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8

Damping implies a loss mechanism. In liquids, where molecules move freely in close proximity, this loss mechanism is a transfer of momentum from one molecule to another. In pure crystalline metals, the position of atoms in the lattice is fixed, and the forces between them are elastic. That is, if an atom is displaced, it will experience a force that puts it ...


8

I'd like to point out the example of cast iron. It is renowned for its excellent vibration-damping properties. It is wrong to reach a blanket conclusion saying that metals are bad for vibration damping. The properties of any solid depends part on the material it is made up of and part on the micro-structure of the material. By micro-structure, I mean the ...


1

Metals are not liquids (I am referring to metals which are in solid phase in standard conditions of pressure and temperature) and have no viscous mechanical dissipation on time scales associated with most mechanical vibrations. This implies that they can transmit transverse waves contrary to fluids. How efficiently they can do so is related to the details of ...


0

The question made me google "water nano structure" and sure enough: Water: Nanostructure and fluctuations S. D. Zakharov Recently a model of local organization of water was experimentally justified, in which tetrahedrally coordinated water clusters of 1–2 nanometers arise and disappear in liquid composed of H2O molecules with partially broken ...


0

Although the average force applied during a collision might be small enough that an object can take it, the peak force applied can be much higher. In physics this is called impulse. Calculating the impulse for real world collisions (like a car crash) is very complicated. This is because cars have many structural members and the materials are not uniform. ...


1

To understand this, use the definition of force $\frac{d{\bf p}}{dt} = {\bf F}$, namely the force is equal to the rate of change of momentum. Something like a collision can be very complicated to model, but the average force is approximately given by ${\bf F}_{average} = \frac{\text{change in momentum}}{\text{time taken}}$. Typically, in a collision, the ...


1

An empirical answer: Metals (often copper) can be used as insulating support structures in superconducting magnets. Compared to ~0 resistivity of the coil, the resistivity of metals makes for very good insulating properties! Or, going in the other direction, a 50-watt VandeGraaff-type power supply may output 500KV at 100uA. Such a supply has an internal ...


0

As is often the case, the answer to this is actually slightly context-dependent. For many everyday purposes, the answer of Cort Ammon that it is all a matter of degree is correct. However, one other context worth mentioning is when condensed matter physicists speak of whether a particular state of matter is a "conducting" (or "metallic") state or an ...


2

I have no experience with either CVD or PLD, but it was interesting to think about this question. In a humble attempt to build on Peter Diehr's answer, here is some theory (at a very heuristic and simplified level). The deposition of each new layer can be thought of as being governed at large scales by some mixture of Laplacian and Eden growth in two ...


0

For what it's worth, lg310's seem to run a little cooler than gxb300's, at least on my roof. See also http://dx.doi.org/10.1016/j.egypro.2014.02.148 but read Table 4 carefully. The differences seem to be small.


8

Another way of distinguishing conductors and non-conductors or insulators is with band gap - for good conductors the fermi level of electrons is inside a band - semiconductors have a small band gap and good insulators have large band gaps... Electrons in solids lie in energy bands, whereas in atoms and molecules they have generally sharp levels. If you ...


20

As you have expected, there is no sharp divide between the groups. The divide is man made. Since all conductors have some resistance, (except superconductors - follow this link to find out more) and all insulators will conduct some current if they are forced to, this means there is no absolute dividing line between conductors and insulators. Since ...


3

I have a lot of experience with CVD and sputtering, but limited experience in PLD; however, several of my colleagues did this all the time in our shared a laser lab. When they were attempting to reproduce a specific result all of the parameters had to be systematically varied, from the laser fluence to the substrate conditioning and temperature, and more. ...


0

I wouldn't agree with your distinction between scratching and denting. There is some difference, but not so simple. If by denting you mean a plastic deformation - one that doesn't go away once you remove the external cause - then this involves some kind of breaking bonds (movement or emergence of impurities, etc.) If you mean an elastic deformation, then ...


1

You can drill a hole with a water jet, or compressed air, or even with light; I've done all of these. Feathers may be a challenge, but with enough feathers, and enough force, you can wear away the wall. A dent is a deformation of the surface; if examined before and after with an AFM, you might be able to find a dent by strain measurements.


0

My proposed explanation is in the thickness. We wish the headphone wires thin and shoelaces thick. The former favors fixing sharp bends by entangling them randomly, the latter favors straightening them. If we tolerated 5 mm thick nylon yarn around our headphones, they would certainly tangle much less.


0

The coefficient of restitution tells you about the energy lost in the collision. Specifically e^2 is the ratio of the kinetic energy after to before the collision in the zero momentum frame. This depends not only on the elastic properties of the material, but also the structure of the body. If you take a very simple example and have 2 springs hit each ...


2

Yes. Is this a useful method in practice to increase the strength to weight ratio of a cable? No, except possibly with the use of superconductors (copper would melt before contributing as much tensile strength as a cable of the same weight).


0

If you are able to make the switch from the conventional scalar representation of material properties to tensor representations, you will be better off. It will open up an entirely new way of looking at things, if you can handle the math.


0

You might want to target a book on $Characterization Methods$ also. John Hudson out of RPI taught an excellent course which was right to the point, but I don't know if he ever published the materials. I'm sure there are books that survey the field.


1

One has to think about what happens in the explosion. In a conventional explosion, a chemical reaction creates a whole lot of hot gas - that volume is initially contained inertially, and it expands as a shock wave travels outward. Any object on the boundary will experience a larger thermal and pressure gradient; the pressure and flow of matter behind the ...


2

short answer no longer answer it depends on what you mean by survive. The exposure to the intense heat is going to melt anything that is small enough to "tape to a Tsar-Bomba-yield nuclear warhead" the temperatures of fission reactions are used to induce fusion temperatures they are going to far exceed any threshold for a small piece of material and the ...


0

This is the engineering stress-strain graph. In this graph, engineering stress calculate by force/initial area. But, what you talk about is real stress that calculate force/instantaneous area. So, there is no wrong. After ultimate strength, the cross section area decreases and strain continues without force increasing. And engineering stress decreases. ...


3

At least a part of this comes from understanding where the stress-strain curve comes from. Normally from a physics background we think of applying a force to a sample and seeing how it responds. Instead, experimental results like what you show are done differently in materials science - the sample is mounted in the testing machine (Instron for example), and ...


1

While their momenta would be the same the bullet is going to have way more kinetic energy and so do more damage.


0

Every material. Well, at a certain angle, at least (the Brewster angle), where p-polarized light is not reflected and hence only s-polarized light is reflected. There is a modern interest in optical metamaterials and such things to engineer polarization reflections over wider angle ranges, though they tend to be limited in frequency.


3

The elongation, or shortening of the materials with temperature depends on the coefficient of linear thermal expansion of the material with which it is made (here: gold) $$\alpha = \frac{1}{L}\frac{dL}{dT}$$ Integrating, $$L = L_0\ e^{\alpha\Delta T}$$ $$L \approx L_0(1+\alpha\Delta T)$$ In your case, $L$ = circumference. Change in radius will be given by: ...



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