New answers tagged

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The stress-energy tensor is a stand-in for whatever forces you want to account for. For example, electromagnetic waves carry energy (as well as momentum, angular momentum, etc.) and you can choose to include these energies in the stress-energy tensor if you wish. Please see the Wikipedia article about the stress-energy tensor for E&M: ...


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Bernouli does not explain wing lift. You can measure an older light plane with a "plank" wing, factor in the wing area, distance over the upper and lower surfaces, cruise speed, and air density, and come up with a total lift figure of about 25% of the aircraft weight. Bernouli equations were published in an aviation text decades ago and the error propagated ...


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There are two reasons why the bottom hole puts out more water. 1, it is exposed for a longer period of time than the two holes above it, add, 2, the velocity of the water coming from the lowest hole in greater than the other two holes above it. See formula below: V = sq. root of 2gh where: V = velocity in ft/s g = accel const. 32.2 ft/ s^2 h = height of ...


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The other way around is more intuitive; if the pressure is lower on the right, the fluid would feel a net positive force in that direction and accelerate toward right. hence it will have higher velocity there. So, lower pressure will result in higher velocity. you can rephrase the above in a way that it sounds as what you may like but is not scientifically ...


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Don't let the size and length of the tube confuse you. What you described can be viewed as a big nozzle with a diameter of 20 meters with a water pressure of 500 atm. entering the back side of the nozzle. The formulas that can be used are: V = sq. root of 2gh where: V = velocity in ft/s. g = acceleration const = 32.2 ft/s^2 ...


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If the parts are perfectly solid and they are at equilibrium, then the pressure is constant (ideally). $$q(x,y) = \frac{\int p(x,y)\,{\rm d}A}{\int \,{\rm d}A} =\frac{\text{applied force}}{\text{area}}$$ But if either the floor or the body are ever so slightly elastic then the pressure distribution is given by a non-hertzian contact which concentrates all ...


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The effect of temperature is included in determining the density as a function of temperature and pressure via the equation of state. For example, in the case of an ideal gas in hydrostatic equilibrium, the density is related at each point in the gas locally to the temperature and pressure by $$\rho=\frac{pM}{RT}$$where M is the molar mass. So the ideal gas ...


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Yes, it is independent in the case to which I believe you are referring, in which one has a container of liquid which is open to the atmosphere here on Earth, and the system is in static equilibrium. However these conditions are merely a special case in which the various effects of temperature cancel each other out perfectly. There are many possible ...


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Differently to what happens for a gas, increasing the temperature of liquid do not increase considerably the mean velocity of the liquid molecules. The molecules are bonded to each other. Molecules in the bulk "pull" the molecules near the walls reducing pressure due to collisions. Therefore the greatest contribution to the pressure of a liquid comes from ...


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PSI does not change with pipe size, only the surface area it is pushing on. Static head pressure is .433 per vertical foot. Water cannot flow over the top of a vessel which is gravity fed. It will only flow as high as the water level in it, because the water in the hose is also pushing back with gravity. A lot of these answers you are reading are ...


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Because hydrostatic pressure depends on depth, the problem requires integration. The pressure as a function of $y$ according to Pascal's Law is: $$p(y)=p_0+\rho g y\sin \theta$$ Where $p_0$ is the atmospheric pressure and $\theta=45\:\mathrm{degrees}$. $y\sin \theta$ is the depth. On an infinitesimal piece of door of length $dy$, at position $y$ and ...


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Short answer: It can be compressed, due to extreme pressure, but only slightly. The Wikipedia page on the Mariana Trench says that the seawater density is 4.96% greater there than at the surface. Also refer to Hydrostatic pressure - doesn't density vary with depth?. Further explanation (and assuming everything you currently know about water is that ...


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Liquids are compressible but in lots of cases the fact that the volume of water decreases by about fifty parts per million for an increase in pressure of one atmosphere can be ignored. A similar approximation is also often made about the compressibility of solids.


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The force of friction is defined as $F_f = \mu N$, where $N$ is the normal force. In the case of a flat surface free of external forces, you can use Newton's laws to determine that $N = mg$, where $m$ is the mass of the object. Notice that we have made no reference to the objects size, or area of contact. This is because in these examples we have ...


