away from the boundary and heads back
into the first medium is <em>reflected</em> light. The process of bouncing is <em>reflection</em>.

You might be interested in

A ball is dropped from rest. how fast is the ball going after 3 seconds

Airida [17]

Answer:

v = 29.4m/s

Explanation:

Since the ball is dropped at rest,

u = 0m/s

a = 9.81m/s²

Using

v = u + at

After 3 seconds,

v = 0 + (9.81)(3)

v = 29.4m/s

5 0

3 years ago

Which is not an example of an external force acting on an object? (1 point)

GrogVix [38]

Answer:

A. a meteor traveling unhindered through space

Explanation:

7 0

3 years ago

Name 3 elements in group one

Vinil7 [7]

Answer:

I don't really understand the question but fire water air and like earth but you said 3

Explanation:

7 0

3 years ago

The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within

zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, $m=9.1\times 10^{-31}\ kg$

Initial speed of electron, u = 0

Final speed of electrons, $v=3\times 10^7\ m/s$

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2}{2\times 0.063}$

$a=7.14\times 10^{15}\ m/s^2$

Now we will find the electric field required in the tube as :

$ma=qE$

$E=\dfrac{ma}{q}$

$E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}$

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0

3 years ago

tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h

Sati [7]

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q is the amount of heat supplied to the object

m is the mass of the object

$\Delta T$ is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

$\Delta T=12^{\circ}$ is the increase in temperature

Substituting into the formula, we find the specific heat:

$C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC$

8 0

3 years ago