What is the diameter of a 5 pound sphere of lead?
152293
1 pound = 0.454 Kg.\nDensity of lead = 11350 Kg/metre cube.\nWe Know, Volume= Mass/Density.\nHere, Volume=0.454/11350=0.00004 m.Cube.\nalso , volume of sphere = (4*pai*radius cube)/3\nso, radius cube= (0.00004*3)/(4*3.14)\n               = 0.0000096\n    Radius     =0.0213 metre\n Diametre      =0.0213*2= 0.0426 metre.\n\nI think , formulae used are correct but some doubts with the calculation, as i didn't have a scientific calculateor with me , at the time of answering.
1013311
according to the International Tennis Federation the official diameter as "more than" 2 and1/2 inches but "less than" 2 and 5/8 inches. The weight is "more than" 2 ounces but "less than" 2 and 1/16th ounces
941899
Tennis balls should have diameter between two and a half inches (6.35cm.) and two and five-eighths inches (6.67cm.)\n\n\nFor golf balls, minimum diameter is 42.67mm
927521
1/2bh\n1/2(8.5)(5.5)\n23.375
1008732
Ksp = [Pb+2]*[Cl-]^2/[PbCl2]\n\nby definition, [PbCl2] = 1\n\nAlso, based on the reaction:\n\nPbCl2 =====> Pb+2  +  2 Cl-\n\n[Cl-] = 2 [Pb+2]\n\nSubstitute this into the equation, and you get:\n\nKsp = [Pb+2] * [Cl-]^2 = [Pb+2]*(2[Pb+2])^2 = 4[Pb+2]^3\n\nNow, since you know that 0.99 grams of PbCl2 is dissolved in your liter of solution, you can calculate the molarity from:\n\n(0.99 g)/ (278.1 grams/mole) = 3.56x10^-3 moles = [Pb+2]\n\nKsp = 4*(3.56x10^-3)^3 = 1.8 x 10^-7\n\nThe text value for the Ksp of lead (II) chloride is 1.7 x 10^-5\n=================\nOkay, now for the second part of your question:\n\nIf you have 0.5M lead nitrate, which is soluble, you will have 0.5M Pb+2\n\nSo, let's plug 'n chug (we'll use the text value for Ksp to show the process):\n\nKsp =  1.7x10^-5 = [Pb+2]*[Cl-]^2\n\nbut, with the lead nitrate in solution, [Pb+2] = 0.5 moles/liter (the amount you'd get from the dissolution of lead chloride will be so small as to be negligible), so:\n\nKsp = 1.7x10^-5 = [0.5]*[Cl-]^2\n\n[Cl-]^2 = (1.7x10^-5)/(0.5) = 3.4x10^-5\n\nso [Cl-] = sqrt(3.4x10^-5) = 5.83 x 10^-3  and since [Cl-] = 2 [Pb+2], we halve that value to get:\n\nsolubility = 5.83x10^-3/2 = 2.92x10^-3 moles per liter\n\n(2.92x10^-3 moles/liter) * 278.1 grams/mole = 0.811 grams/liter\n\nIf you use the value of the Ksp calculated in the first half of the question, then the solubility of PbCl2 in your solution is 3x10^-4 moles/liter, or 0.083 grams/liter
470650
diameter = 9 inches\n\nRadius = 4.5 inches\n\npi * r^2 = 3.14159 * 4.5^2 = 63.61725...
21871
On average, the balls are 2.25" in diameter and all balls weigh 5.5 oz. except or the cue, which weighs 6 oz. According to official BCA equipment specifications, the weight may be from 5 1/2 to 6 oz. with a diameter of 2 1/4", plus or minus .005".
122964
I believe it's 5 1/2 inches
332727
I would assume that lead pellets would be heavier than aluminum. Try placing them in a liquid which is buoyant for aluminum and the lead should fall to the bottom. Thick Veg Oil or something? A magnet might work too. Not sure though.
467744
Circumference between 9 and 9.25 inches and a weight between 5 and 5.25 ounces.
130046
Half of the diameter.\n\nor\n\nMeasure around it.  Divide that number by 3.14.  Then divide by 2.
25489
It depends on what study you look at, but the general consensus is about 5.5 inches.
69568
None, unless you pour something into it, then it is a measure of 1.5 oz. \n\nHope it is something yummy.
2672
About 0.1 microns.  This would be about 0.000394 inches.
14869
Realistic average size is 5.5" to 6.5"
1081689
5 feet 3.7 inches and 110.23 pounds
1055600
the average is 4.5 to 5.5 inches around.