New answers tagged

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Your force will transmit through the cube to the bottom and it will spread like a projection from your pressure point. having sand underneath it will give away a pression mark the size of your cube showing how your force transmitted. the height will no matter, being a solid cube and your force will not deformed it and lose itself in the process.


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So this is an oversimplification of the impact process, but if you model it as a spring damper system, then the deflection response is $$ x(t) = X \exp(-\zeta \omega_n t) \sin (\omega_n t \sqrt{1-\zeta^2}) $$ where $$\begin{align} \omega_n & =\sqrt{ \frac{k}{m}} && \mbox{: natural frequency, rad/s} \\ \zeta &= \frac{c}{2 m \omega_n} ...


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What change in the internal structure causes the transformation from elastic to plastic behavior? It depends on the material. For metals, small elastic strains are just the result of very small changes in the interatomic spacings. When more stress is applied, pre-existing dislocations in the metal start to move, causing re-arrangements in the atomic ...


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Plastic behaviour is characterised by there being permanent (non-reversible) deformations. In terms of molecules held together with springs (the bonds) in plastic deformation the springs are broken and the bonds then might the be between different molecules.


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when the water evaporates it goes up into the sky and comes back down as rain, this is commonly known as the 'circle of nature' no, the water would heat up The chlorine is denser at the bottom of a swimming pool copper would condense, the chlorine and sodium would disolve SOURCE: Jake Burn, Milk Churn, University of South Bradford


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Engineers created that problem. ;) (probably not) Many physics books use $Y$ for Young's modulus (Symon, Knight, Young & Freedman). Taylor's Classical Mechanics uses YM. Halliday, Resnick & -fill-in-the-blank- state that engineers use $E$. I suspect that physicists started using $Y$ for exactly this reason: to highlight a difference in the meanings ...


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Coincidence, nothing deep I'd say. Note that the equation representing the electric field modulus depends on the units you've picked and as such putting so much emphasis on the exact characters appearing in the eq. is senseless. Note that it's possible to form many physics equalities and equations involving 3 characters. E, epsilon and sigma are quite used. ...


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Yeah, that's just a coincidence. The easy way to see this is that $\epsilon$ is a relatively static property of a dielectric but a totally dynamic property of a stretching material.


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Just a coincidence. There are too many quantities and not enough letters. It probably does make a difference that the fields in which these two equations exist (material science and electromagnetism) are well enough separated that you typically won't see them both in the same papers or textbooks; if that weren't the case, people would start using different ...


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To construct a crystal you need a lattice and a basis. The lattice represents the translational symmetry of the system. Namely, graphene has a hexagonal lattice, meaning the two lattice vectors are 60 degress apart. Since the brillouin zone is constructed by inverting the lattice vectors, the brillouin zone is shaped based upon the lattice, but not the ...


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The answer lies in the band structure of the two materials. The band structure describes how the electrons in a solid are bound, and what other energy states are available to them. Very simply, the band gap for transparent diamonds is very wide (see this link): Normally, diamond is not a conductor: all the electrons live in the "valence band", and you ...


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Diamonds are unstable compared to coal (or more exactly, graphite) so high temperature and pressure are required for diamonds to form from graphite. The reason that coal (graphite) is black and diamonds are clear has to do with how the carbon atoms are connected together in the two different forms of carbon. In diamond each carbon atom is bonded to ...


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Yes, all Bravais lattices can be written as linear combinations of 3 vectors with integer coefficients. Your first discovery is that you can write FCC and BCC lattices as a conventional cubic cell with a basis.


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Thermal properties refer only to properties describing aspects of heat exchange whereas thermodynamic properties refer to processes relevant to aspects of thermodynamics. Thermal properties are therefore a sub class of thermodynamic properties. Now heat exchange, as is evident from the above definition refers to a thermal property, and is also a ...


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Wikipedia states that The high frequency response of vinyl depends on the cartridge. CD4 records contained frequencies up to 50 kHz, while some high-end turntable cartridges have frequency responses of 120 kHz while having flat frequency response over the audible band (e.g. 20 Hz to 15 kHz +/-0.3 dB).[5] In addition, frequencies of up to 122 kHz have ...


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Playing a vinyl "LP" implies a 33 rpm motion and 30 cm diameter. The highest frequency recorded will depend on the track velocity and the size of the needle. 30 cm diameter implies a 100 cm track length (roughly - less as you move further in) traversed in about 2 seconds - or 50 cm / second. The radius of the needle is specified in the standard as less than ...


1

Consider that the only frequencies present on the disk are spatial frequencies. The spatial information only gets transformed into the temporal domain when you spin the disk, and the scaling depends entirely on how fast you spin it. How fast can you rotate the disk? If a disk that nominally rotates at $33 \frac{1}{3}$ rpm is able to represent frequencies ...


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Facts : The properties depending of the atomic mass are different. fact 1 : The melting point depend upon the atomic mass. There is no need of a new experiment to relate the hardness to the temperature distance to the melting point : ie heat helps to bend metals or makes other matters more brittle. fact 2 : Moreover, stress diffusion depends upon the ...


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There is one small effect that has not yet been covered in the other answers. When we solve Schrodinger's equation for the electron orbitals we use the so called reduced mass $$\mu=m_e m_n/(m_e+m_n)$$ so the solutions for the orbitals will be slightly different for the case where extra neutrons are added to the nucleus. The electron mass is so much smaller ...


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The nuclear force is a contact force, with potential energy curve $$ V \propto \frac{e^{-r/r_0}}{r}. $$ The range parameter $r_0$ is roughly one femtometer. Nuclei in a solid are typically $10^5\rm\,fm$ apart, so the nuclear interaction between nuclei from different atoms is astoundingly suppressed. If you think of a solid as a lattice of atoms connected ...


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Drawing on my chemistry background and asking a few of my Ceramic Engineering friends, the answer is "No". We had the consensus that the nucleus does not impart any physical property on the hardness or strength but the "electron cloud" and its interactions. The electrons are not affected in any way by the different in isotopes.



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