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3 years ago

Why can light microscopes produce images in their natural color

Physics

1 answer:

Fiesta28 [93]3 years ago

6 0

Answer:

The electron microscopes are uses the electron to produce the image of the any object.The area where the electron passes then that area appeaser white but on the other hand where the electron does not pass that area appears black.But this does not give any information about the color of light it shows only white and black areas.We also know that color is the property of photons ( light).So the microscopes does not produce color of the image.

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Who is most likely to benefit from use of lithium?

Alexxx [7]

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3 years ago

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

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1 year ago

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Ede4ka [16]

Answer:

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marusya05 [52]

Answer:

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Explanation:

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damaskus [11]

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