# Tag Info

31

It is the pressure gradient that is proportional to the local gravitational force. When that force is integrated over a distance, the pressure gradient is integrated to accumulate a total pressure. The maximum occurs at the point towards which gravity is directed in a spherical mass, which is the center. True, gravity at that point is zero, but it and ...

12

The previous answers do a fine job already. But I'll try to add a simple thought experiment. Imagine three objects floating in space, clumping together by gravity: ### ### #A#|c|#B# ### ### Mass A, the negligible mass c and mass B, equal to A. The center is attracted to A and B and their gravity cancels out. However, A presses against c, because it is ...

9

Pressure at the center of the earth is non-zero. You're correct that there's no gravitational force at the center of the earth, but that doesn't mean pressure is zero - the pressure comes from the many miles of rock sitting above the center of the earth. As an analogy, think of a balloon. The pressure inside is higher than ambient because the elastic skin of ...

6

Edit: 1 May 2021 The following procedure uses the less accurate method from page 455 onward from the scanned sections of the book pictured below, from the original answer. The procedure is a multistage process ideally suited to either a spreadsheet or programming code. The equations use SI units. 1. Calculate the saturated vapor pressure at the dry bulb ...

5

So if you have a mass-mixing ratio, you effectively $\frac{ \text{kg pollutant}}{\text{kg dry air}}$. PPBV is parts per billion volume, or number of molecules of pollutant per billion molecules of dry air. Since not all particles weigh the same, you must use a conversion factor. Consider molar mass, which is the ratio of moles (a unit describing the number ...

5

The question may generate primarily opinion based answers but while it is open one would like to present the following information. At the outset most of the people I know in the field work in NWP and the reason people avoid higher mathematics(by that I mean looking for analytical solutions or pure math based projects such as you mention topology or ...

4

No, there isn't, because the area affected depends on so many factors, some of them unknowable. Magnitude, cause, depth, geology (which may be very variable in different parts of the area) etc. Besides which, the easiest way to tell which areas are affected is to monitor reports from observers. In the final analysis, the data that aid agencies want to know ...

4

Math is very important for hydrology. Especially for surface water problems, you require understanding of differential equations while open-water hydraulics like backwater calculations are complex. If you need a great deal of detail, these problems cannot be solved analytical, only numerical. So I recommend a good mathematical basis to study surface water ...

3

Imagine the whole ball being separated into a handful of concentric shells, with the outermost shell being the crust with the surface, and the lower shells ever deeper, hotter, and ghastly regions of the internals of our planet. Now, just for the thought experiment, forget that each individual shell has "stuff" going on in itself, and just imagine ...

3

As a very rough approximation, one could start with equation (9) from Arbab (2009), https://arxiv.org/abs/physics/0304093 to get the effective number of days per year: $$T_{\text{eff.}} = T_0 \left(\frac{t_0-t}{t_0} \right)^{-2.6}$$ with $t$ the time difference between now and then $T_0 = 365.25$ the current number of days per year $t_0=(13.799 \pm 0.... 3$\text{Poulation density} = \frac{\text{Population}}{\text{Surface}}$Re-arranging$\text{Surface} = \frac{\text{Population}}{\text{Poulation density}}$In your case$\text{Surface} = \frac{100,000,000\, \text{inhabitants}}{10\, \text{inhabitants}\, \text{km}^{-2}}=10,000,000\, \text{km}^2$So something doesn't make sense, because those numbers ... 2 Generalizing such relationships is, as suggested above, hard to impossible - but for well studied and understood systems such correlations have been observed. For instance there is a clear, and almost linear, relationship between earthquake magnitude and displacement on Mt Etna. Earlier work with a global focus has observed similar relationships - but with, ... 2 I suspect that this diagram has nothing to do with polar vortices in the meteorological sense. It’s hard to prove a negative, but digging into the source of this image led me in some very non-meteorological directions. The file itself is entitled ‘Growth measure and vortices’ on Wikimedia (note absence of ‘polar’), with a source of ‘own work’ (by the user ‘... 2 This would actually be an interesting research project. The ability to do a 3-D scan and model of a rock's shape in a manner that is fast enough to get a statistically meaningful number of rocks measured is relatively new. Certainly you can easily name things that probably control the shape of rocks (lithology, mode of erosion, mode of transport, history ... 2 Any area south of the northern polar circle will have one peak per day,and the same goes for the southern polar circle any area north of this will have one peak per day,the polar circles is at 66,33 north/south. The areas whitin the polar circles will have only one peak per year but it will last for half a year. A solar panel will only produce significant ... 1 There is no such simple relationship between rainfall and runoff (I wish there was!). How much of the rainfall ends up as runoff depends upon at least the following variables: temperature, humidity, wind-run, hours per day of bright sunshine, latitude, altitude, type and density of vegetation cover, soil type, thickness and consistency, antecedent ... 1 When we say there was 1 mm (or 0.001 m) of rain, it is assumed that it is on a 1 square meter area, so it is equivalent to a volume of water of 0.001$\small\mathsf{m^3}$(or one liter). So first you have to divide your data by 1000. Then you have time, which is easy: there are 3600 seconds in one hour. So to convert L/h in m$\small\mathsf{^3}\$/s, you have ...

1

Yes, using GMPES. E.g., you can look at the USGS shakemap which includes some measure of intensities such as PGA/PGV. Just choose an appropriate cut-off value and you have your area.

1

In addition to other information others have written in comments, gradients measure the rate of change of "a quantity". For example, take a hill. As you walk up the hill your elevation increases relative to the base of the hill. The steeper the hill the more quickly your elevation changes. The slope of the hill is defined as the gradient of the hill. The ...

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