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[Image of Chemical Symbol for Lead]. (n.d.) Retrieved from . com/scene/771788150451208192 b. Solve for the minimum volume of soil that will be excavated in cubic yards? The land owner will have to excavate and remove 183 cubic yards of the lead contaminated soil from the 30 feet by 55 feet area in order to clean up the site. In order to solve for the minimum volume of soil that needs to excavated in cubic yards then you must multiply the length of the contamination by the width of the contamination and then multiply by the given depth. The following formula is: Length x Width x Height = 30 ft x 55 ft x 3 ft = 4950 ft 3 Convert: (1 cubic foot = 0.03703704 yards) = 4950 ft 3 x 0.03703704 y 3 = 183 y 3 of contaminated soil. c. If each dump truck can transport 18 cubic yards, determine how many dump trucks loads will be transported? For calculation purposes, add a 15% "fluff factor" (add to the volume that will be transported). Each dump truck can haul 18 cubic yards of lead contaminated soil with a 15% fluff factor, then it would take 12 dump trucks to haul all of the contaminated soil away from the site.
The following formula was used to determine how many dump trucks would be required to remove the contaminated soil: 183 y 3 x 15% = 27.5 y 3 (fluff factor) 183 y 3 + 27.5 y 3 = 210.5 y 3 to haul away 183 y 3 ÷ 18 = 11.7 It will take 12 Dump trucks d. If the bulk density of soil is 1350 kg/m3 (84.3 lb/ft3), solve for the weight of the soil that will be transported to a disposal site in kilograms. The weight of the soil in kilograms that will be hauled to the disposal site is 189,330 kg. The following formula is used: Density = Mass ÷ Volume (183 cubic yards is converted to cubic feet in order to solve for mass) 183 y 3 ÷ 0.037037 y/f = 4941 f 3 Density = Mass ÷ Volume 84.3lb/ ft 3 = Mass ÷ 4941 f 3 4941 f 3 x 84.3lb/ ft 3 = 416,526 lb 416,526lb ÷ 2.2 = 189,330 kg

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• Fall '10
• GUAN
• Occupational safety and health, Material safety data sheet, United States customary units

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