# Tag Info

## Hot answers tagged density

26

No. The answer is clearly no. This building is 800 meter high. Some comparison: Skydivers are falling more kilometers in free fall. They experience absolutely no damage from the pressure increase. Scuba divers moving fast upwardly or downwardly also don't get any wounds, although 10 meter deep water has the same pressure as there is between the sea level ...

16

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 ...

16

The density does increase with depth, but only to a tiny extent. At the bottom of the deepest ocean the density is only increased by about 5% so the change can be ignored in most situations. If you're dealing with these sorts of depths you also need to take temperature into account because the water temperature changes with depth and the density also ...

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). ...

10

It's actually a surprisingly straightforward differential equation. If you assume that the acceleration due to gravity $g$ doesn't change with altitude (a good approximation if the atmosphere is thin compared to the radius of the earth), Bernoulli's relation tells you the change in the pressure $P$ with height $h$: $$\frac{dP}{dh} = -\rho g$$ Meanwhile the ...

8

Ok, trying my luck with a physics answer. Let's first look at the boundary conditions given in the movie, since we're particularly talking about that here. The water planet is said to have $130\%$ of earth's gravitational acceleration on the surface. So we have $$g_W = 1.3 g_E$$ This is a given and not to be violated. And in fact ...

6

as the depth increases, wouldn't the density of the liquid increase because of the weight of the liquid above it compressing it? No, it doesn't - or at least only negligibly so. At normal pressures, liquids are essentially incompressible. This table gives the compressibility of some liquids, including water. Note that the units are to be multiplied by ...

5

You said the right word: liquid! $P=\rho g h$ holds only if the fluid that you are considering is not compressible, that is a liquid. Try to fill a syringe (without the needle) with some water, then close the hole and try to compress it: you will notice that you cannot do much, indeed liquids are not compressible, this mean that the density $\rho=m/V$ does ...

4

You can rearrange the terms to have any constant as the base of the exponent: $D = 1.25 e^{(-0.0001h)}$ $= 1.25 (e^{0.0001})^{-h}$ $= 1.25 (2^{\frac{0.0001}{ln 2}})^{-h}$ $\approx 1.25 (2^{0.00014})^{-h}$ $= 1.25 \times 2^{(-0.00014h)}$

4

There more sides to this scenario that you're considering. Firstly, if we are assuming that the temperature is the same at sea level and on the high mountains, then the speed of sound doesn't actually change, as a constant temperature will take care of the air pressure-density ratio. $$c = \sqrt{\kappa \frac{p}{\rho}}$$ Where $p$: static air pressure, ...

4

Basically, it has to do with the density of the material as a function of temperature. The density of iron increases as it cools, that is, solid iron is more 'packed tight' than when it is melted. This is understandable, since the kinetic energy of the iron atoms decreases as the temperature drops (ie: the average velocity of the atoms decreases), allowing ...

4

The trouble is that your table, or whatever object it is, will act as a waveguide. That's because the sound waves will (partially) reflect of the wood/air surface then travel back into the table and interfere with other waves. The result is going to be hideously complicated to calculate. As Luboš says in a comment, if the thickness of the table is much less ...

4

You would be wise to somehow determine the exact fluid used by the original manufacturer. Consider that each of the floats has a fixed density, and has a temperature marked on its hanging tag. So you need a liquid which will have the correct, different density at each temperature marked on a tag. In short, the liquid you choose must match both the ...

4

This is an instrument that measures fog density and has an experimental plot, figure 9 . Once you have the relative humidity at the fog appearance at a temperature and pressure , one can use known equations to get the density. This link gives a calculator.

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

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

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

This is because the whole boat, along with the air in the boat, is lighter than the water it displaces. For example, if a small boat will take up 1 cubic meter of water, then it has to be heavier than the weight of 1 cubic meter of water. This is explained in this post by What If here. For the same reason that bowling balls float (because salt water the ...

3

This answer will not make me popular because it gets people up to speed fast on protecting themselves from thieves and levels the playing field for people who like to maintain their advantage over others. There are a few methods we use to determine if gold is bunk or real...Methods that test if your gold is hollow, filled, alloyed (and the alloy percentages ...

3

Assuming you are talking about exoplanets, I'll offer this. To obtain a density you need a mass and radius. Masses come via two methods - either measuring the radial velocity variations of the star it orbits (the bigger the RV variations, the bigger the planet mass), or so-called transit timing variations. This latter works in multiple "transiting planet" ...

3

I'll use this answer to provide some information that's mostly orthogonal to what Phonon said. As Phonon pointed out, the speed of sound depends on temperature, not pressure. It's cold on the top of high mountains, so the speed of sound would tend to be lower. Some mechanisms for sound production have a frequency that depends on the speed of sound, and ...

3

Euler's constant appears naturally in phenomena where the spatial gradient of a quantity (or rate of change with time) is proportional to the quantity itself: $$\frac{\mathrm{d}X}{\mathrm{d}x} = X/x_0$$ ($x_0$ determines the strength of the proportionality, and keeps units straight.) The solution of this differential equation is $$X=X_o e^{x/x_0}$$ $X_0$ ...

2

Yes they would fall differently. For example, according to Stokes' Law the drag force is proportional to dynamic viscosity and the terminal velocity of a falling sphere is: $$\frac{2(\rho_s -\rho_f)}{9\mu}gR^2$$ where $\rho_s$ and $\rho_f$ are the densities of the sphere, $\mu$ is the dynamic viscosity, $g$ is the acceleration due to gravity, and $R$ is ...

2

The dynamics in case of relative motion will be different: e.g., the acceleration and terminal velocity when falling/sinking. But the equilibrium will be unaffected (e.g., floating and waterline). This is because viscosity is at the origin of a frictional force.

2

On paper, a black hole already has infinite density. Two coalescing holes would combine to another object of infinite density. Realistically, we would need quantum gravity to prevent a true singularity from forming,a nd there, we could address, more concretely, what happens when the "masses" in the center of the black holes merge. But until we ...

2

The notion of buoyancy really only applies to fluids or very plastic/glassy solids, so might not be something to look at with the polystyrene beads. However, if we want to model it like a liquid, there are also the factors of "surface tension" and "viscosity" to consider: polystyrene beads are often electrostatically bound to each other, creating an ...

2

If we take neutron star material at say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$. So even in a teaspoonful (say 5ml), there is $1.5\times10^{27}$ J of kinetic energy (more than the Sun emits in a second, or a billion or so atom bombs) and this will be released instantaneously. ...

2

So I am guessing you have a a 2 dimensional density? say, kg/m^2? In this case, you will need to get the area of your box -> say it is 2m^2. The mass of the box is 2 * 0.5 = 1 kg. But the mass of your person is 80kg. So your new mass is 81kg. This means the new density will be 81/2 = 40.5kg/m^2. The item will only float if the item has a lower density ...

2

One should always specify whether one is talking about rest mass per unit rest frame volume, $\rho_0 = m_0/V_0$, rest mass per unit observer-frame volume, $D = m_0/(V_0/\gamma) = \gamma\rho_0$, or relativistic mass per unit observer-frame volume, $(\gamma m_0)/(V_0/\gamma) = \gamma^2\rho_0$.1 (I can't imagine the fourth case, relativistic mass per unit rest ...

Only top voted, non community-wiki answers of a minimum length are eligible