What is the CTE (Coefficient of Thermal Expansion) for muscovite (mica)?
446916
It differs depending on whether you are speaking of parallell or perpendicular to the surface\nand\nat what temperature it is\n\n8 > 36(10-^6/K)   {@ 0 - 1000°C}\n\nparallel and lower temperatures = lower CTE\n\nCoefficient of expansion per °C parallel to cleavage:\n  0° > 200° C *  8x10^-6 >  9x10^-6\n200° > 500° C * 10x10^-6 > 12x10^-6 \n\nCoefficient of expansion per °C perpendicular to cleavage:   \n 20° > 100° C * 15x10^-6 > 25x10^-6\n100° > 300° C * 15x10^-6 > 25x10^-6\n300° > 600° C * 16x10^-6 > 36x10^-6
753296
|t - 78.3| ≤ 38.3\n\nSo either t -78.3 ≤ 38.3 whence t ≤ 116.6\nOR - (t - 78.3) ≤ 38.3\nwhence - t ≤ - 40 ie 40 ≤ t\n\nSo 40 ≤ t ≤ 116.6
1032149
Q=MC (delta) T....(5000 G)(.377 J/G degree C)(200 Degree C) which should be.....i get 377 J, which isn't your answer....i just had an exam on this last week, so i thought for sure i am right....the only thing i am not sure about for certain is the specific heat of coal, i got it from a website.....need more info if i am able to help, surely they provided the specific heat of coal....also if latent heat of fusion of vaporization has a part, the answer above is going to be different....sorry dude
905373
if u are referring to expansion due to heating than the order of expansion of the substances from lowest to largest is solid, liquids, than gases. But a solid's expansion is extremely low, almost impossible to measure unless its a metal.
276057
It's called the "Moho," which is short for "Mohorovicic Discontinuity."  It is named for the scientist who postulated its existence.  No one has ever drilled down far enough to see it, but that layer has certain unique physical properties.
804468
You need to look up the entropy for all compounds, multiply by the entropy by the stochiometric coefficients (e.g. 4 for NH3, 3 for O2). Then subtract the left side from the right side.\n\nVery simple actually, once you find the data.
331051
I listed some expansion coefficients\n\n\nMetal Temperature Range \n(oF) Thermal Expansion \n(microinch/(in oF)) \nAdmiralty Brass 68 - 572 11.2 \nAluminum  68 - 212 13.1 \nAluminum Bronze 68 - 572 9.0 \nBeryllium Copper 68 - 212 9.3 \nCast Iron, gray 32 - 212 5.8 \nCopper 68 - 572 9.8 \nCupronickel 68 - 572 9.0 \nHastelloy C 70 - 200 5.3 \nInconel 68 - 212 6.4 \nIncoloy 32 - 212 8.0 \nIron, nodular pearlitic 68 - 212 6.5 \nIron, pure 68 - 212 6.8 \nManganese Bronze 68 - 572 11.8 \nMonel 32 - 212 7.8 \nNickel Wrought 77 - 212 7.4 \nRed Brass 68 - 572 10.4 \nTin  32 - 212 12.8 \nTitanium 68 - 200 4.8 \nWrought Carbon Steel 70 - 800 7.8 \nYellow Brass 68 - 572 11.3
458863
You could start here:\n\nhttp://webmineral.com/data/Magadiite.shtml
14789
A general form of the equation I beleive you are looking for comes from a basic understanding of thermodynamics.  I am assuming that your ct=(Tfinal-Tinitial)...\n\nSo a general form of the equation would be:\nmCp(Tf-Ti)=Q \n\n"Cp is the specific heat"\n\nWATCH UNITS!!!!!  \n(I am ignoring the conversions in this statement)\n55g*Cp(67degrees UNITS)=343 cal \n\nSolving this equation will give you Cp ... just make sure that you watch units.
960401
You need to know the coefficient of thermal expansion of the brass. This depends on the exact brass alloy that is being used.\n\nAs an example, for brass alloy UNS C35000, the TCE is 20.3 * 10^-6 / °C.\n\nThe basic formula is DL/L = TCE DT where DL means change in length and DT means change in temperature.\n\nYou must expand the sleeve to a diameter greater than the shaft. That means it must expand by at least .02 cm. In general, for an easy slip fit you would want to have about 30 microns of additional clearance - that is .003 cm.\n\nThe fractional expansion is .02/3.195 = 6.3 E-3 for zero-clearance, or .023/3.195 = 7.2 E-3 for 30 microns clearance. \n\nThat corresponds to a temperature difference of 310°C for the zero-clearance case or 355°C for the clearance case.\n\nThese temperature differences are large enough that the assumption of constant TCE will no longer be valid. In particular, you could not achieve this temperature difference by cooling alone, since the projected temperature would be less than absolute zero. \n\nIn reality the amount of interference specified is about a factor of 5 greater than would normally be used.
993423
Don't know exactly, but because milk is mostly water, milk should have  a coefficient of expansion pretty close to that of water.
114715
change in diameter=0.10xcoef.of thermal elongation(350-20)=y\nnew dim.=0.1-(change in diam.)=z\n%age change in area=1-(z/0.1)**2
148711
This is an interesting question because there has been recent evidence found using geodetic surveys in southern and central Lousiana that that area is still falling as a possible result of rebound from the Wisconsin glaciation. \n\nI found a good map of geodetic measurements at Northwestern University.  It shows current measured rates of rebound in New England and Canada. It looks like New England is along the pivot line, with the coastal parts still falliing and inland areas rising. The rates are hard to read from the map but appear to be a few mm/year. \nhttp://www.earth.northwestern.edu/research/geodesy.html\n\nI'm also including some links related to the topic that I ran across while trying to find this map.
249904
Elements of Photonics, Volume II by Keigo Iizuka\nISBN: 0471408158\n\nThe above book has a table of elasto-optic constants.
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
399908
No, it's purely dependent on the concentration of dissolved material in the water (usually limestone/calcium carbonate). The more dissolved limestone, the more rapid growth of stagtites/stalagmites.
364418
Wow, what a question... This was the best I could come up with...\n\n"The characteristics of granitoid-related tungsten deposits hosted in siliceous (carbonate-free) rocks (e.g., Panasqueira, Cligga Head, Pasto Bueno) are reviewed and the ranges of physicochemical parameters of the ore-forming fluids are summarized. The two important tungsten minerals in these deposits are wolframite and scheelite, which were deposited mostly between 200° and 500°C and 200 and 1,500 bars. The salinities of the mineralizing fluids were typically less than 15 wt percent but commonly were significantly higher (up to 55 wt %). The two predominant dissolved components are Na+ and Cl– with subordinate Ca2+, K+, and carbonate species (/HCO3–). The contents of CO2 are highly variable, but XCO2 values typically range from 0 to 0.1. Limited pH and fO2 estimates indicate a moderately acidic fluid with oxygen fugacities between those of the QFM and HM buffers. These parameters were used to guide solubility and speciation modeling of W in hydrothermal fluids in granitoid environments."\n\nThanks for expanding my thinking!