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EleoNora [17]
3 years ago
10

What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified

wavelength encounters the following natural or anthropogenic particles?
Wavelength O2molecules Smoke particles Cloud droplets Rain droplets
(size 10-10m) (size 0.3 mm) (size 20 mm) (size 3 mm)
200 nm
0.6 mm
10 mm
1.0 mm
1.0 km
Physics
1 answer:
oee [108]3 years ago
3 0

Answer:

hello your question is not properly arranged attached below is the arranged table and solution

answer : attached table below

Explanation:

Given data:

02 molecules size = 10^-10m

smoke particles size = 0.3 mm

cloud droplets size = 20 mm

Rain droplets size = 3 mm

Attached below is a table showing the kind of scattering that is expected to occur at various wave lengths

Note : For Rayleigh scattering the wave particle is smaller than the wave length while for Non-selective scattering the wave particle is greater than the wavelength.

and  For Mie scattering the wavelength is the same as the wavelength.

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A source of sound is kept in a jar in a vacuum. Air is slowly introduced in to the jar. What happens to the sound coming out of
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<span>The loudness of the sound increases gradually as the air is slowly introduced in to the jar. This is because sound needs a physical medium and in a vacuum there is none. The air provides that medium and as it is introduced, the transfer of sound energy increases</span>
3 0
3 years ago
Come si esprime il numero 0,00123 in notazione scientifica?
Inessa [10]

Answer:

0,00123 = 1,2*10^{-3}

Explanation:

To write down correctly the number 0,00123 in scientific notation, you take into account that the scientific notation demands that there in only one number after the comma ( , ). Furthermore, it is necessary that you move the comma to the right of the first number different of zero, in this case the number 1. To do this you move the comma three positions.

Then, you have to multiply the expresion 1.23 by 10 with an exponential -3 (because of the movement of the comma in three positions). That is:

0,00123 = 1,23*10^{-3}

But it is mandatory that nly one number can stay after the comma, so, you approximate the number three. In this case, the number is lower than 5, hence, you approximate 3 to 0.

Finally, you have:

0,00123 = 1,2*10^{-3}

3 0
3 years ago
An object is 15 cm in front of a diverging lens with a
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A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

4 0
4 years ago
a gas has a volume of 8 liters at a temperature of 300 K the volume is then increased to 12 Liters what is the new temperature (
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300/8 = 37.5
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6 0
3 years ago
If some electrons are transferred from object A to object B,
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6 0
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