# Tag Info

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The phrase black hole tends to be used without specifying exactly what it means, and defining exactly what you mean is important to answer your question. The archetypal black hole is a mathematical object discovered by Karl Schwarzschild in 1915 - the Schwarzschild metric. The curious thing about this object is that it contains no matter. Techically it is a ...

19

Ice can be denser than water for certain values of $P,T$. Look at these two pictures taken from here: The darker areas in the second picture denotes areas of greater density. So you can clearly see that when pressure is increased, ice becomes denser than water along the coexistence line. For example at $T=400$ K ice VII is clearly denser than water ...

8

Whilst the question is I think largely a matter of conjecture, let me point out that matter exists in basically two phases inside a white dwarf. The outer part of a white dwarf is a gas, albeit a very dense one, consisting of a degenerate gas of electrons and a non-degenerate gas of completely ionised ions. The inner part can (if it has cooled sufficiently ...

8

The image below represents the Sun's density gradient, which shows how the density changes with the radius. The ground we stand on should have a density between 2 to 3 $g/cm^{3}$. That should put you just above the water point on the vertical axis. The corresponding radius is then about 0.45 of the solar radius. Note that the vertical axis is in a ...

6

The obvious interpretation of black hole density is the mass of the black hole divided by the volume inside the event horizon. We need to be a bit cautious about taking this too literally because the volume inside the horizon is not coordinate independant so different observers will measure different densities. However we can easily calculate the density ...

6

Boy, does it ever. To contrast the previous answers I will give a mathematical description and a concrete example to bolster the intuitive understanding. Ideal Gas Law From thermodynamics we know that pressure, $P$, temperature $T$, and density $\rho$ (or specific volume $v=1/\rho$) are related through an equation of state. For suitable gases (including ...

5

10km is about right, 5km is definitely wrong and there are theoretical arguments and also observational evidence that this is the case. The Schwarzschild radius, inside which an object would be a black hole, is $3(M/M_{\odot})$ km. As neutron stars have a typical mass of $1.5M_{\odot}$ then a 5km radius would have them hovering just above disappearing ...

4

The statement "hot air rises" is not in general true, although often used. Instead, Less dense air rises Now usually, locally heated air will expand (because pressure will be similar to the pressure of the surrounding air) according to the universal gas law $PV=nRT$, and less dense air will experience buoyancy from the surrounding more-dense (cooler) ...

4

Yes, of course there is. Addressing the specific question you posed in a comment, of whether underneath it all, in large multi-body systems, there is a real charge density... at least as real as the charge/probability density we get by squaring the wave function for a single electron there is indeed such a density. Given an $n$-particle system in a ...

