At 4/18/08 04:26 PM, BonusStage wrote:
At 4/18/08 04:19 PM, simon wrote:
area = 58.09 m^2
It's a sphere and yet it's meters squared?
Please go on.
i didn't want to embarass the OP, but i know he wants this equation solved for Neopets.
the prize was a Boo Korbat Underwater, pathetic.
Round 258
Date Released: April 17th, 2008.
Date Rewarded: April 24th, 2008. (estimated)
A Chomby was walking down a path one day when Jhudora appeared in front of it. "Answer this question, and you may pass," she said.
Suppose you have a perfectly spherical water tank with an inside diameter of 8.6 metres. If the drain at the bottom of the tank can't handle a hydrostatic pressure of more than 50 kilopascals, what is the maximum volume of water, in litres, that can be contained in the tank? Assume that gravitational acceleration is exactly 9.81 m/s2. Please round to the nearest 10 litre increment, and please submit only a number for your answer. (For example, if you calculate the answer to be 16277 litres, submit 16280 as your answer)
source.
so yeah..
First off, we need to find the maximum height of the water tank that the drain can handle using hydrostatic pressure equation [1]:
P = rho * g * h <=> h = P / (rho * g)
where P = 50k Pa = 50,000 kg/m·s2 is the maximum hydrostatic pressure, rho = 1000 kg/m3 is density of water, g = 9.81 m/s2 is the gravitational acceleration. Remember that do not round h yet.
Then, we can start to calculate the partial volume of the sphere by the following formula (explanation of the formula is at the end of the solution):
V = 1000 * pi * h2 * (r - h/3)
where r = d/2 = 4.3 m, is the radius of the spherical water tank, h is the above maximum height.
We know that: 1 m3 = 1000 dm3 = 1000 liters. We have a 1000 factor in the above formula because the answer should be in liters while we calculated in cubic meters. Now perform rounding as the requirements and you will get the correct answer.
If you are interested to see how the partial volume of a sphere is derived, you can continue to read. Otherwise stop here.
We can consider the sphere with center at origin (above picture). We cut the partial sphere into small slices. Each slice is a circle, perpendicular to the vertical direction. Using Pythagorean theorem [2], we can calculate area of each circle at a certain height x:
Then we do integration on the whole vertical direction to add all area of the slices to get the volume. Since we choose the origin at the center of the sphere, the limit of the integration should be from -r to h-r
proceed clicking for hints.
this guy just wanted this shitty thing vv
