SOLUTION: Using 1 lb. of lead, and allowing 5% for waste, determine amount of shot 1/16 in. in diameter that can be produced. Lead weighs .410 lbs. per cu. in. Lead shot are spheres.

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.
Lead weighs .410 lbs. per cu. in.
Lead shot are spheres.
Sphere volume = 4/3 * pi * r^3
1.333 * 3.1416 * .03125 * .03125 * .03125 = .0001278 cu. in.
5% waste = .95 lb. remaining.
Not sure how to proceed.

Answer by Alan3354(69443) (Show Source): 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.
Lead weighs .410 lbs. per cu. in.
Lead shot are spheres.
Sphere volume = 4/3 * pi * r^3
1.333 * 3.1416 * .03125 * .03125 * .03125 = .0001278 cu. in.
5% waste = .95 lb. remaining.
Not sure how to proceed.
===================================
Sphere volume = 4/3 * pi * r^3
1.333 * 3.1416 * .03125 * .03125 * .03125 = .0001278 cu. in.
Mass of each sphere = .0001278 cu. in. * .410 lbs. per cu. in.
=~ 5.2398e-5 pounds per piece.
===============
0.95/5.2398e-5 = 18,130 spheres

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