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2

Sorry for the enormous delay, I was caught with more work than I thought, then, here it is. @DanielSank, relativity is not necessary, it would help with what you said, since it would pinpoint exactly you are calling momentum in your system. My answer would be an extension of DanielSank comment. When there is the conservation of a continuous quantity, the ...


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The simplest possible answer is that in a closed system the lowest entropy state is one where the temperature is (statistically) uniform.


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As the comments to the question have stated, in real gasses ( contrasted to ideal gasses which just bounce around elastically) there exist both elastic and inelastic scatterings controlled by quantum mechanical interactions. Photons are generated leading to what we call Black Body radiation and an isolated gas volume will lose energy according to the ...


2

Yes you are right. The 10kg piston acts as a force over the area of the piston, increasing the pressure and decreasing the volume of the gas inside. When the set up is tilted, the force no longer acts on the gas, but sideways, so the pressure equalizes.


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The rotation is part of the key to the storm itself. Primarily the pressure and temperature differences are what causes these systems to take the shape and forms that they do. Once a tropical depression starts to form you can already see rotation in the moisture around the low pressure zone, even through it typically looks nothing like a hurricane. Not ...


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I answer my own question and give a good thanks to DavidPh, who has not really gave the answer, but in fact, it was impossible for him to give it. Here is "why": I'm French, so I've many fire hydrant data but from France. And when applying them to the formulas, the result was wrong... In fact, the problem is not the formula but the way we measure the ...


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Like any object moving through a fluid, a high-speed train distorts the air as it moves through it. Broadly speaking, there are three main regions of flow structure around a high-speed train: the upstream distortion, boundary layer and wake. These can be collectively referred to as the slipstream. The effects of the slipstream on a static observer (e.g. a ...


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Therefore could one say that the dust actually has a negative pressure p? In mechanics of solids, negative pressure (positive tension) means the internal forces resist expansion of the body due to external forces. If you have dust (rarified set of particles) in a syringe acting on each other with non-negligible gravitational forces, the gravitational ...


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The discrepancy is that the pressure as measured by the Pitot tube is not just the kinetic energy term of the pressure, but instead is a combinaiton of static pressure and the kinetic energy term. See if pages 16-34 of the following reference are helpful, though not metric: http://www.southsaltlakecity.com/uploads/documents/%5E_Fire_Flow_Calculations.pdf ...


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No, there is no contribution to the pressure from the gravitational attraction between the particles. To see this you need to appreciate that the pressure is an ensemble property, and look at the stress-energy tensor for a single point particle. This is: $$ T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha v^\beta \delta\left( x - x_p(t) \right) $$ where $v$ ...


5

The front of the train compresses air which can blow you away, while at the back of the train air rushes back in after the train has displaced it. This backdraft is especially troublesome in closed areas such as subways, where a train exits a small tunnel near a platform and the displaced air rushes back into the vacated tunnel. Next time you see a big truck ...


2

First and foremost, a primer over what "buoyancy" is is needed. Pressure decreases with altitude. The atmospheric pressure at the top of the balloon is a tiny bit less than the atmospheric pressure at the bottom of the balloon. This pressure difference results in a tiny net upward force on the balloon. The balloon rises if this tiny net upward force exceeds ...


3

Any kind of "funnel" - an area where tall buildings create an obstacle to the free flow of air - acts as an amplifier to wind. That is, even a little bit of air moving from point A to point B will notice the "obstacle" that is a pair of buildings; it will build up pressure in front of the buildings and result in a faster flow of air through the passage. An ...


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The relative humidity of air is pressure dependent. Your method of popping your ears involves increasing the pressure of the air in your mouth. And if you sufficiently compress a volume of air that has a high relative humidity, you can increase the air's relative humidity beyond it's saturation limit, which causes the water vapor in the air to start to ...


0

Higher temperature will make the molecules jitter more and they tend to separate, melt at sufficiently high temperature. Increasing the pressure will counteract that separation, hence the temperature at which the material melts should increase as the pressure goes up.


1

All first order phase transitions have a change of volume. With different pressures you need to consider the sign of the work $P\Delta V$ that needs to occur during the phase change. If $\Delta V$ is positive, the phase change will occur at a higher temperature for higher pressure. If negative, the phase change will occur at a lower temperature. (Note ...


1

I know that air pressure and temperature are inversely proportional. You should not know that. This is the source of your misunderstanding. The ideal gas law, $PV=nRT$, can be rewritten as $P=\frac R m \rho T$, where $m$ is the average mass of a molecule in the gas and $\rho$ is the density of the gas. The first term on the right is a constant for a ...


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If you stick your finger in the water, the weight of the finger is still supported by you and so the weight of the beaker does not go up (it goes down a tiny tiny amount because the water rises and the height of the air above it goes down insignificantly.) If you cut off your finger, the weight goes up by one finger because you are no longer supporting it. ...


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The jump is most likely due to several factors. The main one would probably be similar to what caused the storm in the first place, a high pressure low pressure gradient. Lifting such a cover would require at least Δp=995 Pa pressure difference between sidewalk level and the sewer. Someone already derived 995 pascals would be enough to lift a manhole ...


