# Tag Info

## Hot answers tagged density

31

The speed of sound in a liquid is given by: $$v = \sqrt{\frac{K}{\rho}}$$ where $K$ is the bulk modulus and $\rho$ is the density. The bulk modulus of mercury is $2.85 \times 10^{10}$ Pa and the density is $13534$ kg/m$^3$, so the equation gives $v = 1451$ m/sec. The speed of sound in solids is given by: $$v = \sqrt{\frac{K + \tfrac{4}{3}G}{\rho}}$$ ...

29

When you would enter the water, you need to "get the water out of the way". Say you need to get 50 liters of water out of the way. In a very short time you need to move this water by a few centimeters. That means the water needs to be accelerated in this short time first, and accelerating 50 kg of matter with your own body in this very short time will deform ...

22

It's not the falling that's fatal, it's the deceleration at the end that kills you. Something like water or concrete does this on a sub-meter distance (which requires extremely high forces). On the other hand a gas is much less dense, so it cannot decelerate a falling object nearly as quick. Sometimes inflatable cushions are used as safety nets (think: ...

21

If gas A and gas B are of different density, then the situation sketched is not in equilibrium: the water level on the side of the light gas will be higher. There, the containers are moving down, and you have to push your containers through this net difference in level. You do need to put in energy here, which is probably the piece that you are trying to ...

16

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 ...

14

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 ...

11

The ocean surface is not as hard as the ground but if you drop from a plane, you would hit it with such a high velocity that the pressure would most likely kill you or cause very serious damage. Considering air resistance, the terminal velocity of a human, right before reaching the water, would be at most some $150\text{ m/s}$. If you weigh $70\text{ kg}$, ...

11

Consider jumping into a swimming pool. Do a barrel-roll (sorry I mean cannon ball, that just kind of slipped out). It's fun, you enter the water nicely and make a huge splash, probably soaking your sister in the process (that'll learn her). Now do a belly flop. Not as fun. You displace exactly the same amount of water in the same time, but this time there is ...

11

Let's look at this another way: you're just moving from one fluid to another. Sounds harmless, right? By specification of the problem, we're at terminal velocity when we hit the water. The force of drag (in both mediums) is roughly: $$F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A = \rho \left( \frac{1}{2} v^2 C_D A \right)$$ You can imagine that ...

10

Gaseous hydrogen and helium are lighter than air. Hydrogen, helium and air are close approximations to ideal gases, and for an ideal gas the volume of one mole of gas is about 22.4 litres. That means the density of an ideal gas is inversely proportional to its molecular weight, so hydrogen ($M_w = 2$) and helium ($M_w = 4$) are lighter than air (average $M_w ... 9 The observation that old windows are sometimes found to be thicker at the bottom than at the top is often offered as supporting evidence for the view that glass flows over a timescale of centuries. The assumption being that the glass was once uniform, but has flowed to its new shape, which is a property of liquid. However, this assumption is incorrect; ... 9 Send an ultrasonic pulse through the gold bar and analyse the returning wave. This technique is actually used to detect impurities in gold bars. To quote this article: Where the wave encounters a region of material with different physical properties – particularly the density and elastic constants – to the rest of the metal, the beam is affected in a ... 8 The answer is No and the reason is the equivalence principle which says that there exist natural units in which the gravitational mass (the mass$m$in$F=GMm/r^2$) is equal to the inertial mass (the mass$m$in$F=ma$) for all objects in the Universe. This is equivalent to the statement that all objects, regardless of their composition, density, and other ... 7 "The speed of sound is variable and depends on the properties of the substance through which the wave is traveling. In solids, the speed of transverse (or shear) waves depend on the shear deformation under shear stress (called the shear modulus), and the density of the medium. Longitudinal (or compression) waves in solids depend on the same two factors with ... 6 A proton is a bound state of three quarks. The quarks themselves are (as far as we know) pointlike, but because you have the three of them bound together the proton has a finite size. It doesn't have a sharp edge any more than an atom has a sharp edge, but an edge is conventionally defined at a radius of 0.8768 femtometres. Protons are spherical in the same ... 6 The article's preprint Mayer H. C., Krechetnikov R. "Walking with coffee: why does it spill?," Phys. Rev. E 85, 046117 (2012). is available from the UCSB site. From a glance of the article the phenomenon is not specific only to coffee. The authors make use of the next formula: The natural frequencies of oscillations of a frictionless, ... 6 Here is a table I made for you listing the elements with a density higher than 10 g/cm$^3$and their approximate price per kg: I couldn't find any prices for Einsteinium or Actinium and some of the other prices might come from poor sources, but take it as a rough guide. Now you only have to figure out how much you need and your budgetetary constraints, ... 6 The density of mercury (13.534 g/cm^3) does not imply high intermolecular forces. It simply reflects that the mercury atom is much more massive than a water molecule. The atomic weight of mercury is 200.6, while the molecular weight of water is about 18, so mercury atoms take up$\frac {200.6} {18\cdot 13.534}=0.823$as much volume as a water molecule. ... 6 I'm not a physicist. So I am treading very carefully attempting to answer a question here... :) A physical example that may help explain this is rock skipping. When you skip a rock, it will 'bounce' off of the water when at high speeds. Eventually it slows enough to no longer bounce but 'sink' into the water. Picture your body doing the same thing. Your ... 5 Have a look at http://en.wikipedia.org/wiki/Speed_of_sound#Basic_formula for info on how the speed of sound depends on the medium it's passing through. Generally the important factors are the stiffness of the medium and it's density. To get sound to travel faster you need a stiffer lighter medium. For ordinary matter you'll never get speeds at anything like ... 5 Black holes are really hard to get a density. Basically, they are so dense that there is no known mechanism for providing sufficient outward force to counterbalance the inward pull of gravity, so they will collapse into an infinitesimally small size. Of course, that doesn't seem likely, it seems likely there is something that will keep the volume from being ... 5 The only mechanically static situation is that at bottom of water column temperature is$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 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$\rho_w. This is because the weight is \begin{align} mg = \rho_\text{avg} V_\text{object} g \end{align} while the buoyancy force is \begin{align} F = \rho_w V_\text{displaced} g, \end{align} whereV_\text{object}$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: ... 5 In addition to Bernhard's answer, just because three gases (Gas A,B and air - which is itself a mixture of nitrogen, oxygen, and other gases) have different densities, it does not mean they will remain seperated when in a container. In fact, as entropy of the system increases over time, Gas A, B and air will make an even (if heterogeneous) mixture. 5 In the moist air of the clouds, the water condenses on dust particles. At the altitude where this happens, it is usually below the freezing point of water, so it quickly freezes. If winds and updrafts keep these particles of ice in the moist air, they collect more water. Eventually, the weight of the ice particles overcome the updrafts and fall to the ... 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 ... 4 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 ...

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