SOLUTION: Problem- I have 50,000 gallons of water. If the water is stored in a sphere, what must the radius of the sphere be? I'm trying to figure out the steps and conversions necessary to

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Question 156475: Problem- I have 50,000 gallons of water. If the water is stored in a sphere, what must the radius of the sphere be? I'm trying to figure out the steps and conversions necessary to answer this problem.
Found 2 solutions by gonzo, Earlsdon:
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
from the internet it was determined that 7.48 gallons of water equals 1 cubic foot in volume.
this translates to .13368984... cubic feet per gallon of water.
you have 50000 gallons of water so the volume would be 50000 * .13368984... = 6684.49491979... cubic feet.
to put it in a sphere you then take the formula for the volume of a sphere and works backwards to get the radius of the sphere.
formula for volume of a sphere is 4/3 * pi * r^3.
formula becomes volume = 6684.4949... cubic feet = 4/3 * pi * r^3
becomes r^3 = 6684.4949 ... / ( 4/3 * pi )
becomes r^3 = 1595.804911 cubic feet.
becomes r = cube root of 1595.804911
becomes r = 11.6858... feet.
pi is a constant with value of 3.141592654...
radius is 11.6858... feet.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
I would first convert the 50,000 gallons to cubic inhes.
1 gallon = 231 cubic inches. So...
50,000 gallons = 50,000(231) cubic inches
50,000 gallons = 11,550,000 cubic inches.
Now if you want to store all that water in a spherical container, it will need to have a volume of 11,550,000 cubic inches, right?
The volume of a sphere is given by:
V+=+%284%2F3%29%2Api%2Ar%5E3 Solve this for the radius,r.
r+=+root%283%2C3V%2F%284%2Api%29%29 Substitute V = 11,550,000 and evaluate.
r+=+root%283%2C3%2A%2811550000%29%2F%284%2Api%29%29 Use pi+=+3.14 as an approximation.
r+=+root%283%2C%2834650000%29%2F%2812.56%29%29
r+=+root%283%2C2758757.96%29
r+=+140.25inches, approximately, or...
r+=+11.69feet, approximately.