1

Well this happens mainly due to two factors- The pressure difference inside the drain and outside. there is high pressure inside the drain which is balanced by lifting the lid. The excessive increase in water level in the drain. These two factors result in the water gushing out after the air from the air gaps come out. This may also happen due to ...


3

This phenomenon has been well-researched. If the underground pipe is almost full with rapidly-flowing water the waves in the pipe will cause the pipe to be completely filled in some places. The waves are irregular and the air in between the wave crests gets compressed - when it passes the manhole access the air is vented. 1000Pa is actually a very small ...


2

The jumping manhole lid caused by the Bernoulli`s principle sounds quite reasonable at the first glance. However, I have seen a few storms in my life, but no jumping manhole covers that I can recall. Also, I noticed, that in the video the jumping does not match the wind gusts (observe that small tree). In the absence of trolling sewer workers, the lid ...


1

There's static pressure (P) and dynamic pressure ($\frac{1}{2} \rho v^2$). A pitot tube placed in the center section of your drawing will measure the stagnation pressure, which is the sum of the two terms. Physically, the dynamic pressure will be larger at the narrow section, and the static pressure will be smaller there. The sum will be constant ...


2

As mentioned in the other answer, additional air is being entrained. This principle can be used to produce considerable flows and considerable pressure differences. Examples In a steam locomotive, the exhaust steam from the cylinders is directed to a nozzle that points upwards towards the exit of the funnel. This draws the smoke through the boiler and ...


10

Very interesting observation. I think what you are observing when your mouth is away from the tube is MORE than just the air you exhale - because the air in the vicinity of your breath is being "dragged along" by something called "entrainment". This is the principle behind an ingenious fan, called the Dyson Air Multiplier: A detailed explanation of how it ...


2

I think there is a difference between the amount of pressure exerted on a flat surface by the fluid in a wind tunnel versus the amount of pressure that makes down your ear canal to your ear drum. The wind in a wind tunnel is not directly incident on your ear drum. I imagine if one could directly apply a 15 m/s flow onto the ear drum, it would hurt. ...


0

But what if the pressure in the balloon increases? Doesn't it make sense that the balloon would want to expand? That is, that as pressure increases, volume increases. This seems to contradict Boyle's Law. In simple words: If you increase the pressure in the balloon and let it expand, then the pressure in the balloon is not really increasing, as you are ...


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The two scenarios you mentioned both are correct, “the pressure $p$ has different sign from other generalized force, if we increase the pressure, the volume increases, whereas if we increase the force, $Y$, for all other cases, the extensive variable, $x$, decreases”.[1] There is no conflict between the two scenarios. [1] L.E.Reichl, A Modern Course in ...


1

Why can't we measure the pressure at 1000 m for different temperatures? Meteorologists certainly can do that, and in fact do do that, all the time. They use weather balloons, sounding rockets, and all other kinds of instrumentation to measure conditions in the atmosphere. The resulting picture is rather complex. Conditions vary with place, the seasons, ...


1

For aviation purposes, standard atmosphere is considered to be dry air at mean sea level, at 15 degrees C (59F). It is true that pressure decreases with increasing altitude, and temperature usually does, but not always. It is not a simple relationship, because it depends on humidity, heat transfer from above and below, vertical circulation, horizontal ...


0

For filling balloons - whether for parties or research - it's not so important to control pressure. In fact having a regulator in line would only slow down the process. There are compatible gas regulators for nearly every gas including He and H. Some require special materials - such as Viton O rings for oxygen compatibility. Regulators are used more often ...


1

Usually it's not quite true that there is no regulator - typically these types of cylinders have a pressure reduction valve that doesn't look like much. There's nothing to "regulate" but they reduce the pressure from 100+ bar to something that makes sense for filling balloons. A similar thing is used in scuba diving: the "primary regulator" takes the ...


1

One could probably measure the light pressure of a photon gas on a pair of parallel mirrors. Experimentally that's on the order of 1e20+ particles in a relatively small setup with two dielectric mirrors and a laser. The resulting force should be around 1e-7N, which is easily measurable. Would you accept that as an implementation, or are you looking for an ...


2

When the cap bursts off the bottle the air inside it will expand rapidly and adiabatically, so its temperature will fall. If there is enough water vapour in the air inside the bottle, and if the temperature reduction takes the temperature below the dew point, the water vapour will condense giving the fine mist that you see. In this case it looks to me (it's ...


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I can only assume that 25 cm of vacuum refers to a vacuum which is at 25 cmHg of pressure. I do think going down to 25 cmHg would mean ~33,330 Pa; that's within the range of a low vacuum.


0

While it could represent the pressure of that much water, my first thought is that it is a pressure difference equal to 25 cm of Hg, or 250mmHg. 25cm of water would be a quite weak vacuum. Since it's a vacuum, the internal pressure would be ambient minus that amount. $P_{abs} = 1013mbar + (-333mbar) = 680mbar$



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