SOLUTION: A 3-inch diameter hole is drilled 6inches deep into the center and normal to the top face of brass cube 8 inches on the side. The hole is filled with lead. If a brass weighs 525lb/
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Question 1181029: A 3-inch diameter hole is drilled 6inches deep into the center and normal to the top face of brass cube 8 inches on the side. The hole is filled with lead. If a brass weighs 525lb/ft^3 and the lead weighs 705lb/ft^3. Find the total mass of the two metals Found 3 solutions by mananth, htmentor, ikleyn:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! mass in cylindrical hole = pi * r^2 * h * density of lead in ounce/cu.in
Mass= Density * volume
1lb/cft = 0.009 ounces/cu. in
705 lb/cft = 6.35 ounces / cu. in
pi*1.5^2 *6*6.35= 269 ounces
Now volume of remaining cube after drilling
8^3 -269= 243 in^3
weight
243 * 4.86 =1181 ounces
You can put this solution on YOUR website! The mass of brass is equal to the volume of the cube minus the hole times the
density of brass
The mass of lead is equal to the volume of the hole times the density of lead
The volume of the cube = 8^3 = 512 in^3
The volume of the hole = pi*r^2*l = pi*(3/2)^2*6 = 41.412 in^3
Convert the densities to lb/in^3. 1 ft^3 = 1728 in^3
525 lb/ft^3 = 525/1728 = 0.30382 lb/in^3; 705 lb/ft^3 = 0.40799 lb/in^3
Thus the mass of brass = (512 - 41.412)*0.30382 = 142.974 lb
The mass of lead = 41.412*0.40799 = 16.896 lb
