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3 years ago

In case of frequency of light what changes

Physics

1 answer:

maks197457 [2]3 years ago

6 0

Answer:

frequency remains constant when light passes through different mediums . only speed changes and due to it , wavelength also changes. one way of changing the frequency is by Doppler effect, where the apparent frequency changes due to motion of the source ,observer or both of them.

Explanation:

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What is the equation for frequency, wavelength, and speed of a wave?

Nastasia [14]

Answer:

λ = v/f

Explanation:

frequency=f

wavelength = λ

speed of a wave=v

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3 years ago

Activity 1 more or less, pls answer this ASAP

AURORKA [14]

Answer:

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5 0

3 years ago

A light wave traveling through medium 1 strikes a boundary with medium 2 at at a 45 degree angle. the light then enters the seco

Alona [7]

Here light ray strikes to interface at an angle of 45 degree and then refracts into other medium such that it will bend towards boundary.

So here the angle of incidence will be less than the angle of refraction as light moves towards the boundary after refraction which mean it will bend away from the normal

here it can be said that medium 2 will be rarer then medium 1

So here the possible options are

1. Water

Air

2. Diamond

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8 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

4 years ago

A drone flies 8 m/s due East with respects to the wind. The wind is blowing 6 m/s due North with respects to the ground. What is

vekshin1

Answer:

B. 10m/s

Explanation:

If a drone flies 8 m/s due East with respects to the wind and the wind is blowing 6 m/s due North, the speed of the drone with respect to the ground is its displacement.

Displacement is calculated using Pythagoras theorem.

d² = 8²+6²

d² = 64+36

d² = 100

Square root both sides

√d² = √100

d = 10m/s

Hence the distance of the drone with respect to the ground is 10m/s

Option B is correct

8 0

3 years ago

In case of frequency of light what changes​

In case of frequency of light what changes