Answer:
load factor = 0.782
Shrink Factor = 0.833
no of truck is 62500
Explanation:
given data
soil weighs in situ condition = 2,520 lbs/CY
soil weighs in loose condition = 1,970 lb/CY
soil weighs in embanked state = 3,025 lbs/CY
average volume = 16 LCY
soil from a borrow pit = 1 million CCY
solution
first we get here Load Factor that is express as
load factor = [tex]\frac{1,970}{2550}[/tex]
load factor = 0.782
and Shrink Factor will be as
Shrink Factor = [tex]\frac{2520}{3025}[/tex]
Shrink Factor = 0.833
and
no of truck will be
no of truck = [tex]\frac{1000000}{16}[/tex]
no of truck is 62500