# Tag Info

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For most substances, increasing the pressure when a system is in equilibrium between liquid and solid phases will increase the phase transition temperature. Water is one of a few special substances for which the pressure lowers the temperature of transition. The basic reason is that water actually expands when it goes from the liquid to solid phase. In ...

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Can throw out the ideal gas law as it is not a gas You do work on the system and from conservation of energy that work is in the system. It will result in a temperature increase. You calculate it based on the work performed on the system. Speed of the work is not a factor. Assume the walls are insulated (adiabatic). Since a liquid is very ...

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Acid Jazz is quite right. Generally any instability in air medium can result in sound waves radiation. Vortices causes local pressure instabilities and the medium is "able to propagate the image of it" (well, the Wave equation). Kármán vortex street is nice (and most of all solvable!) example of vortex shedding. But in certain domain of the Reynolds number ...

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I like BowlofRed's answer, but I think there's more going on here than that. I also don't believe that straight forward cooling would suck in the egg that fast. The egg gets sucked in very quickly. Certainly as the burning starts, air is escaping. It's impossible to see precisely when the egg stops vibrating, as some vibration might be quite small and ...

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the air [pressure] is lowered by heating the air inside the bottle. I think you need to watch the video again. The air pressure is raised by heating the air, not lowered. This increase in pressure forces air out of the bottle, which is visible by the egg vibrating on the rim. The egg acts like a one-way valve. When the pressure inside is greater, it ...

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When the thermos is closed, the air at the top of the thermos will be warm relative to the ice-cold water below it. As time proceeds, the two materials that are in contact with each other at the surface of the water will exchange heat due to their different temperatures. This will cool the air. When gasses are cooled they move more slowly and they don't ...

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As a simplification, you can consider that you have a 2D viscous flow between two boundaries that approach each other. Assuming that the flow is symmetrical about the line (with the line along the Y direction), you can simplify this further to "no flow at x=0". What you are left with is a pressure distribution $p(x,t)$ whose integral in $x$ should equal the ...

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There is a continuous pressure gradient vertically along a cup or tube like shown. The pressure gradient means more force on the matchstick on the bottom side then on the top, which has a net-upward force. For a light object, this net upward force isn't compensated by the object's own weight, so it rises. The experimenter does work when he lifts the water ...

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A force does not require a constant input of energy to exist. Energy is only required to perform work, which is exerting a force over a distance. $$W = \mathbf F \centerdot \Delta \mathbf x$$ That distance is key. In your example, if the size of the container does not change, no energy is expended no matter how long the force lasts. If the force is used ...

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Seems to me that the "meat" of your question is not about Bernoulli's principle, but rather a related concept called "conservation of mass" . I'll address that first, and get to Bernoulli's principle later. Conservation of mass The idea you are looking for is Conservation of Mass. If you assume the fluid doesn't have any "gaps" or spaces (and real fluids ...

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Short answer: The major difficulty lies with the definitions themselves, and none of the possibilities given has a real physical meaning which can be univoquely related to stress in non extensive systems in its conventional mechanical original meaning. The long one: What kind of systems does this apply to? This is not answered by referring to systems with ...

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First you half fill the pipette with hot water. So at that point it's filled half with hot water in the lower part, half filled with air at room temperature in the upper part. Then you block the top end of the pipette with the tip of your finger and turn the pipette upside down, so the hot water starts flowing down to the blocked side. However, the presence ...

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Rubber is, to some degree, porous but it's more likely that most air is escaping at the bead of the tyre, where it's in contact with the wheel rim. There will be some relative movement between the tyre and rim as the two rotate, the tyre deforms, and the seal is compromised. That would explain why the stored tyre doesn't lose so much air (anyone who has ...

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You don't want to use compression ignition in gasoline engines. If you attempt to compress air and gasoline vapors together to the point where they ignite, the ignition point will not be "controllable". Ideally, the ignition should occur 10-15 degrees before "top dead center", meaning that the ignition starts when the crank shaft is "under rotated" 10-15 ...

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When the piston compresses the fuel air mixture, it heats up due to high pressure until in some point it combusts the mixture with the spark plug. What if engines dont have spark plugs, the piston will keep compressing the mixture until it gets very hot to combust itself, as I think it will do, will that really happen? If a gasoline fueled car could ...

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One is an Otto cycle (with spark plugs, spark ignition), the other one a Diesel cycle (compression ignition). The one producing less green house gas emissions and particles for the same vehicle is probably better.

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Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube. The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube. Supporting means to keep from falling. (Forget the air pressure - that's ...

