# Tag Info

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It's all density. Gravity should start to immediately pull it down. The density of the cloud nor atmosphere can hold it up. More than up draft would be needed

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Water vapor pressure is a function of temperature (and pressure as well). It increases with the temperature. When vapor is produced on the heat surface, bubble forms. The size of the bubble is determined by the buoyancy and the water pressure at the location. When the size is large enough, the buoyancy force dominant. The bubble will not stay but rise. Due ...

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Yes, the solution you posted in your last comment is good. With the sole criticism, that it is a little too precise. If the density of steel has only two digits, then the result should not have 4 as in 10.37kN But this is very advanced criticism :) Also minor points, to be even more precise: for the circumference volume of the mantle you could equally ...

5

Bubbles are formed when the pressure in the fluid is less than the saturated vapor pressure of the liquid at that temperature, modified by surface tension effects. Surface tension actually increases the pressure inside a gas bubble in water; the approximate increase in pressure is $\Delta P = \frac{2\sigma}{r}$ (see for example this earlier answer). The ...

1

At the equator the effect on the local value of $g$ due to the rotation of the Earth and the non-spherical shape is only a few parts in a thousand. So even planting a tree at the top of a high mountain at the Equator probably will not have much of an effect on its ultimate height. I found this Scientific American article interesting although it did not ...

1

If you can't read a phase diagram, a good set of steam tables can answer your question. BTW, water can only exist as a vapor and a liquid in equilibrium when conditions place it below the saturation line. For a pressure of 15 bar (1.5 MPa) absolute, the saturation temperature is 198.3 C. At this pressure, only superheated vapor exists at temperatures ...

1

The real answer is quite complex; I think we should break it into a couple of different pieces. First - the static case. If you submerge an open pipe into water, the pressure inside and outside will be the same at a given height, and the water level inside the tube will settle at the same height as outside. If you add the effect of surface tension, it is ...

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Before valve-A is opened, the pressure in the tube is atm. Pressure since the top of the tube is opened. When the valve-A is opened, assumed instantaneously, the water will rush into the tube at: $$V = \sqrt{2gh}$$ where: $g$ = gravitation const (32.2 ft per sec per sec.), $h = 100\ \mathrm m$ = Height from bottom of tube to tank ...

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First, please avoid spamming comments, this will not help at all to increase your response rate. Secondly, Leon's answer is not completely accurate. For floating bodies, you have 3 options. The object is more dense than the liquid -> It sinks (to the bottom) The object is less dense than the liquid -> It floats The object has the same density than the ...

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The lid will float. The shape of the box does not matter. If the lid has a true perfect fit water will not be able to move from under the lid. so as long as the lid is placed in straight and level it will float.

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A possible explanation is simply that the air and water have not equilibrated due to the high heat capacity of water. It's the same reason you can jump in a lake in the early summer and it still be cold. So if you filled up a glass from the sink and measured it 15 minutes later, it's entirely possible that it would be cooler. A more detailed description of ...

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This could be explained by evaporative cooling. That would match all of the (extremely scant) information you have provided about your experimental setup and results. If you would like a more thorough analysis of possible causes, you are going to need to write at least a couple sentences (preferably even more than that!) about your experimental setup. ...

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There is a pressure differential. When the falling water hits the partial obstruction in the middle of the straw, the pressure around the edges of the straw increases. This increased pressure accelerates the water into the smaller passage below (and also tries to decelerate the water coming from above). This effect is not consistent over the period in ...

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The formula to use for the nozzle velocity is: V = (gpm * .321) / A where: gpm = gallons per minute V = Velocity in ft/sec A = Area of nozzle in sq. inches. A 2mm nozzle = 0.0787 inches. Area A = .785(0.0787^2) = 0.00486 in^2 Flow of 60 L/M = 15.7 gpm. Nozzle velocity V = (gpm * .321) / A ...

2

Drying happens when dry air comes into contact with a wet surface. There are a lot of variables in that. The dryness of the air, the dampness of the material, different temperatures and sources of heat, currents in the air, flows in the water... Nothing is more unlikely than "all things being equal" in the instances you have observed. Anyway, let's imagine ...

