All categories

3 years ago

A light wave is traveling through the air and then enter water. How will the light change?

Physics

1 answer:

Bogdan [553]3 years ago

8 0

Answer:

A) It will move at a different speed (i.e slower)

B) A vacuum

Explanation:

For A), Light moves at a different speed as it starts to interact more with the particles in said medium (in this case, water) than the earlier medium.

For B), Light propagates using its own particle, i.e the photon whereas sound is more of a transmission of energy from the particles it originates from. In a vacuum, there are no particles to transfer energy to, that is why light can travel in a vacuum and sound cannot

You might be interested in

Why is an object's density expressed as a combination of two units

miss Akunina [59]

Well, first of all, EVERY physical quantity is measured in a combination
of 2 or more units, except for mass, length, time, and electric charge.
ALL other units are made out of these. So it should not surprise you.

[ Example: Speed = (length) / (time) ]

Density is not the mass of a substance. It's the mass of a substance in
a standard volume of it. So the density is made of the mass in any lump
and the volume of that lump. That way, no matter how much of a substance
you have, you can always compare the lump you have to all other substances.

6 0

3 years ago

A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi

Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

$= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

= 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

$=\frac{1}{2} \times 1.225 \times 200^2$

$=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

= 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

= 270 N

Therefore the net drag = 270 x 2

= 540 N

7 0

3 years ago

HELP ME! One airline limits the size of carry-on luggage to a volume of 40,000 cm^3. A passenger has a carry-on that has an area

EleoNora [17]

You must times the area by the volume, look at it as if the area is just one of 23 layers that makes up the volume.
1960x23=45080
so no it cannot be carried as it is 5080cm^3 over the limit

8 0

3 years ago

Read 2 more answers

What is the force if the mass is 75kg and the acceleration is 24.5m/s^2

Snowcat [4.5K]

Given;

mass m = 75 kg

acceleration a = 24.5 ms²

F = ma 

F = 75 kg * 24.5 ms²

= 1837.5 kg ms².

4 0