3

To derive the Bernoulli equation for inviscid fluids, the plan is to rewrite the Euler equation in such a way that we have gradients. I'll write the Euler equation with gravity here $$\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \vec{\nabla} \vec{u} = -\frac{1}{\rho} \vec{\nabla} p + \vec{g}.$$ Recall $g = - \vec{\nabla} \Psi$, and $\vec{u} \cdot \... 3 The important thing about a density is that its integral over any volume, which represents the total charge (or whatever it is a density of) inside that volume, is finite. At$r=0$,$\theta$is not defined since the polar coordinate chart for$\mathbb{R}^n$covers everything but the origin. However, since the origin as a point is a set of zero Lebesgue ... 3 In the standard model of particle physics which fits the data up to now elementary particles entering the lagrangian are point particles with mass. The electron, for example is one of the elementary particles, and it does have a mass and the fit gives it 0 volume. There are experiments which try to set limits to how small the volume of the electron is. ... 3 Your question states that We think we know that matter is anything having mass and that it occupies space but in fact, we know better than that. We have good reason to believe that fundamental particles are point-like. In other words, they have no internal structure, size, or volume. And they indeed have mass. We have a theoretical understanding (in ... 3 For non-experts it is worthwhile mentioning the simpler 'Newtonian' interpretation of a black hole. Given the fact that most theorists believe that the General Theory of Relativity (GR) is actually inapplicable in the vicinity of a black hole singularity, that viewpoint is also not necessarily disfavored. (This point seems to meet some criticism; please see ... 3 Most of the books which I looked at give approximately 10 km as the radius of a neutron star. Just yesterday I looked at a book by Dave Goldberg titled The Universe In the Rearview Mirror (2013) which says that they're "only about 5 km in radius" [p.225]. Is this true, [i.e., is there some recent evidence for this], or did he make a mistake here? These are ... 2 As @KyleKanos point out, "the answer is a google search away", but it's not quite as simple as he suggests. The mean density doesn't answer your question (the mean density turns out to be about 1.4 times the density of water by the way). An ill-defined idea in your question is the "surface" of the sun. Where is the "surface" of the sun given that none ... 2 It does seem odd that a star that isn't a black hole can explode, and therefore presumbly lose mass, and still form a black hole. The explanation is that to form a black hole requires a high density not just a high mass. Even a small object such as, well, you or I could form a black hole if compressed enough, though obviously in practice that level of ... 2 Since a car is made up of many different materials, which all likely have their own different densities, the density of the car is, therefore, not the same everywhere. The average density is the density such that, were the entire car to be that density, it would have the same volume and mass. It is very easy to figure out. The total mass of the car divided ... 2 As long as the outlet tube has greater vertical depth than the inlet tube, the weight of falling gas in the outlet tube should maintain an area of decreased pressure at the top of the siphon which should keep the gas in the inlet tube from sliding back into the source pool, and a flow should be maintained. I don't see why this wouldn't work. The Wikipedia ... 2 Suppose you have two masses M1(=M) and M2(=M) with volumes V1 and V2, respectively. Then the total density is the total mass divided by the total volume. So$\rho_{mix}$=2M/(V1+V2). V1=M/$\rho_1$and V2=M/$\rho_2$so$\rho_{mix}$=2M/(M/$\rho_1$+M/$\rho_2$), which after canceling the M's and simplifying the expression is equal to what you wrote above. ... 2 1) The relation$\frac{dF}{dS}=d\left(\frac{F}{S}\right)$is certainly incorrect as @Floris has mentioned in the comment. As the simplest counter-example, consider a linear function,$F(S) = \alpha \, S$, with$\alpha \neq 0\$ as a proportionality constant. Then, one could easily see that $$\frac{dF}{dS}= \alpha \neq 0 = d\left(\frac{F}{S}\right) = d\left(\... 2 As stated above, the mass of the whole system (sugar + water) doesn't change. In addition, with "ideal" mixing, the total volume of the water plus the total volume of the sugar equals the total volume of the mixture. However, this is not a sure bet, and there are many cases of a volume of one material mixed with a different volume of water, and the total ... 2 One mole of oxygen is 32 grams. So the figure you cite tells you that 23.5 cm^3 of solid oxygen weighs 32 grams making the density about 1.36 g/cm^3 or 1360 kg/m^3. For comparison the density of air is about 1.225 kg/m^3 and only 23% (by weight) of the air is oxygen. So the density of the oxygen in air is around 0.28 kg/m^3, which is a factor of ... 2 A few clarifications on this thread in case anyone is reading in the future and is getting confused. I know however, that the temperature did in fact change, hence it's not an adiabatic process. That isn't really how an adiabatic process is defined. The temperature can change within a system (and often does) and it still be adiabatic. An adiabatic ... 2 The main reason is hydrogen bonding. Have a look at the following diagram (^*): Hydrogen and oxygen have significantly different values of electronegavity (the tendency of an atom to attract electrons) which causes molecular bonds between them to be permanently polarised. The oxygen atoms (shown in red) in the \mathrm{H_2O} molecule have a permanent ... 2 You have not specified a frictionless float, but it doesn't matter. In any case (assuming the air temperature remains constant), the volume of ball above the ice will remain constant. As the ice gets thicker, the volume below the ice will get less and less. A lake or any stationary body of water freezes from the top down. This creates a sheet of ice with a ... 2 Very interesting question. As you wrote yourself in your Edit it is hard to describe water via the ideal gas model. You have to introduce at least two important improvements of your ideal gas: Dipole - Dipole - Interaction instead of no interaction. Let's call this pair potential V_d and note that for two given molecules V_d is not only distance but ... 2 The interstellar medium (ISM) has many components that vary in densities (see the table in the link to the ISM Wiki page). For the regions with densities in the n_\text{ISM}\sim10^0\text{ cm}^{-3} range, the temperatures vary from between (50,\,10^4)\text{ K}. Using simple Boltzmann statistics, the mean thermal velocities of particles in the ISM could be ... 2 Obviously, it does change sign when r_{0} = \sqrt{\frac{3}{8\pi\rho_{0}}} Perhaps a more interesting question is "why is this?" Well, we know that the exterior of the star has to be the Schwarzschild Metric, since this is the unique spherically symmetric vacuum solution with zero charge. Consequently, we know, at the surface radius R:$$1-\frac{2M}{R}...

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