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Physical things (solid, liquid, gas, plasma) both absorb and emit energy in the form of electromagnetic radiation of a wide range of frequencies. How fast they radiate and the strongest frequencies of radiation depend on the absolute temperature. How fast they absorb depends on the temperatures of objects around them. Therefore, the net intensity (energy per ...


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The colour you see in the sky on cloudy nights is due to the reflection of city lights off the clouds. In rural areas, a cloudy night is, as you expected, significantly darker. However, the massive amount of light given off in urban areas reflects back to Earth when there is cloud cover. And so, you see a red-orange hue, similar to the overall colour ...


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The accepted airglow answer might be technically true, but it does not answer the question! The existence of an additional and very faint source of green light in the atmosphere does not explain the absence of the green light in the sunset sky gradient. I wasn't satisfied with other answers either. The only satisfactory answer I could find is this one. ...


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Pressure of a gas comes from the average, collective change of momentum as its particles bounce against each other and the walls of anything encountering it. What gravity does is provide a counter-pressure - the weight of the gas itself - which stops it from expanding freely into space. The weight of the air above Earth's surface averages about 101,325 ...


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"Downshear of the vortex"? And your link is just a google search for these words. -But as Vortex doesn't have too much shear stresses, it must mean this bottom flow which connects/feeds the low pressure vortex center with fluid. Please note that it doesn't make any difference if the fluid is going away from the bottom or from the top of the vortex. ...


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Stratification is density variations causing the fluids to separate to different layers. Stratified Flow -lecture


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If you would just calculate the Reynold number to Troposphere, say, 10 000 m thick. Kinematik viscosity of air is, say, 1.2 x10^-5 so you found that with Re 4000 you would have just need to have a 0.0000048 m/s velocity In praxis, though, this is not the case, but it's a stratified fluid system; ...


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Your drawing shows an empty box . If it is at room temperature and not air tight the inside and outside air pressure are in equilibrium, the arrows balance against the rigidity of the wallss If the box is air tight and the inside is a vacuum then it will be crushed. Nobody makes vacuum tubes from polysterine. Why we can have vacuum tubes ( incuding ...


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It looks like you don't take into account elastic forces in the rigid cube. The pressure on the cube's faces will tend to bend them inwards, the force will be passed to the ribs, so the faces and the ribs will get compressed, and they will resist compression. Perhaps it is easier to understand what happens if you consider an evacuated spherical shell, rather ...


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I am reading your question as wondering why the total force on the (smaller) inside surface area of the cube is not crushed by the (larger) outside surface area. Perhaps exaggerating this will make it a bit clearer. Yes the total force on the outside is larger, but much of that force does not reach the interior of the cube. If you imagine the force from ...


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You have to compare the most probable speed of the atoms of He with the escape velocity from the planet. So you are working out whether $$ \left(\frac{2kT}{m} \right)^{1/2} > \alpha\ \left(\frac{2GM}{R}\right)^{1/2},$$ where $T$ is the local temperature, $m$ is the He atom mass, $M$ is the planet mass and $R$ is the radius at which you are considering ...


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Read the Wikipedia article on air pressure, the section on Mean Sea Level Pressure: http://en.wikipedia.org/wiki/Atmospheric_pressure "When barometers in the home are set to match the local weather reports, they measure pressure adjusted to sea level, not the actual local atmospheric pressure." So weather reports do not report the actual local air ...


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Following up on your comment: If you initially have $5.7 \cdot 10^{-4} \mathrm{kg/kg}$ of carbon dioxide, and the total mass of the atmosphere is $10^{18} \mathrm{kg}$, you can compute the mass of $\mathrm{CO_2} = 5.7 \cdot 10^{14} \mathrm{kg}$. Adding $10^{14}\mathrm{kg}$ to that does indeed give a $\frac{1}{5.7} = 18\%$ change as you calculated in your ...



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