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Concerning your wording "force is transmitted (and maybe decreases because of loss of energy)" - no, no, the decrease of force is not easily connected to the loss of energy. Force can be decreased because there is friction, but this does not imply a loss of energy (not if nothing moves). And also energy can be lost (plastic deformation of the rope) without a ...


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You can use the same explanation : "The first molecule pushes its nearest neighbours (which also push back), the nearest neighbours push their nearest neighbours, and so on until the force is transmitted throughout the fluid." The total force transmitted to or by a surface is in proportion to the number of molecules pushing, which is proportional to the ...


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Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container. To explain this, the hydraulic jack is a better example. Car jack works on mechanical forces. A mechanical jack employs a screw thread for lifting heavy equipment. But hydraulic jacks use force ...


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Have a look at the answers to Pressure and altitude as they explain how the pressure:altitude equation is derived. There is nothing wrong with our working, but you have assumed that the temperature is constant and in reality the temperature falls with altitude (in the troposphere at least). That means the pressure falls more rapidly with height than your ...


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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 ...


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As mentioned by @Chester, Bernoulli isn't a good approximation for viscous flows which blood flow is. Instead you should use the Hagen-Poiseuille law which relates the average volumetric flowrate and the pressure gradient in the pipe. From it we find that the flowrate $Q$ is proportional to: $$Q \propto R^4 \Gamma$$ where $R$ is the radius of the pipe and ...


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This has nothing to do with air pressure since the air pressure is exactly the same at all of the holes (including the top one). Regarding pressure only differences can cause stuff to move, the absolute pressure is only relevant for density and such. The concept you should read more about is called hydrostatic pressure and given by a very simple formula. ...


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The Bernoulli equation is a good approximation only if viscous flow resistance is not important. In blood flow through arteries, veins and (particularly) capillaries, viscous flow resistance is very important.


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I am not sure if the same laws apply to the heart as that of mechanical pump, but for a given flow rate, say X gallons per minute, the mechanical pump must develop a pressure P to overcome pipe friction and any other force trying to retard flow. If the pipe in a system is reduced in size, to pump the same flow rate a higher pressure will be required. The ...


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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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Sorry, didn't complete my answer above. The area A of the 20 m exit hole you mentioned is 313 m^2. To determine that calculate the area of a circle with a 20 m diameter. Example: A = .785(20^2) = 313. If you want to calculate the vol. of water mult. The V by the A as follows: Vol(m^3) = 26(m/s) * 313(m^2) = 8,112 (m^3/s)


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So we have a tower (tube) 5000 m (16500 ft) tall. The bottom of the tower has a hole 20 m ( 66 ft ) in diameter. To determine the pressure the column of water (assuming specific gravity = 1.0) exerts on the bottom of the tower at point of water exit is: P = (H * SG)/2.31 where: H = height of the liquid in feet. SG = 1.0 ...


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BP = 101.325 kPa STP = 101.325 kPa, 0°C NTP = 101.325 kPa, 20°C Molar mass of oxygen = 31.9988 g [1] Molar volume of gaseous oxygen at STP = 22.41 L [5] Molar volume of liquid oxygen at BP = 28.04 cm³ - Molar volume of metallic oxygen = 23.5 cm³ [4] Density of gaseous oxygen at STP = 1,429 g/m³ [2] Density of ...


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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 ...


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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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The bottles are filled at sea level. There is a small bit of air in them, so the inside pressure is also at sea level, along with quite a bit of (presumably) incompressible fluid. Bottles are plastic and thus flexible. Assuming leaks occur when travelling to 2300 feet, outside pressure drops from 760 mm Hg to 702 mm Hg. This would flex the bottle and ...


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As I understand it, you have two theories. the leak is caused by the drop in temperature of the liquid the leak is caused by the increase in pressure when screwing on the top I think we can probably rule out both of these, and here's why: Temperature drop As we know from the ideal gas law, $$\displaystyle \frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f}$$ ...


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chemistry better describes the difference between adhesion atoms in mercury vs hydrogen bonding in water. Mercury has boiling over 600f vs only 212 f for water shows the amount of heat energy to break bonds hold the respective particles together.


