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2 years ago

Why would a flare be observed in visible light, when they are so much brighter in X-ray and ultraviolet light?

Physics

1 answer:

raketka [301]2 years ago

3 0

Answer:

Because it radiates a spectre of frequencies, this means that it radiates in a continuous range of frequencies, ones with more intensity (like x-ray and ultraviolet) and others with less intensity (like the visible light). While most of the radiation is not in the visible range, there still is a part of the radiation in the visible spectre. And while this part is not the most intense part, the radiation is so large that we can see a very bright visible light.

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Any organism that has a skeletal sytem

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A length of copper wire carries a current of 11 A, uniformly distributed through its cross section. Calculate the energy density

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Answer:

a) $0.983 \frac{J}{m^3}$

b) $u_E =7.329x10^-3 \frac{J}{m^3}$

Explanation:

The energy density is "the energy per unit volume, in the electric field. The energy stored between the plates of the capacitor equals the energy per unit volume stored in the electric field times the volume between the plates".

A magnetic field is a "vector field that describes the magnetic influence of electric charges in relative motion and magnetized materials".

Part a

For this case we can assume use the equation for the magnetic field in terms of the energy per unit of volume.

$B=\sqrt{2\mu_o u}$

Where μ0 represent the permeability constant, also known as the magnetic constant. If we solve for u we got:

$u=\frac{B^2}{2\mu_o}$

We also know that the magnetic field can be expressed in terms of the current and the radius of action R like this:

$B=\frac{\mu_o i}{2\pi R}$

Replacing this on the formula for u we have:

$u=\frac{1}{2\mu_o}(\frac{\mu_o i}{2\pi R})^2$

And simplyfing we got:

$u=\frac{\mu_o i^2}{8\pi^2 R^2}$

Replacing the values given we have:

$u=\frac{(4\pix10^{-7} \frac{H}{m} (11A)^2}{8\pi^2 (0.0014m)^2} =0.983 \frac{J}{m^3}$

Part b

The density current is given by this formula $J=i/A$ and the resistance by $R=\frac{\rho l}{A}$

If we use the equation for the energy density we have this:

$u_E =\frac{1}{2}\varepsilon_o E^2 =\frac{\varepsilon}{2}(\rho J)^2=\frac{\varepsilon}{2}(\frac{iR}{l})^2$

And replacing the values given we have:

$u_E =\frac{8.85x10^{-12}\frac{F}{m}}{2}(\frac{11A(3700\frac{\Omega}{m})}{l})^2 =7.329x10^-3 \frac{J}{m^3}$

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