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I had a similar question then realized that I could use graphine to do this. Now I think that if the airtight sealer was strong enough, instead of creating a vacuum in the way you described, you could instead make a collapsible frame of graphine inside as well, that would lock in the way your knee does. Then you could expand the frame with servo motors and ...

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Yes, and without a lot of extraneous information, simply do it in zero gravity Spacelab, with a water spraying apparatus. Procedure: Melt a floating glass blob by means of a couple gas jet burners, and by hand using jet aerodynamic force to keep the blob relatively fixed in space. Directing a spray of water droplets from several water nozzles perhaps ...

3

The steam pressure in the boiler will be greater than atmospheric, so the steam temperature will be above 100C. A typical ready-built boiler boiler for steam powered models (sold with a pressure test certificate!) would have a maximum working pressure of 4 or 5 bar, which would give steam temperatures around 150C. Many common plastics will start to soften at ...

2

I think I understand where you are stuck - it appears (I don't "speak" Matlab) you're trying to solve u directly, without adressing that u goes into Re. Here's what I'd do: Guess $u$ for the smallest diameter, calculate $Re$ from this $u$ to arrive at a new $u'$. Repeat/Reiterate by calculating $Re$ and a new $u'$ until the difference between $u$ and $u'$ ...

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I'm curious what material they currently use in these cellphone holders? The recrystallisation temperature of tin is about 30C, so conceivably it could work as a non-toxic alternative in lieu of lead.

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Generally I would recommend against heating a plastic container with a resistance heater (like a hot plate or mug warmer) unless you know for a fact it has a hardware shutoff that will kick in well below the melting point. Hardware failures do happen, and I've personally seen a runaway hot plate almost melt an unattended plastic container (full of alcohol!)....

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In this case, we say that $\epsilon_{axial}$ ($z$-direction in your diagram) is positive by convention. Then, for a normal material with a Poisson's ratio $\nu \ge 0$, $\epsilon_{transverse}$ ($x$- and $y$-directions in your diagram) will be negative. The fact that the material is isotropic means we have no way of distinctly labeling one direction $x$ and ...

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