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0

In a simple way the pressure will decrease because there's less air above you, so there's less mass pushing you.


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Your reasoning is indeed rubbish, and potential energy has absolutely nothing to do with stress-energy or pressure. The following reference is a good source about the origin of the pressure term in the stress-energy tensor: "Momentum due to pressure: A simple model" by Kannan Jagannathan in American Journal of Physics 77, 432 (2009); ...


0

If the balloon was completely isolated from the outside atmosphere and contained a certain gas (in this case helium), then the helium will exert a pressure outwards, attempting to expand the balloon. The atmosphere outside will also exert a pressure inwards, attempting to keep the balloon in a small shape. These two forces compete, and the balloon will ...


0

Well, if I'm not mistaken, it's pretty straightforward. Let $p(r, \theta, t)$ be separated in two functions with variables of time $T$ and spatial variables $\Theta$ (I'm not using $R$, cause it's already defined): $$ p(r,\theta,t)=\Theta(r,\theta)T(t) $$ then: $$ T = e^{i\omega t} $$ $$ \Theta = i\frac{Q\rho c k}{4\pi R}e^{-ikr} $$ $T$ is given ...


1

How would you solve this for a single small hole? What happens if you now move that hole down by a small amount? What if you add up the contributions of all these holes? Congratulations, you just integrated the expression for the flow rate over the aperture.


4

The pressure at any point is the total weight of everything above an area of a square metre. So for example the atmospheric pressure at Earth's surface, 101325 Pa, means the total weight of atmosphere above a square metre at the surface is 101325 N. So when you ask for the pressure at the centre of the Earth, the way to calculate this is to work out the ...


0

In reviewing the answer supplied by troy, The explanation of his increased water pressure on the larger pipe is as follows: The pressure from the tank is based on the height of the tank. A tank on a 25' tower will supply at least 12.5 pounds per square inch. (we don't know the height of the surface of the water.) The 3/4 inch pipe has an area of .44 sq in. ...


1

If the temperature of the gas is kept constant during the compression then the bulk modulus of an ideal gas is just equal to the pressure. The definition of the bulk modulus is: $$ K = -V\frac{dP}{dV} \tag{1} $$ For an ideal gas $PV = RT$, so $P = RT/V$. If the temperature is constant this gives: $$ \frac{dP}{dV} = -\frac{RT}{V^2} \tag{2} $$ and ...


1

Take an empirical approach; go to a oil-change place. Put the car up on a rack/lift (the kind that supports the frame, not the tires - allows tire rotation during oil change). Have them adjust tire pressure while car is up in the air; have them remeasure when the car is back on the ground (tell them why - give them "scientist for a day" certificate). Because ...


0

I would say that the question is still not very well defined, as it is important if in the high-temperature (classical) limit spin is conserved or not. If spin is conserved (think strong magnetic fields $B\gg T$), then in the classical limit spinful quantum gases become a mixture of classical gases, as coherent superpositions between spin states are not ...


1

Static pressure in a compressible flow depends on the density but not the speed (not directly). Speed and geometry may affect the density. For isentropic flow (neglecting gravitational potential): $$ {p \over \rho^\gamma} = constant, \gamma = {c_p \over c_v} $$ which could be turned into this: $$ {p \over p_0} = ({1 \over 1+{(\gamma-1) \over ...


0

In order for water to boil, it must become a vapor. Bubbles rising from boiling water contain water molecules that have enough kinetic energy to separate themselves from the less-energetic liquid molecules that remain in the liquid state. It's easier for vapor to form in a low pressure environment because there is less total kinetic energy outside the ...


1

The required change in pressure for a change in melting point can be found from the phase diagram of water. The typical variations in atmospheric pressure are negligible, just as you neglect the additional water pressure experienced in the lower parts of the ice bath. I will leave it up to you to find the factor involved - it is the slope of the water/ice ...


0

You have to find out what kind of statistical ensemble you are dealing with. As soon as you know that, you can get the corresponding thermodynamical potential from the knowledge of the partition function. When you know the potential, you know everything! EDIT: Since you don't know to which ensemble this partition function $\mathcal{Z}$ belongs to you ...


