What is the approximate weight of oil in a container 3 feet high and 9 inches in diameter? (Oil weighs approximately 55 pounds per cubic foot.)
45 pounds.
82 pounds.
53 pounds.
73 pounds.What is the approximate weight of oil in a container 3 feet high and 9 inches in diameter?
Area of bottom of container = 蟺r虏
Converting this to square feet, area will be:
A = 蟺( (9/2) / 12)虏 = 0.4418 square feet
Volume is area x height
V = (0.4418 sq ft)(3 ft) = 1.325 cubic feet
Weight = volume * density
W = (1.325 cu ft)( 55 lbs / cu ft) = 72.9 lbs
Your answer will thus be
73 poundsWhat is the approximate weight of oil in a container 3 feet high and 9 inches in diameter?
73 pounds, assuming the container is cylindrical from the reference to its diameter.
The diameter is 9 inches or .75 foot. Halve it (to get the radius), square that and multiply by PI, to get the area of the base in square feet. Multiply that by 3 to get cubic feet, then by 55 pounds to get the weight of the oil.
So you need to calculate the volume inside the oil container.
The formula is Volume is
V = (pi) h r^2
you have d = 9 inches so first convert it to feet
d = 9 in * (1 ft/12 in) = (9/12) ft = (3/4) ft
r = d/2 = (3/4)/2 ft = (3/8) ft
V = (pi) (3 ft) (3/8 ft)^2
V =(27(pi)/64) ft^3
Weight of the oil = V * 55 lb/ft^3
W = (27(pi) ft^3)/64 *55 lb/ft^3
W = ~ 72.85 lb
9 inches = 3/4 feet diameter, so 3/8 feet radius
Volume = pi * r^2 * h
V = pi * (3/8 ft)^2 * 3 ft
V = pi * 9/64 square feet * 3 ft
V = 27/64 * pi cubic feet
V = approx. 5.3 cubic feet.
55 pounds/cubic foot implies approx. 71.5
The closest of your possible answers is 73 pounds.
Weight = Volume * Density
where
Volume = (pi/4)(9/12)^2(3) = 1.325 ft^3
Density = 55 lbs/ft^3
and substituting values,
Weight = (1.325)(55) = 73 lbs. --- this is the fourth option in your choices.
http://en.wikipedia.org/wiki/Cylinder_(g鈥?/a>
EDIT: I calculated with diameter instead of radius. I'm an idiot.
pi(.375)^2*3=1.325
1.325*55=72.89
So 73 pounds
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