Asked by Jacob
A pile of sand that is stockpiled for winter road treatment is in the shape of a perfect cone. The beginning diameter is 110ft and the height at the center is 24ft. Assume wet sand weighs 3400 lbm/yd3. a) How many cubic yards of sand are in the pile? b) What is the mass o the sand in the pile? c) If the pile keeps the same conic shape during drawdown (ratio of center height to diameter stays constant, so the diameter shrinks by a constant ratio as the pile is drawn down), how many cubic yards are left in the pile when the center height is 10ft? d) If a dump truck holds 6 cubic yds, how many trucks could the pile fill as it is drawn down from 24ft to 10 ft? Assume the trucks are loaded to 90% full.
answers to a) V= 25342.18 yd3
b) Mass= 8618341.15 lbm/yd3
Cant figure out c). If i could then i could figure out d).
answers to a) V= 25342.18 yd3
b) Mass= 8618341.15 lbm/yd3
Cant figure out c). If i could then i could figure out d).
Answers
Answered by
Reiny
a) volume of cone (1/3)π r^2 h
= (1/3)π(55^2)(24
= 76026.54 cubic feet
= 2815.798 cubic yards , (I divided by 27)
b) Mass = 2815.798(3400) lbs
= 9573712.7 lbs
c) the volume of similar solids is proportional to the cube of their sides
volume of new/76026.54 = 10^3/24^3
new volume = 5499.605 cubic feet
or
203.689 cubic yards
check:
new radius/55 = 10/24
new radius = 22.916666..
new volume = (1/3)π(22.91666.^2)(10)
= 5499.6 as above
my answers are correct so far
I will leave the rest of the question to you.
= (1/3)π(55^2)(24
= 76026.54 cubic feet
= 2815.798 cubic yards , (I divided by 27)
b) Mass = 2815.798(3400) lbs
= 9573712.7 lbs
c) the volume of similar solids is proportional to the cube of their sides
volume of new/76026.54 = 10^3/24^3
new volume = 5499.605 cubic feet
or
203.689 cubic yards
check:
new radius/55 = 10/24
new radius = 22.916666..
new volume = (1/3)π(22.91666.^2)(10)
= 5499.6 as above
my answers are correct so far
I will leave the rest of the question to you.
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