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The reason is electron degeneracy pressure. The cores of giant planets are dense enough that the electrons in the gas occupy about $h^3$ of phase space each. The Pauli exclusion principle means that they cannot all occupy low energy/momentum states. This means that even at relatively cool temperatures the gas can still exert considerable pressure due to the ...

11

Although we don't have a quantum theory of gravity, we think we have some reliable knowledge about the properties of black holes from general relativity. One thing we think we know is the so-called "No-hair conjecture", which says that black holes can be described by just three numbers: mass, charge, and angular momentum (i.e. how much they are spinning). ...

9

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 ... 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 ... 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 ... 4 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 ... 4 It is commonly believed that the speed of sound at high densities is bounded from above by c/\sqrt{3}, where c is the speed of light. Calculations of this quantity in many theories, ranging from QCD to systems with scale invariance, have all shown it to either stay below or exactly saturate the bound. See the introduction of this paper for a recent ... 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 Far away from a black hole, spacetime is curved only a little bit, and many different things could curve it like that out there. It's like if you had a dollar in your pocket, and it's been there for a long time, and you can't remember if you got it from your boss or from your friend. But a dollar is a dollar. So you could have a massive star, or a black ... 3 Your intuition is right: the density of the string goes down a little bit when you increase the tension. HOWEVER: the wave in a string is a transverse wave which depends on the tension and the mass per unit length. If you double the tension the mass per unit length goes down by a small amount (the string gets a bit "thinner" because it gets longer) . Both ... 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 ... 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 ... 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 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 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 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 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

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

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 ...

2

Consider this. I have an Earth-sized quantity of water that I throw into space. Naturally, it will assume the shape of a ball. Only if it is non-rotating and is not being influenced by a tidal field. So is there some sort of simple formula that ties these two properties together? Yes. It is called the equation of hydrostatic equlibrium.  ...

2

In a static situation, the water would have a larger density when it has a larger pressure, so you could sink until you reach a level where the density matches. In reality, the water can have a different temperature and a different salinity (both of which affect density) and if can be flowing (up/down, east/west, north/south) and so it might never settle ...

2

The article you refer to is talking about the speed of sound (or speed of longitudinal wave vibrations) within a material and how that relates to volumetric mass density. If you were transmitting sound from one end of a guitar string to another, this would be relevant. But the sound produced from a string is not related to the speed that it travels within ...

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