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"Perceived" is the operative word here. Perception is subjective and pressure is hard to perceive objectively without a measuring device. Hydrostatic pressure experienced by a body submersed to a height $h$ under water is given by: $p=p_0+\rho gh$, with $p_0$ atmospheric pressure, $\rho$ the density of the water and $g$ the gravitational acceleration ...

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I think it is partially a perceptional problem. The pressure, i.e. force per area is depdendend on the depth of water in this case, and it doesn't matter much, if you are naked or if you wear waders. But if you wear waders, the pressure is implied on the surface of your clothes, not on your skin. Your clothes will now dent to give in to the pressure, until ...

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If you remove the bottom you no longer have a container. The pressure is atmospheric and you just have gravity at work.

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The simple answer, which is what I think you're hoping for is the following: At constant volume, the system (by definition) is not able to do work on the surroundings because work involves a change in volume. All the heat you put in is spent raising the temperature (internal energy). At constant pressure, some of the energy you put in goes into raising the ...

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If you add mass keeping the volume constant, the pressure would change. When a source says that pressure is independent of mass, it means that pressure is independent of mass if you keep the density constant, not the volume. The idea is that if you take two identical copies of a system, then combine them to make a system twice as large, extensive ...

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First law of thermodynamics $$\delta Q=\delta E+p\delta V$$ $$\therefore\frac {\delta Q}{\delta T} = \frac{\delta E}{\delta T}+nR$$ As $p\delta V$ can be written as $nR\delta T$ When volume is constant $$\delta Q= \delta E$$ $$\therefore\frac {\delta Q}{\delta T} = \frac{\delta E}{\delta T}$$ $\frac {\delta Q}{n\delta T}$ is Molar heat capacity. ...

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Well, the temperature profile of an object on the macroscopic scale is governed by, $${{dT} \over {dt}}=\kappa \cdot \nabla^2 T$$ Where T is the temperature, t is time, and $\kappa$ is the thermal diffusivity. It's the constant that matters here. The larger the constant, the faster the temperature changes. The equation itself is called the Heat Equation. ...

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Yes space is a vacuum which means no pressure and air and no gravity. As you said ice melts quickly in high pressure so in space ice will take a longer time to melt if it's not so hot there.

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Only the symmetric stress tensor is physical, since thermodynamics demands a symmetric stress tensor. Since the symmetric stress tensor is unique, your option 2 is the correct one. (Canonical versions may be simpler but need not be physical; cf. the canonical momentum, which is often different from the physical momenetum.) This is completely unrelated to ...

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Ok, here's one possible explanation. First, if you're graphing pressure vs payload, you should switch your axes--I was confused! Anyway, the payload obviously is going to increase the pressure in your tires. However, your tires are also elastic. So, they can "stretch" a little, to increase their volume in response to a change in pressure. However, most ...

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@LukeAllen, you don't just need to increase the pressure in your bottle. If you increase the pressure as stated in some of the above responses, you will not be increasing the PARTIAL PRESSURE of carbon dioxide, so you will still have a difficult time dissolving carbon dioxide into the water in your bottle. I have done exactly what you are trying to do. To ...

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are you filling the bottles with CO2? or regular air. If your bottles are air tight, you could add a small bit of dry ice to each bottle then seal it tight and then refrigerate it, as CO2 absorbs better in colder water. If you plan on drinking this, I'd do more research cause I'm not sure dry ice is safe for consumption - so check that out if this is a ...

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You could just apply external heat with the cap sealed tight

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You can use electrolysis (two wires from a battery through a cork in the bottle - water will dissociate into oxygen and hydrogen, creating high pressure). However, beware: this is dangerous (as it is almost always the case with high pressure), as the bottle can explode and harm you or other people. Please be reasonable. EDIT: And high pressure is not the ...

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The cheapest way will probably be to submerge the bottle. Approximately for each 10 meters of water the pressure will increase 1 atmosphere. So your 4 atm will be reached at about 30 meters below the surface. Just go to the sea, a deep river or even a big swimming pool, drag your bottle, tie a long cable to it and let it sink.

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This problem contains a functional form that is very difficult to work with. A numeric integration is one approach that will give an approximate (but good) answer. To do this, the following procedure can be used: 1) Start at pressure P1, and establish a "small" value for dP 2) Use a trial-and-error method to calculate V 3) Multiply V by dP and keep ...

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When high pressure is applying on a gas, it gets converted into liquid form. Similarly, if more pressure is applied to the liquid, force of attraction increases so that the liquid is converts into solid state. As the pressure increases the rate of crystallization also increases. i.e., the freezing point also increases.

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The variable $h$ in this equation always refers to a vertical distance from the point where you want to know the pressure up to the fluid surface in contact with the air. In the tilted cup you would still measure straight up to the fluid surface. The bit of tea in the middle of the cup doesn't "know" anything about the walls of the cup. Their orientation ...