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What is important to understand here is that the system is always going to try to attain equilibrium. So if you're temperature surroundings are the same as before , your tank will still attain the same temperature as it did before. However it is necessary to mention the time taken to reach this temperature or at least whether the temperature varies very very ...

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If you are using a good temperature controller it should hold that temperature at the set point almost regardless of external atmospheric temperature. As the external temperature rises it would use less power to maintain the fixed set point. As external temperature fall it will compensate by adding more power to match the heat loss from the tank.

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According to what I know, if the temperature of the room is increasing that is, heat is being input into the system, the water will gain heat from the air regardless of the temperature difference between the water and the Room. according to the equation H=M x c(Specific Heat capacity) x change in temp. However due to the difference in the final temperatures ...

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Your understanding of the degradation mechanism is correct. However, you can compensate for this to some degree by training your pupils to contract underwater, thereby reducing the refractive variation of light transmitted to your retina. This famous study by Anna Geslen describes this in detail, and also showed that the remarkable underwater visual acuity ...

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Even with a constant flow of 60 litres per min you need to provide at least the nozzle entry area(one other than the 2mm one) or the nozzle inlet velocity. Otherwise you will be left with one unkown in the Bernoulli's equation. If it is a constant area nozzle then the inlet and the outlet velocity will be the same because of the continuity equation(assuming ...

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It's rarer than that, more like 1 in 6000. Because each molecule of water has two atoms of hydrogen, then about every 2 in 6000 (1 in 3000) has a single atom of deuterium (DHO). And would be closer to 1 in every 6000$^2$ for a molecule of D2O. The linked question Deuterium density in seawater gives sources that show deuterium is well-mixed in the ocean. ...

1

Let's ignore the planet in the middle for now and just consider a big drop of water. The mass of water is proportional to the radius cubed: $$M = \tfrac{4}{3} \pi r^3 \rho$$ and the surface gravitational acceleration is given by: $$a = \frac{GM}{r^2}$$ so substituting for $M$ we get: $$a = G\tfrac{4}{3} \pi \rho r$$ So the surface gravity ...

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It's irreversible. The reason can be easily understood when you look the molecular properties of water The presence of a charge on each of these atoms gives each water molecule a net dipole moment. Electrical attraction between water molecules due to this dipole pulls individual molecules closer together, making it more difficult to separate the ...

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Let x be the instantaneous mass fraction of original water remaining in the lake at time t, w be the mass recharge/discharge rate, and M be the total mass of water in the lake (assumed constant). Also assume that the water in the lake is well-mixed. Then, for a mass balance on the original water, we have: ...

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Note that the instructions do not claim that the volume of water is changing measurably over the course of the procedure. The volume of water is not changing noticeably. What this procedure allows one to observe is the fact that a salt+water solution has less volume than the total volume of its ingredients. Rather than a demonstration of water changing ...

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Very interesting experiment you did! 1. The effect of surface tension due to the water/rubber and rubber/air interfaces is negligible. For a radius of $R = 10$ cm, and taking a typical surface tension of $\gamma = 70$ mN/m, the increase of pressure in the balloon because of the interfaces is typiclally of the order of the Laplace pressure \$\Delta p = 2 ...

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Water conducts heat away from the body 25 times faster than air. Probably most liquids with a comparable density would fell the same.

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Two things to consider: You do not have temperature sensors sticking out into the environment. What you feel is not the temperature of the air or water; it is the temperature of the cells around the spot(s) where the temperature sensors are buried in the skin. That's why your face can "see" a hot stove element. The radiant energy from the hot metal ...

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Firstly, to make a valid comparison between how water and air 'feels' on your skin, two conditions would need to be met: Both water and air would have to be at exactly the same temperature. That temperature would have to be lower than human body temperature (strictly speaking skin temperature). If those conditions are met then water would certainly feel ...

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Water couldnt stay in liquid form, because without gravity there wouldnt be pressure, and a liquid needs pressure to stay in liquid form

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