So, the first question is: how many meters are 10 nm?
1nm =<span>0.000000001 m.
So 10 nanometers are </span><span>0.00000001 m!
Now, how many milimeter are those?
let's start with meters, 1 meter are 1000 milimeters.
so </span>
0.00000001*1000=0.<span><span>00001</span> m!
now, micrometers .1 micrometer are 1000 nanometers.
so 10 nanometers are 0.01 micrometers! (1 nanometer is 0.001 micrometers)
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A/c/d. that is the anser hope it helps
Answer:
An object on the moon would weigh the LEAST among these. So correct answer is B.
Explanation:
- Weight of an object on any place is given by:
W = Mass * Acceleration due to gravity(g)
- It means when masses of different objects those are in different places are same, the weight of those objects depends upon the 'g' of that particular place.
- As we know, acceleration due to gravity on surface of moon (g') is 6 times weaker than the acceleration on surface of earth (g), which is due to the large M/R^2 of the earth than the moon.
i.e. g' = g/6 so W' = W/6
- And in the space between the two, the object is weightless.
Answer:
1. The lowest stratum of the atmosphere
troposphere
2. Most of the earth's ozone is located in the
stratosphere
3. Earth coldest temperature occur in the
mesosphere
Explanation:
Answer:
a much larger slit, the phenomenon of Sound diffraction that slits for light.
this is a series of equally spaced lines giving a diffraction envelope
Explanation:
The diffraction phenomenon is described by the expression
d sin θ = m λ
Where d is the distance of the slit, m the order of diffraction that is an integer and λ the wavelength.
For train the diffraction phenomenon, the d / Lam ratio is decisive if this relation of the gap separation in much greater than the wavelength does not reduce the diffraction phenomenon but the phenomena of geometric optics.
The wavelength range for visible light is 4 10⁻⁷ m to 7 10⁻⁷ m. The wavelength range for sound is 17 m to 1.7 10⁻² m. Therefore, with a much larger slit, the phenomenon of Sound diffraction that slits for light.
When we add a second slit we have the diffraction of each one separated by the distance between them, when the integrals are made we arrive at the result of the interference phenomenon, a this is a series of equally spaced lines giving a diffraction envelope
When I separate the distance between the two slits a lot, the time comes when we see two individual diffraction patterns
