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Gold will compress to about half of its volume at atmospheric pressure if you compress it to 2 million atmospheres at room temperature, which is something that I'm sure has been done with diamond anvil cells. For many metals, the atomic lattice will also undergo structural phase transitions from one lattice type to another at certain pressures, but I don't ...


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I would say that gold will remain gold and do not go into nuclear fusion. However I am not saying that it is impossible. You may have heard of neutron stars and black holes, when the matter is under immense pressure the electrons around the nucleus could not withstand it and they fall into the nucleus and convert everything in a giant neutronic mass. ...


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$$\% Ionic Character = 100(1 - e^{-({\frac{\Delta{\chi}}{2}})^2})$$ The factor of 0.25 comes out when you simplify because of the electronegativity difference $\Delta \chi$ is halved before squaring. The formula is stated empirically by Pauling and has NO proof.


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Consider Fick's second law of diffusion, in one dimension, where $u$ is the concentration of the diffusing gas (in $\mathrm{mol/m^3}$) and $D$ the diffusion coefficient: $$\frac{\partial u}{\partial t}=D\frac{\partial u}{\partial x}$$ If we assume the concentration of gas outside the container to be much smaller than inside (a reasonable assumption), then ...


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The short answer would be the empirical Matthiessen's rule: the total resistivity of a crystalline metallic specimen is the sum of the resistivity due to thermal agitation of the metal ions of the lattice and the resistivity due to the presence of imperfections in the crystal (scattering). There are of course deviations from that rule: it assumes that ...


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Dielectric materials have a loss tangent : at a given frequency, the response to EM radiation will have a phase lag relative to the incident wave. For pure dielectric materials, the phase angle is zero; when there are losses in the material, the loss angle will be non-zero. For a given loss angle (usually given as the tangent of the angle) the amplitude of ...


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While friction plays a good role, the molecular structure of each metal plays a greater role. Just imagine how energy released by dry wood in the form of sound differs to the same wood if it is green when suddenly cracked! Aluminium is more malleable than iron. It is something to do with crystal structures, when the bonds are abruptly broken through applied ...


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Atomic concentration is the number of atoms that can fit into a given volume and therefore has the dimension of inverse volume. To calculate the atomic concentration one needs to know two things: the lattice constant to calculate the volume of the unit cell and the number of atoms that can fit into the lattice. The number of atoms that can fit into a ...


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The results you get in the experiment depend on the dimensions of the sample. Are T and W much smaller than L? The shape of the sample is also important in another way. Is it a dog bone shape, with a short middle section length being used as the sample length? If T and W are small compared to L, then the cross sectional area used in calculating the true ...


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Your question is short on details, so to answer your question I am assuming a particular configuration. Hope it helps with whatever your actual configuration is. A closed porous container consists of a gas at partial pressure $p_1$ which is less than partial pressure $p_2$ of that gas in the ambient. We shall assume that $p_2$ is constant and to further ...


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The equation you wrote is for steady state flow through a porous wall of thickness d. Q is the volumetric flow rate. Q/A is the so-called superficial velocity. The equation inherently assumes that the pressures on both sides of the wall are constant, and not varying with time. The only way that you would get an increase in pressure difference with time ...


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After more reflection, I realized that this fact can be easily demonstrated for Poisson Ratios in the range $ -1 \leq \nu < 0.5$: We are given $ \nu=-\frac{\Delta{W}/W_0}{\Delta{L}/L_0}=-\frac{\Delta{T}/T_0}{\Delta{L}/L_0}$ where $ -1 \leq \nu < 0.5$ Let $ \frac{\Delta{L}}{L_0}=\alpha > 0$ then $ \begin{cases} \Delta{W}=\frac{W_0*\nu*\Delta{L}}{...


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Heat passes between the steel and the paint by conduction, which is typically much faster than radiation. The limiting factor is the radiation at the object:air interface i.e. you need the emissivity of the surface at the object:air interface. So the emissivity you need is the emissivity of the paint. This also explains why your colleagues are correct to ...


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Your analysis for an incompressible material is just done incorrectly. It should read $$\frac{T}{T_0}=\frac{W}{W_0}=\sqrt{\frac{L_0}{L}}$$. So, for small strains, $$\frac{\Delta T}{T_0}=\frac{\Delta W}{W_0}=-\frac{1}{2}\frac{\Delta L}{L_0}$$. So the Poisson ratio of an incompressible material is equal to 1/2.


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Your calculation makes no reference to the properties of any real material. Therefore the conclusion you make is the consequence of your assumptions, not any properties of matter. Those assumptions (constant volume and surface area) can be wrong, as you acknowledge.


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$A \approx \frac{W_0*L_0}{L}*\frac{T_0*L_0}{L}=\frac{W_0*T_0*L_0^2}{L^2}$ If my interpretation is correct, you are assuming that $W*L=W_0*L_0$ and $T*L=T_0*L_0$ That would make the volume: $W_0*T_0*L_0^2/L$, which decreases when you stretch the material. For small strain, the Poisson ratio would approach 1. Poisson ratio should be between -1 and 0.5 for a ...


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I think some cargo holders generally aren't concerned about maintaining pressure, most things survive, it's only when pets and people are involved that pressure becomes an important aspect. If the air pads (I guess you're talking about large bubble wrap) were filled with enough air to be tense on the ground the reduced air pressure at 10,000ft would have a ...


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First, I'd like to point out that at this point the Dyson Sphere is purely theoretical in nature. Second, currently all plans to build a Dyson Sphere are, in the words of this website, far beyond humanity's engineering capacity. Portions of the technology involved in the Dyson Sphere have been developed, however, such as solar sails (a method of ...


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Hard though it is to believe, pH does have an effect on sound absorption in water. There are some reactions that are affected by pressure, that is pressure changes their equilibrium. One example is the equilibrium between boric acid and the borate ion: $$ B(OH)_4\,^- + H^+ \rightarrow B(OH)_3 + H_2O $$ Increasing pressure pushes the reaction over to the ...


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The author should have been more specific on the matter or at least provide a reference. The speed of sound in water depends on the bulk modulus and density of the water, so in the open oceans the factors that most affect the speed of sound are salinity, depth (pressure) and temperature. I was an ocean engineering major and have taken courses in physical and ...


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Abrasive is intended to increase friction. You want your brake pads to have (and retain) a certain shape, so you use some structure material that resists well to changes in temperature. Not sure what performance material is supposed to be. In order to limit the costs, in places where no structural integrity is threatened, and no friction exists, you can use ...


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I tend to agree with Sanya in that I am not sure about the universality of this. There might of course be instances where this is the case. A pure metal has a periodic lattice of ions. There is then a conduction band of electrons that fills the space between the ions. These electrons have wave vectors in the reciprocal space. In space the occurrence of ...


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First of all, I want to see an (experimental) proof that any metal has a higher resistance than any alloy (at any pressure, temperature and volume). What I presume your teacher might have wanted to hear is something along the following lines: a perfect, perfectly static crystal would be, if I remember correctly, perfectly transparent to an electron, so there ...



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