approximate mass of uranium in 1 bundle of fuel rods?
963632
Fuel rods come in different lengths.  You need to specify which size you want to know about.  Also, do you want to know about one rod or an entire bundle?  There are various configurations of bundles, also.  Need more specifics.\n\nAlso, the use of the rod needs to be known.  If it is a rod used by the military (navy nuclear) it will have a different pellet arrangement than a commercial (power plant) rod.\n\nDid you know that a fuel bundle can be handled before it is used?  Nuclear fuel needs to be inspected before it is placed inside of a reactor or a storage pool.  It is stood straight up on an inspection platform and technicians (wearing gloves of course) look the bundle over for any foreign material or obvious imperfections.  Quite cool, actually.
890320
Thermal conductivity: \nCopper: \n9.2*10-2 (kcal/sec)/((m^2)(Cdegree/meter))\nAluminum: \n4.9*10-2 (kcal/sec)/((m^2)(Cdegree/meter))\n\nWe need to assume the rods don't loose any heat to the surroundings. Therefore the kcals conducted through both rods is equal. To calculate the heat conducted per second by the copper, multiply its conductivity, \n\n9.2*10-2 (kcal/sec)/((m^2)(Cdegree/meter)),\ntimes (A*50Cdegree/.88 m).\n\nSet that expression equal to aluminum's conductivity times A*50Cdegree/L. Simplify and solve for L.\n\nEdit: I don't know why the conductivity lines are truncated. Maybe Yahoo is making sure it is compatible with any screen resolution. Anyway, the truncated lines, that I see, should end with \n")(Cdegree/meter))".
116087
keeping with the same unit fractor harry is correct.\n \n .2 ml
268969
not regurgitating from wikipedia, which isn't always correct OR understandable OR contains little information on the topic, here I go.\n\nUranium, as it's been said is a big atom which contains a whole lot of little pieces. There is something called "binding energy" that keeps all these little pices together. When a big atom undergoes all of this radioactive decay it goes from a smaller mass to a large mass. THe numbers are really small, but using Einstien's E=mc^2 any change of mass no matter how small will produce some meaningful amount of energy. Now, for 238 grams of Uranium, there are 6.023*10^23 atoms. So, you multiply the little miniscule energy emission by a very large number and you get a pretty big amount of energy.\n\nSo, short answer, becuase it is a big atom, it contains a lot of binding energy that is released when the atom splits.
26899
go to encarta.msn.com and search for keys mass/solid structure, surface Structure
754037
The moments of inertia for a rod of length L and mass m is given by:\n\nI = (mL^2)/3\n\n(a) I(total) = 3 x (mL^2)/3 = mL^2 = (240)(5.2)^2 = 6489.6 kgm^2\n\n(b) KE = (1/2)Iω^2, where ω=350 rev/min or 36.65 rad/sec\n\nKE = (1/2)(6489.6)(36.65) = 118,928 joules
442120
Whoa, man, why do you want to know? Do you know how suspicious you look after 9/11? \n\nthe critical mass for U-235 is 15 kilograms.
3627
one estimate is about 10^52 kg, that is:\n\n\n100000000000000000000\n 00000000000000000000\n 00000000000000000000\n 000000000000 \nkilograms\n\nWithout about 90% being the "invisible" missing mass stuff.
850091
I think that the answer must be YES.\n\nLet's assume that  these are dry cell batteries (so they are not losing any matter by evaoporation or gaining any by using oxygen in the atmosphere.\n\nSo no particles get in or out of the battery.  However, the battery, once spent, has lost some energy (perhaps by powering a torch which then gives out light and heat energy).\n\nAccording to Einstein's famous equation, E=mc^2.  So this amount of energy must be equivalent to (an extremely small) quantity of matter and the mass of the battery must have reduced by this amount.  It will not be measurable in ordinary circumstances and I am sure that other factors (eg dust particles settling on it, etc) would FAR outweigh this mass of the released energy.  But theoretically, yes, it would weigh less.\n\nOne gets a similar kind of balance in radioactive decay; a uranium nucleus decays into some other bits and pieces but the mass of the other bits and pieces is slightly less than the mass of the original uranium nucleus in order to account for the energy that is released (in the form of EM radiation).\n\nHope this makes sense.
1001801
The atomic number for plutonium is 94.  The weighted average value for the atomic mass of plutonium is 244.0642 grams/mole.
941617
The Specific Gravity of water is 1
109271
This is from Mr. Jeffy:\n\nIf you know that 60.4% of the naturally occuring isotopes have an atomic mass of 68.9257 amu, then you know that the other 39.6% much have an atomic mass such that the averge is 70.9249 amu.\nSo,\n(60.4% of 68.9257) + (39.6% of x) = 70.9249\nSo solve for x.\n============================================================\n\nI think he merely switched a couple of terms here, so I think the equation will look like this:\n\n(60.4% of 68.9257) + (39.6% of 70.9249) = x\n\nSolving:\n\n41.6311228 + 28.0862064 = (approx.) 69.72\n\nYou have gallium (Ga), my friend.
813427
The size of the metal would be dependant on many factors such as plane design and location of the explosion to name a few.\n\nthe problem with elemental rare earth metals is they are very reactive with water, making handling them in such a manner virtually impossible.\n\nAs far as the realative reactivity, the ammount of energy released can be calculated from the Gibbs free energy for each reaction.\n\nFor the reaction of sodium with water at room tempreture\n\nSodium with enough water would release 9260 KJ per pound of sodium while Potassium would release 2350 KJ per pound.\n\nI would suspect that at least 10 pounds of sodium to do any significant damage to a plane and 50 pounds of potassium.\n\nThis is a very unreasonable amount, although the cost would be less than 1000 dollars for this ammount of sodium, there are very few places that a large quantity could be obtained.\n\nAdditionally, sodium is so reactive, that I do not think that anyone could ever get it onto a plane before it blew. Lastly, that large of an ammount of sodium would fill a carry on bag, and would never get thru even the worst of TSA screeners.
883072
volume of the rod=3.14*1^2*20\nmass=50g\ndensity=mass/volume\n=50/(3.14*20)\n=0.796 g/cc
1054376
me=[ (Vx2)/450240] x W in grains = .2 ftlbs
9026
The fuel rods in a reactor are uranium-235.  Graphite is used in the control rods that slow down the reaction.
446055
Electrons per shell 2, 8, 18, 3