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3 years ago

Why does the sky change colors at sunset?

2 answers:

8 0

Atmospheric refraction is the deviation of light or other electromagnetic wave from a straight line as it passes through the atmosphere due to the variation in air density as a function of height. ... Refraction not only affects visible light rays, but all electromagnetic radiation, although in varying degrees.

So in short, the answer is D.

(My answer got deleted because it didnt explain which is dumb)

Ne4ueva [31]3 years ago

5 0

Answer:

because the atmosphere reflects light

Explanation:

When light travels from sun to Earth then the path of light will change continuously due to change in the optical density of the medium.

As we know that as air mass density increases then its optical density also increases and this will bend the light from sun to earth.

Now as light comes towards us from sun it will have multiple refraction and each time while light refracts the different components of light will bend in different directions and this will disperse the light in all directions.

So here while sunset the position of sun will change with time and due to this light of different colours will disperse at each positions of sun.

Due to this phenomenon the colour of sky will change

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Evaluate the scenarios and select the one that demonstrates the training principle overload.

matrenka [14]

Answer:

B :)

Explanation:

:) JUST TRUST ME I GOT IT CORRECT

5 0

3 years ago

Read 2 more answers

The ammonia molecule (NH3) has a dipole moment of 5.0×10?30C?m. Ammonia molecules in the gas phase are placed in a uniform elect

Neko [114]

Question (continuation)

(a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E S from parallel to perpendicular?

(b) At what absolute temperature T is the average translational kinetic energy 3/2kT of a molecule equal to the change in potential energy calculated in part (a)?

Answer:

a. 9.0 * 10^-24 Joules

b. 0.44K

Explanation:

Given

Let p = dipole moment = 5.0 * 10^-30 Cm

Let E = Magnitude = 1.8 * 10^6 N/m

The charge in electric potential = Final Charge - Initial Charge

Initial Charge = Potential Energy

Initial Energy = -pE cosФ where Ф = 0

So, initial Energy = - 5.0 * 10^-30 * 1.8 * 10^6

Initial Energy = -9 * 10^-24 Joules

Final Energy = 0

Charge = 0 - (-9.0 * 10^-24)

Charge = 9.0 * 10^-24 Joules

Absolute Temperature

Change in Kinetic Energy = Change in Potential Energy = 9.0 * 10^-24

Change in Kinetic Energy = 3/2kT where k is Steven-Boltzmann constant = 1.38 * 10^-23

So,

9.0 * 10^-24 = 3/2 * 1.38 * 10^-23 * T

T = (9.0 * 10^-24 * 2)/(3 * 1.38 * 10^-23)

T = (18 * 10^-24)/(4.14 * 10^-23)

T = 0.44K

6 0

3 years ago

On Earth, friction can be _____ but not removed.

Whitepunk [10]

Answer:

reduced

Explanation:

The use of bearing surfaces that are themselves sacrificial, such as low shear materials, of which lead/copper journal bearings are an example

4 0

3 years ago

What force must be used to do 224 Joules of work on an object over a distance of 32 meters?

cestrela7 [59]

7.625 Newtons

work = force× distance
Newtons is an accepted value for force

so take the total 224 joules and decide by distance 32 meters to find force in Newtons

4 0

3 years ago

Blue light of wavelength λ passes through a single slit of width d and forms a diffraction pattern on a screen. If we replace th

ololo11 [35]

Answer:

We can retain the original diffraction pattern if we change the slit width to d) 2d.

Explanation:

The diffraction pattern of a single slit has a bright central maximum and dimmer maxima on either side. We will retain the original diffraction pattern on a screen if the relative spacing of the minimum or maximum of intensity remains the same when changing the wavelength and the slit width simultaneously.

Using the following parameters: y for the distance from the center of the bright maximum to a place of minimum intensity, m for the order of the minimum, λ for the wavelength, D for the distance from the slit to the screen where we see the pattern and d for the slit width. The distance from the center to a minimum of intensity can be calculated with:

$y\approx\frac{m\lambda D}{d}$

From the above expression we see that if we replace the blue light of wavelength λ by red light of wavelength 2λ in order to retain the original diffraction pattern we need to change the slit width to 2d:

$y\approx\frac{m\lambda D}{d} =\frac{m2\lambda D}{2d}$

7 0

3 years ago