Hot answers tagged density
14
If something is infinitely dense, must it not also be infinitely massive?
Nope. The singularity is a point where volume goes to zero, not where mass goes to infinity.
It is a point with zero volume, but which still holds mass, due to the extreme stretching of space by gravity. The density is $\frac{mass}{volume}$, so we say that in the limit ...
13
Well, it can't (float), since a Black Hole is not a solid object that has any kind of surface.
When someone says that a super massive black hole has less density than water, one probably means that since the density goes like
$\frac{M}{R^3}$
where M is the mass and R is the typical size of the object, then for a black hole the typical size is the ...
9
The observation that old windows are sometimes found to be thicker at the bottom than at the top is often offered as supporting evidence for the view that glass flows over a timescale of centuries. The assumption being that the glass was once uniform, but has flowed to its new shape, which is a property of liquid. However, this assumption is incorrect; ...
8
You have hit on the major explanation of the unusual thermal stability of surface-frozen lakes. The deep earth is temperature stable, since the surface seasonal fluctuations can't penetrate the heat by diffusion more than some meters into the deep ground. So the deep ground is at a temperature which is stable all year.
Advection only raises heat to the ...
6
The article's preprint
Mayer H. C., Krechetnikov R. "Walking with coffee: why does it spill?," Phys. Rev. E 85, 046117 (2012).
is available from the UCSB site.
From a glance of the article the phenomenon is not specific only to coffee. The authors make use of the next formula:
The natural frequencies of oscillations of a frictionless,
...
6
Here is a table I made for you listing the elements with a density higher than 10 g/cm$^3$ and their approximate price per kg:
I couldn't find any prices for Einsteinium or Actinium and some of the other prices might come from poor sources, but take it as a rough guide.
Now you only have to figure out how much you need and your budgetetary constraints, ...
5
Black holes are really hard to get a density. Basically, they are so dense that there is no known mechanism for providing sufficient outward force to counterbalance the inward pull of gravity, so they will collapse into an infinitesimally small size. Of course, that doesn't seem likely, it seems likely there is something that will keep the volume from being ...
5
The only mechanically static situation is that at bottom of water column temperature is $T_\text{bottom} \le 4^\circ \text{C}$ and at top of the water column temperature is $0^\circ\text{C} \le T_\text{top} \le T_\text{bottom}$, with continuous drop between them. Of course, there will still be heat transfer due to thermal conductivity of water and ice. You ...
5
Have a look at http://en.wikipedia.org/wiki/Speed_of_sound#Basic_formula for info on how the speed of sound depends on the medium it's passing through.
Generally the important factors are the stiffness of the medium and it's density. To get sound to travel faster you need a stiffer lighter medium. For ordinary matter you'll never get speeds at anything like ...
5
The nature (and glory) of the dirac delta function is that the volume integral
$$ \int_{\Delta V} dV' \delta ( \boldsymbol{r-r'} )
= \left\{
\begin{array}{cc}
1 & \text{if } \Delta V \text{ contains } \boldsymbol{r}\\
0 & \text{if } \Delta V \text{ does not contain } \boldsymbol{r}
\end{array} \right. $$
Using this function, you can write the ...
5
Simply if the average density $\rho_\text{avg}$ of the sphere + helium (or your horse, for that matter) is less than the density of water $\rho_w$.
This is because the weight is
\begin{align}
mg = \rho_\text{avg} V_\text{object} g
\end{align}
while the buoyancy force is
\begin{align}
F = \rho_w V_\text{displaced} g,
\end{align}
where $V_\text{object}$ is ...
5
I see this is a follow-up post to Suppose a hollow metal sphere filled with helium is dropped in a body of water
Well, the situation you are describing is possible if the object in question can change its average density while in the water. It will stop sinking when $\rho_\text{average} = \rho_w$.
In fact, there's a vessel that uses this exact principle: ...
4
The key's is the Bernoulli's equation for the compressible flow:
$$\frac{v^2}{2} + \frac{p}{\rho} + u = \text{const}$$
$u$ is internal energy per unit mass, or using enthalpy $h$ per unit mass:
$$\frac{v^2}{2} + h = \text{const}$$
The other equation to find $v$ you'll get from the definition of the volumetric flow.
You have two equations to solve the ...
3
It's actually more extreme than you think.
The short story is this:
Associated with any amount of matter, there is an associated radius know as the Schwarzschild radius.
There is theorem in General Relativity that essentially states that if ever all of the matter is contained within the associated Schwarzschild radius, that matter must collapse to ...
3
A proton is a bound state of three quarks. The quarks themselves are (as far as we know) pointlike, but because you have the three of them bound together the proton has a finite size. It doesn't have a sharp edge any more than an atom has a sharp edge, but an edge is conventionally defined at a radius of 0.8768 femtometres. Protons are spherical in the same ...
