document.write( "Question 1170913: Using 1 lb. of lead, and allowing 5% for waste, determine amount of shot 1/16 in. in diameter that can be produced.\r
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document.write( " Lead weighs .410 lbs. per cu. in.\r
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document.write( " Lead shot are spheres.\r
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document.write( " Sphere volume = 4/3 * pi * r^3\r
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document.write( " 1.333 * 3.1416 * .03125 * .03125 * .03125 = .0001278 cu. in.\r
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document.write( " 5% waste = .95 lb. remaining.\r
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document.write( " Not sure how to proceed.
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Algebra.Com's Answer #795810 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! Using 1 lb. of lead, and allowing 5% for waste, determine amount of shot 1/16 in. in diameter that can be produced.\r \n" ); document.write( "\n" ); document.write( " Lead weighs .410 lbs. per cu. in.\r \n" ); document.write( "\n" ); document.write( " Lead shot are spheres.\r \n" ); document.write( "\n" ); document.write( " Sphere volume = 4/3 * pi * r^3\r \n" ); document.write( "\n" ); document.write( " 1.333 * 3.1416 * .03125 * .03125 * .03125 = .0001278 cu. in.\r \n" ); document.write( "\n" ); document.write( " 5% waste = .95 lb. remaining.\r \n" ); document.write( "\n" ); document.write( " Not sure how to proceed. \n" ); document.write( "=================================== \n" ); document.write( " Sphere volume = 4/3 * pi * r^3\r \n" ); document.write( "\n" ); document.write( " 1.333 * 3.1416 * .03125 * .03125 * .03125 = .0001278 cu. in. \n" ); document.write( "Mass of each sphere = .0001278 cu. in. * .410 lbs. per cu. in. \n" ); document.write( "=~ 5.2398e-5 pounds per piece. \n" ); document.write( "=============== \n" ); document.write( "0.95/5.2398e-5 = 18,130 spheres \n" ); document.write( " \n" ); document.write( " |