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They are wrong. There are no solid-state microstructural transformation processes active at those temperatures. If the knives you have are stainless steel, then there are no corrosion processes active at those temperatures either which might dull the cutting edge of the knife. If your knives are made from carbon steel (NOT stainless) then the knife will ...


3

They are relatively inexpensive, easily formed into useful shapes, nontoxic, and they conduct heat well. Please note that aluminum belongs on this list too, and that certain combinations of metals work even better (copper-cladding on the bottom of a stainless steel pan, for example).


2

Firstly, if the objects are made of the same material but are different colours then presumably they are painted or dyed or otherwise surface-pigmented. If that is the case then, unless the paint is very thin, the absorption of IR radiation will depend more on the paint than on the underlying material. Different paints can have very different absorption (...


1

The properties of semi-conductors arise from the structure of the energy levels that electrons can occupy in the material. Put simply, electron energy levels can be divided into lower energy valence band levels, where electrons are attached to a particular atom, and higher energy conduction band levels, where electrons can move throughout the material. ...


1

One can show that the attenuation of an E&M wave as it propagates in a material depends basically on the ratio $\sigma/\epsilon \omega$. The attenuation factor, usually denoted by $\alpha$, has a somewhat complicated expression but, in the case for instance of a low-loss dielectric, $\alpha\approx \frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}}$. Thus, when ...


1

It depends on the specific context. If we were talking about driving an electric current through metal, then the losses would be the energy dissipated by this electric current. This I think is the premise behind the question. However, if we talk about interaction with light, then a conducting media will incur losses, while a media with zero conductivity - i....


1

TLDR's point on the temperature-dependence of the diffusion coefficient is indeed an interesting point of departure: substrate temperature alters the crystal surface onto which you deposit, for instance it alters the diffusion coefficient D(r, T) 'experienced' by the arriving species. This means that, in straightforward terms, you can control the growth ...


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