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Keep in mind that pressure is not force. Also, pressure is not mass per area. Pressure expresses the ratio of differential force exerted normal to a surface per differential area, and is expressed in units of force per area. A pressure of 100 kPa exerted on a surface means that on 1 cm$^2$ of that surface there is a total force of 10 N normal to the ...


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I think you misunderstand the definitions. The critical temperature is the temperature above which no amount of pressure will cause a gas to liquefy. The critical pressure is the pressure which will cause a gas to liquefy at its critical temperature. A supercritical fluid is another state of matter. A liquid and a gas phase have been subjected to ...


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I think that there is a confusion because there are two forms of the ideal gas equation. In terms of moles the ideal gas equation is $PV = nRT$ where $R$ is the molar gas constant and is a constant independent of the gas or gases which are being considered. In this equation there is no mention of the mass of the gas or the composition of the gas other ...


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An ideal gas is one that follows any of a number of linear relations such as $PV = k$ or $PV = nRT$. Many gases near room temperature and atmospheric pressure are nearly ideal and show this relationship pretty closely. All gasses deviate from the ideal as pressure increases and temperature decreases. As an example, if you cool the gas so much that it ...


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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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A self-contained, careful derivation of (4.10): We consider a thermodynamic system whose state can be characterized by the macroscopic variables $(S, V, N)$, then starting with the fundamental relationship $\mathrm dU = T\,\mathrm dS -P\,\mathrm dV + \mu\,\mathrm dN$, and noting that $\beta = 1/(k T)$, one can deduce the following useful expression for the ...


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Their claim that atmospheric pressure would increase if moon is over head is wrong. Atmospheric pressure at same height is same. This follows directly from Pascals law. So, even if moon is directly upwards or whatever is the shape of bulge out. It does not matter. If there was an increase in atmospheric Pressure , it will create a pressure gradient. This ...


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An ideal gas or near-ideal gas such as air at about atmospheric pressure and room temperature has a bulk modulus which is the same as its pressure. That can be readily confirmed by taking the ideal gas equation $PV=nRT$ and substituting it into the equation for the bulk modulus $B=-V \frac{dP}{dV}$. Now for what P-V equations-of-state does the bulk modulus ...


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It was a good thaught of you and was similar to mine,I do not have any theoretical answer but according to my understanding : An air bubble inside water will experience a lot of inward pressure than what it can exert outward as it is inside a denser medium , and a water droplet in air will exert a more pressure outwards than what it experiences from air ...


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Pressure is force per unit area. If you have 5N/m^2, it means that a force of 5N acts at every 1 m^2 area of the surface. Suppose you have a 100m^2 area. Then the total force distributed over the entire surface is 500N which corresponds to about 50Kg.


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If the pressure is uniform, there is no problem, because you just multiply the pressure by the total surface area of interest. However, if the pressure is varying on the surface, pressure should be regarded as a point function of location, and the contribution to the total force on the surface at a differential element of area on the surface dA is equal to ...


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It means 5 newtons acts on one meter square. Not exactly a point, any patch of any shape having 1 meter square area will have 5 newtons. If area is more then the force would be more if area is less force less. F=PxA


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A and C are not identical; that's where the thought experiment breaks down. Consider the pressure in the fluids around the opening between the long thin neck section and the wide base section. In C, you have one continuous fluid, and the pressure is the same both above and below the neck (and equal to $\rho g h$ where $h$ is the height of the next and ...


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Sure, as pressure increases so does boiling point and the amount of work needed to be done against the atmosphere you would assume. But this is not the case. Work done by gas = change in volume x atmospheric pressure. If the atmospheric pressure is indeed higher the amount of expansion of the vaporized gas required in order to reach pressure equilibrium ...


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If the space you're pumping air into is six feet below the ocean surface, then the minimum pressure is rho * g * h = 1030 kg/m^3* 9.8 m/s^2 * 1.83m = 18.5 kPa (or, 2.68 psi) If the 'tube' has to be emptied of water before it moves, that means you need to add its depth dimension (40 inches) to this, and that'd call for 4.2 psi. Those are gage pressures, ...



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