0

If you raise 1.2 liter of water (mass 1.2 kg) by 55 cm (h) in 3 minutes (t) against a gravitational acceleration $g$ of 9.81 m/s2, the average power needed is $$\frac{m\cdot g\cdot h}{t} = \frac{1.2\cdot 9.81 \cdot 0.55}{180} < 40 \mathrm{\;mW}$$. But if the depth of your water is 30 cm, then on average the water needs to be raised just 40 cm, not 55 ...


1

The force would be proportional to the area, so the force on the hole would be 10^(-5) N


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The bubble will get bigger. This is because in a pipe, when you look at a cross-section the same amount of liquid will flow through in a certain amount of time, even if it widens or narrows. This means that current must speed up at narrow points. Bernoulli's principle states that when a fluid increases speed, its pressure decreases. This means that, when the ...


2

I can't speak to what sources you were reading, but the air under a hovercraft is higher than ambient air pressure, not lower. Wikipedia: Hovercraft use blowers to produce a large volume of air below the hull that is slightly above atmospheric pressure. The pressure difference between the higher pressure air below the hull and lower pressure ...


1

It sounds like you are not asking about Bernoulli's principle, which describes energy conservation in a fluid, but about why fluids move faster in the thin section of a pipe. This is not Bernoulli's principle, it is just something someone might have mentioned when talking about Bernoulli's principle. Suppose the liquid moves the same speed everywhere in the ...


-1

233 atm, according to the following BBC link: http://www.bbc.com/travel/story/20150415-the-quest-for-the-perfect-pad-thai


3

The can is acting as a compact heat pipe. When you grasp the can, the liquified gas inside the can absorbs the heat from your hand and expands, carrying the heat away in convective currents that dissipate the heat to other parts of the surface that are at a lower temperature than your hand. Since it draws heat from your hand it feels cool to the touch.


1

The differential amount of force is given by: $$ dF = \Delta p* dA $$ You do need to add up the vertical components, which are given by: $$ dF_\downarrow(\theta) = \Delta p * dA * \sin(\theta) $$ This goes from $0 \leq \theta \leq \pi$. At each value of $\theta$ the differential area is the same: $$ dA = rd\theta * dx $$ This gives that: $$ ...


1

Initially, I agreed with Olaf Chujko's answer to this question; however, on further reflection, I think the most accurate answer is 'it depends': Firstly, from the schematic that is given, when switch A is pressed, the cylinder volumes above the two cylinders will be vented to a reservoir at ambient pressure (trust me, I work in Oil & Gas and I look at ...


0

By "gas supply", you mean gas supply to residences? It depends on standards in your country. In UK for example, it's 21 mbar. Putting "21 mbar in mmhg" in Google yields 15.8 mmHg.


1

Probably you have your question answered already, however, let me point out that: You are incorrect. You can't think of a hydrostatic system as it was an electric circuit. In an electric circuit what you (or the source) supply(ies) is the voltage and the result of that voltage acting upon the resistor of given resistance is the current. In hydrostatic ...


2

An additional consideration is that sea water has a higher density than pool water so that the under water barometer under the sea will read a higher pressure than the higher altitude under pool barometer, for two separate reasons. Higher water pressure due to salt water density+Higher air pressure due to lower altitude is greater than fresh water pressure + ...


0

My guess is that the pressure difference you observe is caused by other factors, rather than a tire being used or not. Most probably by heat. When you drive your car your tires get heated up. When exposed to sunlight, they get heated up again. Given the (approximately) steady volume of the air inside the tire, temperature increase would result in ...


1

At sea level the average atmospheric pressure is 29.91" Hg. That is equivalent to 760mm or 76.0cm. If 33' of seawater is 1 atmosphere then 33' of seawater equals 760mm Hg. It follows that 1' of seawater would equal 23mm Hg (760/33=23) 1 Atm (at the surface) x1 liter should= 2 Atm (at 33') x 1/2 liter 1 x 1 = 2 x 1/2 So it follows that any variation to the ...


3

Yes the reading will change. The barometer is telling you the total weight of all the stuff above it - specifically the total weight of stuff per unit area. If the amount of stuff changes then the reading on the barometer will change as well. So the barometer in the pool will read a different pressure from the barometer in the sea. Both have the same ...


0

The basic idea you're reaching for here is called hydrostatic equilibrium. That is, gravitational forces are balanced by pressure gradients. In a context like geophysical fluid dynamics, you'll encounter hydrostatic equilibrium written as $${\partial P\over\partial z} = -\rho g, $$ but as you point out when you bring up Bernoulli's equation, that won't work ...


2

Your overall method for determining the equilibrium for a star is correct. We use a combination of an assumed 'polytropic relation': $$P = K \rho^{1+\frac{1}{n}}$$ and the requirement that this balances the gravitational pressure at a radius r (for a spherical shell of thickness $dr$): $$4\pi r^2\frac{dP(r)}{dr} = -\frac{GM(r)}{r^2}4\pi r^2\rho(r)$$ where ...


0

One cause of pressure loss in tires is permeation of air molecules right through the rubber, as gas inside the tire tends toward equalizing pressure with the outside. This will happen regardless of whether a tire is in use or being stored. But when a tire is in use, frequently rotating under load and at speed, a great deal of external air pressure is ...


2

If you have an object immersed in air, then you can calculate the forces on it using Archimedes' principle. There are two forces to consider. Firstly you have the weight of the object, which is simply: $$ F_g = mg $$ where $m$ is the mass of the object and $g$ is the acceleration due to gravity. This force acts downwards. Secondly you have the bouyant ...


0

The walls of the structure, in this case .2m of polystyrene, are themselves solid and rigid. They provide pressure which fills up the space between the inside surface and the outside surface, effectively adding to the atmospheric pressure inside the cube. If the polystyrene is foamed, there would be pressure from the gas inside the polystyrene bubbles ...


1

The Naïve and Wrong Approach Use the ideal gas law, $PV=nRT$, can give you a first-order approximation. This fails, though, because water is not a gas. It does not need to observe the ideal gas law. (In fact, many gasses do not follow the ideal gas law.) Other equations of state could yield a more accurate answer. The application of these equations to this ...


0

It is converting units: \begin{align} 196.9\,\bar{V}\,[m/s]&\longrightarrow \bar{V}\,[ft/min]\\ 0.0040\, P \,[pa] &\longrightarrow P \,[in\,w]\\ 0.0624\,\rho \, [kg/m^3] &\longrightarrow \rho\,[lb/ft^3] \end{align} So your factor is just $$ 196.9\sqrt{2\times\frac{0.0624}{0.0040 }} $$


3

The argument is that the air was flowing through the hole at around 700 mph, so the air inside the aircraft had a substantial velocity in the direction of the hole. The air velocity inside the plane would have been less than 700 mph because the flow was converging on the hole, but the speed of the air would still have been hundreds of mph. When the hole was ...


3

First of all, it's not universally true. Particularly with solid phases that only appear at high pressures. But the reason why it does often occur, e.g. with ice, is that as you increase the pressure, you force the atoms closer together which causes the enthalpy of the system to increase. Increasing the pressure of a liquid phase, on the other hand, does ...


1

Air pressure exists because if we place something in a gas, then the molecules/atoms flying around will keep banging into it, and in this way produce a net constant force per unit area. As explained by @Chris2807 in the neat formula $P=n k_{B} T$, this is proportional to how many particles there are (since this is proportional to the amount of "banging" in ...


-1

Brionius has the right answer, but there is more to be said. Water at room temperature in air will slowly evaporate. Water at room temperature in a vacuum will boil, as is shown here. So these mini torpedos can prevent damage to chemical bonds. Water molecules are polar. The O's are a little negatively charged. The H's are a little positive. The H's and ...


4

In some sense yes. Let me explain a little. If we were to take a sealed container of gas and put it into free space far away from other bodies so that the gravitational force on the box is negligible would you agree that there would still be some pressure in the container? If we assume we have an ideal gas then the pressure is simply given by $$P=nk_{B}T$$ ...


3

In general, air pressure in the Earth's atmosphere is hydrostatic pressure, caused by the Earth's gravitational field. If there was no gravity then there wouldn't be any centripetal force and all the air molecules would just float away into space. This is why there is no atmosphere on the moon - because it doesn't have enough gravity to sustain one.


-1

In fact, they do!! Watch what happens to an ice cube that is left in the air... trillions of particles of its exterior are torn out of their stable arrangement, and soon they cascade down the sides—a microscopic waterfall! So in this case you are right, but it is just the very exterior surface of an object that is exposed to the air and thus affected ...


-1

Another way of looking at this is that things that would be destroyed by the environment (be it heat, light, etc) have already been destroyed (like ice on a hot summer day). The things that you see around you are the ones where the bond energy was high enough that they survived.



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