# Tag Info

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http://en.wikipedia.org/wiki/Degenerate_matter#Degenerate_gases http://imagine.gsfc.nasa.gov/docs/science/know_l2/dwarfs.html carbon-oxygen white dwarf stars By what modality do you plan to compress the diamond? Graphite is much cheaper and gets you to the same end. Fermi degenerate matter is stable. Squeezing electrons into protons requires Type II ...

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Is it correct that the zero atmospheric pressure occurs at A? Yes, approximately. In a real system the pressure at A would equal the vapor pressure of the liquid. Your "other resources" are referring to the pressure of air in the atmosphere. In the problem, the diagram represents a situation like mercury in a barometer, where the weight of the mercury ...

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How does this varying pressure conform with the constant pressure/density obtained from the equation of state? It doesn't conform. It is a contradiction to reality: you cannot have an incompressible material. However, that doesn't mean the approximation isn't a useful one under many circumstances. For example, consider water which as a density of ...

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This can be treated by use of Bernoulli Equation, which says that the sum of geodetic, dynamic and static pressure is constant for every flow cross section of a frictionless fluid system. On top of that, friction can be accounted separately and represent the other side of the Eqn.. For the fluid part, friction factors depend on Reynolds Number and geometry. ...

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I don't think that the size of the hole has an influence on the energy required. Friction is a complicated matter here, not easy to calculate. Try to calculate the pressure imposed by the water on the piston. The force working against that pressure will require most energy. Edit The pressure on the piston isn't uniform along its height in this case. At ...

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Hint: What potential energy does the water have with respect to the heigh of the container valve?

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Using Bernoulli you get: $$\frac{P_1}{\rho} + \frac{1}{2}v_1^2 = \frac{P_2}{\rho} + \frac{1}{2}v_2^2$$ Using your formula: $\displaystyle{c_s^2 + \frac{1}{2}v_1^2 = c_s^2 + \frac{1}{2}v_2^2}$ and this implies: $v_1 = v_2$ From mass continuity: $v_1 \times A_1 = v1 \times A_2$, so $A_1=A_2$, which is a false. There is clearly a misinterpretation. ...

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This is quite humorous. In an 1883 offical US military publication, "Weather Proverbs" by 1st Lt. Dunwoody, at page 107 it is stated "When coffee bubbles collect in the centre of the cup expect fair weather. When they adhere to the cup, forming a ring, expect rain." This is the converse of the lifehack proverb! In 1997 Dave Thurlow, using a grant ...

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Normally when compressing a gas the temperature increases. If you assume adiabatic compression, the law is $PV^\gamma=k$, where $\gamma=\frac {C_P}{C_V}$ is the ratio of specific heats and is usually about $1.4$ for air. Then, as shown here $\frac {T_2}{T_1}=\left(\frac {P_2}{P_1}\right)^{\gamma-\frac 1\gamma}$ This assumes you don't leak heat to the ...

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A force is a vector. It represents a value and a direction. Pressure applied on a small surface results in a force applied in the direction perpendicular to the small surface and with value $F=P\, A$ where $P$ is pressure and $A$ is a small area. If you have an object, where low pressure is applied on a large area on one side, but high pressure is applied ...

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In physics, pressure, $p$, is a Force, $F$, per unit Area, $A$. Where: $p=F/A$ We can see from the simple formula that to increase pressure we can increase the force or decrease the area. The example which you gave of a paper pin demonstrates mechanical pressure, by lowering the surface area of the end of a pin we can more easily penetrate the paper ...

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There is no stupid question when it comes to genuinely trying to understand physics. Pressure is the result of molecules smashing against the walls of their enclosure. Except at zero Kelvin, there is always some kind of erratic movement which increases with temperature. For a given number of molecules in an enclosure, the smaller it gets and the more impacts ...

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Consider a ball of water floating in zero G, as demonstrated on the ISS. Ignoring for the moment the surface tension of the water (I'll come back to that) the pressure inside the water is the same as the pressure of the air around it. This is simply because without any forces, like gravity, acting on the water there is nothing to cause a pressure gradient. ...

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The reason is, quite simply, that the walls of the container exert a force on the fluid given by the pressure at each point, and orthogonal to the container's surface. In your second container, the wall is slanted and therefore provides an upward force which helps stabilize the fluid elements next to it. If you draw a free-body diagram of the bit of fluid ...

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