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4 years ago

How does the scattering of tiny particles in the air affect the colors of the sky at sunrise and sunset

Physics

2 answers:

Zepler [3.9K]4 years ago

6 0

Answer:

In the sky, the scattering of light takes place. The comparatively smaller particles such as the tiny droplets of water, dust particles in the air and other molecules have the ability to scatter higher number of photons that have a shorter wavelength. The sky usually appears to be blue in color but it appears to be reddish during sunrise and sunset as the sun is at a very low horizon, and presence of more number of molecules allows more scattering to take place, eliminating the violet light and blue light away from our naked eyes and allows the red light to our eyes which has the highest wavelength.

This is how the scattering of tiny particles and molecules present in the air affects the color of the sky during sunrise and sunset.

timofeeve [1]4 years ago

4 0

It affects the colors of the sky at sunrise and sunset because only the longer-wavelengths colors are visible.

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Which of the following statements about the force on a charged particle due to a magnetic field are not valid?

Ratling [72]

Answer:

this is from quizlet and is similar hope it helps.

Explanation:

Which of the following statements about the force on a charged particle due to a magnetic field are not valid?

Check all that apply.

a. It depends on the strength of the external magnetic field.

b. It depends on the particle's charge.

c. It depends on the particle's velocity.

d. It acts at right angles to the direction of the particle's motion.

e. None of the above; all of these statements are valid.

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2 years ago

Certain suv ads claim that their vehicles can climb a slope of 45?. what is the minimum coefficient of static friction between t

kotykmax [81]

First, you make a force body diagram to illustrate the problem. Then, apply Newton's laws of motion.

Summation of forces along the y-direction:

F = 0 (bodies at equilibrium) = Normal force - weight*cos45 = 0
Normal force = wcos45

Summation of forcesalong the x-direction:

F = 0 (bodies at equilibrium) = Frictional force - weight*sin45 = 0
x*Normal force - weight*sin45 = 0 (let x be the min coeff of static friction)
x*weight*cos45 = weight*sin 45
x = sin45/cos45 = 1

Therefore, the minimum coefficient of static friction must be 1.

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3 years ago

What is any motion in which an object is moving along a curved path

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4 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }