# Tag Info

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 ...

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 ...

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

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 ... 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 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 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 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. 4 If it's a cuboid, then I think a simpler way would be to measure its resistance. Use Suitable contacts Measure resistance Use R=ρL/A and find out resistivity ρ. Either use the table found here and check if it matches up You can also use similar method for other properties like -Thermal conductivity: Heat one end and monitor how quickly the other end's ... 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 ... 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 ... 3 Different objects with the same mass but different sizes will have different gravitational pulls at their surface, which is probably what you heard. But that is only because their surfaces are in different places. If you compare the gravitational pulls of different objects with the same mass at a fixed distance away from their centers, you will find that ... 3 I don't think the term electron probability cloud has a precisely defined meaning. It's more of a metaphor meant to show that the electron does not have a well defined position. Like any quantum particle the electron does not have a position until you interact with it e.g. scatter another particle off it. The interaction takes place around some position (I ... 3 It does not work for the following reason. Let's look at the right side, where the containers float to the top. When a new container enters at the bottom (from water to gas B), it pushes some gas B away, so it can occupy the space. The pushed away gas B has nowhere to go, but up. Gaining height means potential energy. That is energy used, to make the ... 3 I think you're asking if there is some special cutoff density after which spacetime "collapses" and forms a black hole. If this is your question then the answer is no, there is no specific cutoff. Density unites are$\frac{\mathrm{mass}}{\mathrm{volume}}$but the size of black holes is dependent on the mass and the size is not proportional to the volume ... 2 The mass difference of the air it would have contained and the Helium it does = the volume of the balloon multiplied by the difference in density of the helium and air. Suppose the balloon is spherical and 12" in diameter (physicists can only do the arithmatic for spherical objects, and preferably in a vacuum) . That gives it a volume of$\frac43 \pi r^3$... 2 I) Well, Gaussian integrals $$\tag{1} \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^t A x} ~=~ \sqrt{\frac{(2\pi)^n}{\det A}}$$ are easy to calculate exactly, where the matrix${\rm Re}(A)$is positive definite. II) But if OP just wants to confirm that the power$p$of the determinant$\det A$on the rhs. of eq. (1) is$p=-1/2$(as opposed to some ... 2 This is a great question, although unfortunately it turns out to be very difficult to interpret it in a way that allows a definite answer. The question is ambiguous because of the way mass is defined in relativity. From the way the question is posed, I assume the OP doesn't have a lot of technical background in relativity. However, there is no way to resolve ... 2 You can take a Reissner-Nordström solution for the charged non-rotating black hole, and put its mass$m=0$. Then it would become a so-called naked singularity. More precisely, singularity is a point where some value ends at infinity, while density of mass being just one option. For more thorough consideration of Reissner-Nordstrom and Kerr-Newman solutions, ... 2 I just read that statement from Wikipedia, and when they were mathematically calculating the density, they were using the Schwarzschild radius (aka radius of event horizon), not the singularity. The singularity itself, theoretically, has infinite density and exists in one point (radius approaches 0), so no, it won't float on water. Always remember when the ... 2 The laws of physics are local so if you change something on one side of your celestial body, it won't immediately affect other parts of the celestial body. In fact, your friend won't experience anything special even after the signals about the turning on of your machine get to your friend. He will just find himself on the surface of a collapsing neutron star ... 2 Uniform acceleration is not realistic in this problem. Why even include the text about the expansion of the balloon, if it is not going to be taken into account? The only way the balloon could continuously accelerate would be if it experiences no drag from the water. But even if we go with this greatly simplifying assumption, it is still far, far from ... 2 $$\rho=\frac{n*M}{V}$$ With$n$the number of moles of the gas,$M$the molar mass of the gas, and$V$the volume of the gas.$n$and$M$are constant, so: $$\rho_1=\rho_0*\frac{V_0}{V_1}$$ For$V$we have: $$V=\frac{n r T}{p}$$$n$and$R\$ are constant, so: $$V_1=V_0*\frac{T_1}{T_0}*\frac{p_0}{p_1}$$ So: $$\rho_1=\rho_0*\frac{T_0}{T_1}*\frac{p_1}{p_0}$$ ...

2

Wikipedia quotes Other substances that expand on freezing are silicon, gallium, germanium, antimony, bismuth, plutonium and also chemical compounds that form spacious crystal lattices with tetrahedral coordination. EDIT:The same paragraph says silicon dioxide also exhibits this property.

2

Multidimensional Dirac delta is usually defined as $$\delta_{3D}(x,y,z)=\delta(x)\delta(y)\delta(z)$$ You can find its Fourier transform as a convolution of constants, which will still appear constant. Alternatively, you can just use the definition and find triple integral, which will give the same result: ...

1

If I take a heavy lead object of arbitrary shape and mount it in a cardboard sphere such that the center of mass of the object coincides with the center of the sphere, you would never determine the lead object's shape nor mass density to be anything other than that of a point source using Newtonian mechanics or external field measurements. Referencing your ...

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