3
This question (v1) is discussed near eq. (8) in Ref. 1.
The simplest regularization is to truncate the variables $x\geq\ell_x$ and $p\geq\ell_p$ at cut-offs $\ell_x$ and $\ell_p$, respectively, in such a way that the product $\ell_x \ell_p = h$ is Planck's constant.
In an $(x,p)$ diagram, the truncated area under the hyperbola $p=\frac{E}{x}$ reads in ...
3
The black hole would float in water, if you could make a large enough pool to submerge it, and with enough replenishment to replace the water that the black hole will sucks up. The black hole will remove water from its surroundings, but the water below will come into the horizon at higher pressure than the water above, so the velocity inward will not be ...
3
As far as I know, there are a lot of kinds of density - you mentioned volume density, but there is linear density (amount of mass for unit of length), for example. If your string is very-very thin, there is no sense in definition that density = mass / volume, cause very thin string doesn't have volume and you should use linear density = mass / length or ...
3
1/3) As Newton pointed out way back in the Principia, the gravitational attraction due to a spherically symmetric mass distribution is, assuming you are outside of it, the same as if all the mass were at a point at the center. Thus the gravitational acceleration at the surface of a sphere is determined solely by the total mass $M$ and the radius $R$. What is ...
3
Solid metals are crystals, not liquids.
The way any crystal plastically deforms is by motion of crystal dislocations.
Every crystal obeys a stress-strain curve, where stresses up to a certain amount do not result in permanent deformation.
Higher stresses do result in permanent (plastic) deformation because dislocations move.
If a metal is "soft", all that ...
2
I think it is actually misleading to make the claim that is puzzling you. "Density" suggests that the mass is distributed more or less uniformly within the black hole, and this is non-sense. The black hole is mostly empty, and all the mass is concentrated within a tiny region (clasically a point) in the center of the black hole.
If you ignore this and ...
2
The Schwarzschild radius scales with mass as $r~=~2GM/c^2$. What might be defined as a Schwarzschild volume would then be $V~=~4\pi r^3/3$ $=~(32/3)\pi(GM/c^2)^3$. So the density of matter defined by the horizon is $\rho~=~(3/32)(c^2/G)^3M^{-2}$. So density scales as the inverse square of the mass. A 10 billion solar mass black hole has a radius about ...
2
The "edge" of the sun that we see (the photosphere) arises not so much from any feature in its density profile, but from the properties of how light travels through the sun as the density drops.
The photosphere is the point where the density drops enough that photons can begin to free stream away without interacting any more with the gas. Formally, this is ...
2
Yes the density of water changes with temperature in a non-linear way (which is important if you want life on your planet).
It has a maximum density at 4deg C and is unusual in that it expands (lower density) as a solid - see http://en.wikipedia.org/wiki/Water_(properties)#Density_of_water_and_ice
As a gas it's density varies with temperature and pressure ...
2
You must specify wether the temperature is constant or if there is some heat exchange. Then you just have to use the isothermal compressibility or adiabatic compressibilty factor.
http://en.wikipedia.org/wiki/Compressibility
2
Assuming your pipette has a volume of 15ml, the error in measuring the volume of your sample is 0.02 in 15 i.e. 0.133%.
The weight of 15cc of ethanol is 0.789 $\times$ 15 = 11.835g. The error in measuring the weight is 0.002 in 11.835 i.e. 0.025%.
Hedge physicists like me would immediately note that the error in the weight is a lot lower than the volume, ...
2
I'm guessing that you're asking this because you've heard that atoms are mostly empty space. The trouble is that your question doesn't have an answer because how much empty space you think there is in an atom depends on how hard you're willing to press on the atom.
As chance would have it, I've just answere a question that deals with exactly this question ...
2
You ask how atoms can be that tightly compressed. Atoms are made of electrons and quarks (the protons and neutrons are made of quarks) and as far as we know electrons and quarks are point like i.e. they have no size. So in principle they can be compressed to infinite density if you squeeze hard enough. At this point someone is going to point out that all ...
2
Any cross section of your wall is supporting the weight of all the wall above it. In a first approximation, every cross section will be in a state of pure axial compression. The most heavily solicited cross section will be the one at the very bottom, which will be supporting a compressive pressure of $\rho h g$, where $\rho$ is the density of the ice, $h$ ...
2
The book I am reading defines the position of the com of a two-particle system to be $x_{com}= \Large\frac{m_1x_1+m_2x_2}{m_1+m_2}$ I'm sorry if this seems like a trivial question, but could someone explain to me the interpretation of this definition? Perhaps even why they defined it to be this way.
It's a weighted average of the position of the ...
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