# 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 ...

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 ...

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 ...

7

Density is a 3-form, since you would write it as $$\omega:=\rho\text dx\wedge\text dy\wedge\text dz.$$ In special relativity it remains (the time component of) a 3-form. More specifically you have a current density $J$ of the form $$J = \rho\text dx\wedge\text dy\wedge\text dz + J_x \text dt\wedge\text dy\wedge\text dz+ J_y \text dx\wedge\text dt\wedge\text ... 6 It depends on how the quantity in question transforms. Almost always, densities in the form of "stuff per unit volume" and generally the "stuff" (like a charge) is a scalar (a number of things - number of elementary charges), but the volume it is contained in is observer dependent, owing to the Lorentz contraction. Therefore the density is ... 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 ... 5 It's true. Special equipment and a long time is required to mix helium and nitrogen. According to one study, a mixture of 2.7% He, 93.3% N at 800 p.s.i.g. required a special cradle to repeatedly upend the cylinder, and 20.5 hours to reach equilibrated gas, which then remained mixed: http://pubs.acs.org/doi/abs/10.1021/je60005a002. The helium repeatedly ... 5 Sometimes I feel Wikipedia is a funny place... In the article you quote they provide a calculation from our patent application (see, e.g., http://akhmeteli.org/wp-content/uploads/2011/08/vacuum_balloons_cip.pdf ) proving that a homogeneous shell made of any existing material cannot be both light enough to float in air and strong enough to withstand ... 5 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 ... 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 ... 4 I've just remembered that there was once a suggestion to use a mixture of xenon and oxygen under high pressure to allow people to float/fly/swim in it. It was also stated that water could be lighter than such a mixture. According to Smithsonian Physical Tables the critical point for xenon is 16.6\,\text{C}^{\circ},\quad ... 3 The black hole event horizon is not a thing i.e. not a physical object. It is just a surface in spacetime from which light can never escape to infinity. Also, if we take the Schwarzschild description of a (non-rotating) black hole then it is a point mass hidden away behind the event horizon. You can't spaghettify a point mass. When two black holes merge, ... 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 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 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 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 ... 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 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 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 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) = ...

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

Since you have not specified the "real world" size of the ship, let's take a 74-gun ship of the line https://en.wikipedia.org/wiki/Seventy-four_(ship) as the desired type, firing a 36-pound cannon. The bore on such a cannon https://en.wikipedia.org/wiki/36-pounder_long_gun was about 175 mm, with a shot weight of about 39 lb. Shrinking this cannon to a bore ...

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 ...

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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 ...

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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 $R$: ...

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Its just a figure of speech to emphasising the difference in density between the materials of the planet and the density of the water. Nothing will float forever. Even polyethylene, which is obviously lighter than water will go into breakdown in billion years.

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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

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 ...

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