# Tag Info

7

The concept you are looking for is the attenuation length of light in water. This is the quantitative measure of how strongly different wavelengths get absorbed in water. The absorption of light is fairly simple to describe: it mostly depends on the material, but also if the light transverses a longer length of material, the attenuation will be stronger, ...

5

You are mistaken. Actually, you can melt ice by applying pressure. This is why ice is so slippery, when you step on a frozen lake, you are melting the very first layer of water, and thus creating a very good instant lubricant for you to slide on. It is a common knowledge false fact, see comments. Ok, granted, at very high pressures water does become solid. ...

5

Reading a discussion on this topic on the XKCD forums, it seems there is a whole book dedicated to this question. If you check e.g. page 51, the graph there indicates a much more detailed structure than an uniform one. To be more precise, there are several mechanisms in play: evaporation ice formation ice melting influx of fresh water Several quotes: ...

4

This is really just a footnote to Ross' answer as Ross is quite correct and the link he provided contains the information you asked for. In the open sea waves are normally produced by the wind. When the wind hits the sea surface it creates essentially random patterns of pressure variation and these lift some parts of the surface up and press others down. ...

3

Just to clarify the text: Integrating (5) gives (7) (the very unstable conditions) Integrating (3) gives (8) (moderately unstable to marginally unstable) Integrating (2) gives (9) (marginally to moderately stable) Integrating (6) gives (10) (very stable). Equations (1) and (4) are definitions that are used. So for example, to get (9): $$\phi_m = ... 3 The unit langley is amount of energy distributed over an area found in solar radiation, named after Samuel Pierpont Langley. The conversion is straight-forward:$$ 1\frac{\rm ly}{\rm min}=697.3\,\frac{\rm W}{\rm m^2} $$So you'll have to do some more unit conversions to get the ly/day to W/m^2 conversion. See also this site for other unit conversions ... 3 First, note that \delta(z)\neq\frac12 but$$ \int_0^h\delta(z)\,dz=\frac12 $$which is different than your assertion that \delta(z)\simeq\frac12. If you insert Equation (1) into Equation (3), you get$$ \frac{dT}{dt}+\frac{\partial}{\partial z}\left(\overline{W'T'}\right)=\beta Se^{-\beta z}+2B\delta(z) $$Now integrate this over z (using z' to ... 2 The first term on the right is \beta S^{*}e^{-\beta z} This means that the amount of light penertrating into the ocean decreases exponentially with depth z. The second term on the right is 2B^{*} \delta \left(z\right) This takes into account heat transfer processes taking place only at the surface, such as evaportion. 2 The question is simple but the reason why it probably has not been answered yet is that the answwer is extraordinarily complex. Basically if you consider a steady state with a velocity field V(x,y,z), you are asking to : find the horizontal and vertical components Vh(x,y,z) and Vv(x,y,z) integrate Vh².dV and Vv².dV all over the globe This is a matter of ... 2 Have a look at this compilation of optical properties of water. If I take for example this recent measurement of the absorption coefficient, the minimum absorption coefficient is 0.0000442 /cm at 417.5 nm (which is blue light). The next question is where you consider the cutoff for vision to be. The Wikipedia article on daylight gives the ratio between ... 1 I like your photograph, but as I'm sure you realize that is not the open ocean. (I didn't notice the painted background the first time I looked at it!) In the pictured situation, the waves are going to be highly influenced by the nearby man-made structures, the shore, and the bottom (if this area is relatively shallow.) Wave reflection, refraction, and the ... 1 Previous answers make it clear that there are many factors that could lead to temporary non-uniform distributions of deuterium atoms in the majority of "normal" hydrogen atoms in sea-water. So, if one were to pour a beaker of D_2O carefully and slowly into a beaker of normal water, the heavy water would be found concentrated at the bottom of the beaker. ... 1 Therefore, it seems that the climatic system has a length-scale. Where does it come from? Let us not forget that the weather system is a classical case of chaotic dynamics: several interacting differential equations are at work, not just Navier Stokes. Think of tides, think of seasons, think of clouds/albedo etc. But mainly it is the boundary ... 1 Part of your scale comes from the observer, the time and space averaging of the observations. You say the weather 10 meters distant is pretty similar to here, but look a bit closer, and it is not. The wind 10 meters from where you are can be quite different from where you stand. A sunny rock can be dry and hot 1 meter away from a shady moist spot with moss ... 1 Emerson & Hedges' Chemical Oceanography and the Marine Carbon Cycle (Google book link, you'll have to scroll to page 87) has a table that gives the value \beta_0=2.71\times10^{-2} for air in seawater at 20^\circC and one atmosphere. This website has a table of values at multiple temperatures from a few different published sources--the value at ... 1 Start by assessing the relative importance of each Q term. Assume some order of magnitude values for the unknowns (based on experience and/or intuition) and see how big Q_{lin} is compared to e.g. Q_s. Do that for all Q terms and hope some will be negligible so they can be discarded. My guess is that Q_{lin} and Q_e are insignificant. You are ... 1 From a physics perspective the main factor here is "pressure differential". As long as the pressure inside the fish and outside is the same, there no pressure differential and therefor no net force on the shell of the fish and there is no crushing. So the key here is proper pressure equalization as the depth changes. 1 The light intensity at a depth d is given by the Beer-Lambert law:$$ I = I_0 \exp\left(-\frac{d}{\ell}\right)  where $I_0$ is the intensity at the surface and $\ell$ is the depth at which the light intensity has fallen by 73%. The depth $\ell$ is dependant on wavelength as seawater absorbs red light more strongly than blue light. I did a quick Google ...

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