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The pressure in the water at a certain level ($z$-coordinate) is everywhere the same. This implies that the pressure at the water surface outside the beaker, i.e. the atmospheric pressure $p_0$ has to be equal to the sum of the pressure created by the water column $\rho g h$ and the air pressure in the beaker $p_b$. $\rho$ is the density of the water, $g$ ...

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Start off with beaker filled with water and completely submersed in the bath, with the beaker upside down (as in your schematic). Now, via some conduit, you introduce 10 ml of air into the beaker. Then move the beaker upwards until the water levels inside and outside of the beaker coincide exactly. The pressure of the air inside the beaker is now exactly the ...

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Here is an interesting article that shows how water was frozen at high temperature under pressure http://www.azom.com/news.aspx?newsID=8016 Here is an extract Sandia Convert Water to Ice in Nanoseconds Published on March 19, 2007 at 1:15 AM Sandia’s huge Z machine, which generates temperatures hotter than the sun, has turned water to ice in nanoseconds. ...

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By the way, it's customary to use primes or some other indicator inside the integral to be slightly less confusing. I.e. $$W(t) = \int_{t_0}^{t} P(t')Q(t')dt'$$ Be careful about how you think about Q(t). You've described it as a "flow rate", but really, you are saying $Q(t) = \frac{dV}{dt}$, or the rate of change of the volume with time. Now, the term ...

1

The other answers already mention pressure and heat. A bomb sets nearby bodies in motion with a speed depending on the strength of the explosion, the distance to the body, and how much surface area of the body was facing the bomb. While - as explained in the other answers - being set in motion is rarely lethal, being smashed against a wall can easily lead ...

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The blast overpressure of the explosion is a very strong shock wave which can kill humans. There are a number of ways an explosion without shrapnel can do harm to people: Rupturing of the hollow organs due to rapid compression and expansion by the shock wave. The body can get thrown through the air if a strong detonation occurs nearby. Impact of the body ...

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This is a fun question and has about 4 or 5 different factors at play: Types of forces involved: Pressure vs. Inertia Types of resistance involved: Rigidity vs. Plasticity Objects involved: Soldiers vs. Buildings Scenario: Damage from a bomb blast (energy wave) vs. shrapnel impact/penetration (kinetic collision). How damage is applied with: Blast - the ...

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Blast can definitely kill you, although it is only lethal at much shorter ranges compared to shrapnel. A building can be destroyed by 5psi overpressure while a Human can withstand up to 45psi and live. Some data here: A 5 psi blast overpressure will rupture eardrums in about 1% of subjects, and a 45 psi overpressure will cause eardrum rupture in ...

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I think this picture covers the gist of it. You need rubber or plastic, a mailable material that flattens when pressed against the object, but that wants to return to it's original shape when released. Think of, as part of measuring the weight of an object, you measure the pressure all around it. If you have more air pressure on the bottom than the ...

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Suction cups work by having lower air pressure inside of the cup and more on the outside. The strength of a suction cup is dependent on both the difference in air pressure between the outside and inside of the cup and the area the cup covers, if the force applied to lift an object exceeds this amount then the cup will come free.

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Force due to pressure over a given area is $F=AP$. A bubble will have three forces to consider. 1. Air pressure from air inside the bubble pushing the wall of the bubble outward (trying to make the bubble bigger) 2. Air pressure from air outside the bubble pushing the wall of the bubble inward (trying to make the bubble smaller), and 3. a surface tension of ...

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The boiling water will boil at a constant temperature. As long as the can is in the boiling water, you will not need to worry about the temperature of the stove. If this experiment is run at sea level, and you are using pure water, the can will remain at 100 deg C throughout the experiment. If you are at an elevation higher than sea level, and you want to ...

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Measure first, think later. The best tunnel for experiencing ear discomfort around here is a train tunnel under a canal, the Drontermeertunnel. The graph shows the air pressure recordings I made in the train. A and B are the gates of the tunnel. Orange indicates when I experienced ear discomfort (although it wasn't easy to decide when discomfort started and ...

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The idea gas law work; more specifically Boyle's law is a reciprocal law. Pressure and volume are inversely related. When you increase volume by a factor of 1.0666 (by 6.66%), you decrease pressure by $\frac{1}{1.0666}$. Pressure is inversely proportional to volume. $P \propto \frac{1}{V}$ So when you increase pressure by a certain factor, volume ...

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OK the first thing to notice is that in a gravitational field the pressure of a gas is not constant, but decreases with altitude. This means simply asking "what happens when we change the volume of the gas?" is not a well defined question; the amount of work done in the expansion is going to depend on exactly how we change the shape of the container. We ...

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