A cylindrical brass sleeve is to be shrink-fitted over a brass shaft whose diameter is 3.215 cm at 0°C.?
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.
450194
Take 3" dia pvc and cut to 18" length. Add screw cap to one end and 1 1/2" x 30" pvc pipe to other end. Get gas grill push button lighter (Wal-Mart/Lowes/Home Depot) and drill hole in 3" pvc about 4" from screw cap and the same circumference as lighter body. Apply some pvc pipe glue around drilled hole and insert lighter. Allow to dry and then apply nut on underside of lighter body inside of pipe(nut supplied with lighter). Sharpen edge of 1 1/2" pipe end.Shove lemon or potato of larger diameter down pipe and tamp down about 18". Unscrew cap and spray hairspray into pipe. Re-cap and put on shoulder like bazooka or hold at side and push button. Can shoot well over 100ft. Pvc $5 Lighter $8 Glue $3 Hairspray $1.50 Lemons $2 Scaring the crap out of your neighbors...PRICELESS.
612367
Volume of cylinder = Area of the base X height = πr2h.
1065648
square inches doesn't matter. 3/4" is thicker than 1/2"
401461
permitter would be 83 cm\narea would be 415 cmXsq
562654
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
579127
what is the diameter....? Just multiply the diameter times pi. (3.14)
195018
Draw all this out so you can see it - it won't be to scale, but don't let that fool you.....\n\n\ndiameter = 50cm (radius = 25cm)\nchord = 20cm from diameter (thus it's 20cm from the center)\n\nOK, you'll need to assume that the line segment forming the diameter touches the chord at point B on the circle - i.e. we'll say the diameter is line segment AB and the chord is line segment DB. \n\nC is the center of the circle, so we have radius CB = 25cm. \n\nAlso, we'll have line segment CE drawn perpendicular from the center (C) to point E on the chord, forming a right angle and thus *bisecting* chord DB.\n\nYou have the hypotenuse of triangle ECB (radius CB) as 25cm, and side EC as 20cm (distance from the center to the chord - always perpendicular).\n\nLine segment EB will be the second side of a right triangle, so your formula:\n\nEC^2 + EB^2 = CB^2\n20^2 + EB^2 = 25^2\nEB^2 = 25^2 - 20^2\nEB^2 = 625 - 400 = 225\nEB = 15cm\n\nSince EB = half the length of the chord DB, DB = 2*15cm = 30cm.
409033
a vernier caliper should do the trick.  it can measure the inner and outer diameter
329557
Try this....\n\nhttp://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm
217916
the diameter cannot be 9cm when the radius is 12cm because Raduis (r) = Diameter (d)/ 2 = 9/2 = 4.5 .\nRadius is half of diameter
39411
Crank shaft with a bigger throw and shorter conecting rods.
175566
The only safe way do do it is to make a chamber cast and measure the results. There's a product called Cerrosafe, which is a low-melting metal, that you can melt in a double boiler (melts at about 180F) and pour into the chamber. As it cools it will shrink, at which point you can tap it out. Once it gets to room temperature, though, it will regain its original dimensions. Once it's completely cooled, grab a caliper or micrometer and start measuring.\n\nThe problem is that just because you know the caliber, you don't necessarily know what cartridge it takes. .38 special, .357, 9x19, 9x21, 9x23 and a few other calibers all use the same diameter bullet. Better to be safe than sorry by getting accurate information.\n\nSee the link below and search for "cerrosafe." It's on sale this week at Midway USA.
886387
half an inch---- 0.5 in.
795677
cooling it or pressurized it
301282
decreasing the heat to avoid degredation of the PVC
960889
Circumference of spool = 2PiR\n\nSince the length of the cable is l, the number of times the cable can be wrapped around the spool = l/(2PiR)\n\nEach "wrap" will cover the spool by 2r.\n\nTherefore the total length that can be covered = 2r*l/(2PiR)\n=rl/(PiR)