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28

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 ...


3

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 ...


3

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 ...


2

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. ...


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 ...


2

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 ...


2

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 ...


2

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


2

I am sorry to sound discouraging, but you do not seem to be familiar with some of the common sense of physics, namely, the conservation of energy. The conservation of energy states that among a closed group of systems, (which you can understand as a room that is shut in every way so nothing can get in or out) energy is conserved. If you try to put water as ...


2

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: ...


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 ...


1

To follow up on user289661's excellent answer. You can think of water as burnt Hydrogen. The Hydrogen fuel has already burnt to create water some time in the distant past - releasing energy. You can split water back into hydrogen and oxygen and burn it again - but that would obviously take the same energy to split it as you get burning it, less any ...


1

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 ...


1

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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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 ...


1

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 ...


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 ...


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

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 ...



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