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seattleite

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A good way to think about it is that the oceans rarely freeze, even when sea temperatures get below 20 degrees F. Because the salt content of oceans is relatively high, the freezing point of the water is greatly reduced.

I completely forgot chemically why this happens, but I think it has to do with the balance of water molecules with salt compounds.
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[center][spoiler=Dark's question]Imagine a sphere with radius 5 meters. At any given point on the sphere, the density is (5 - x) kilograms per meter cubed, where x is the distance from the center. This means at the center, the density is 5 kg/m^3, and all around the outside, the density is 0. Now, find the mass of the sphere in kilograms.[/spoiler]
[spoiler=My first attempt](The formula for finding the mass of a sphere is 4/3Ï€r^3d)
4/3 x π x 5^3 x 5 kg/m^3
(5^3 = 125)
4/3 x π x 125 x 5 kg/m^3
(4/3 x 125 = 166.66...)
166.66... x π x 5 kg/m^3
(round 166.66... to 167 for approximation)
(167 x 5 = 840)
Mass = 840Ï€ kg/m^3
(left in pi form)
2,637.6 kg/m^3
(approximated)[/spoiler]
[spoiler=Second attempt(double checking)]I did everything the same but instead of rounding 166.66... I made it into a fraction, and got this answer;
2,617.993878
I don't know which one is right, I checked with my science teacher who taught Calc once, and my math teacher. They triple checked my work and found no errors.[/spoiler]
Dark you bastard.[/center]
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It kind of needs to be exact. I could have estimated it myself.

[quote name='42.42.564' timestamp='1299085770' post='5045866']
integral(from 0 to 5) of (5-x)*(4/3)pi*(x^3) dx maybe.

name: 42.42.564
expertise: good at nothing..
[/quote]

That doesn't really take into account the density at each point.
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one clarification would be nice: At any given point on the sphere, the density [b](Of the hallow sphere the point is on, or just the point?) [/b] is (5 - x) kilograms per meter cubed, where x is the distance from the center

I think the set up would be entirely different if it is actually the density of a "point"
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Of the entire sphere. You are basically adding spheres from radius 0 to radius 5, each sphere having an increasingly lower density. When I say point, I am referring to any point that lies on the sphere with a certain density inside of the radius 5 sphere.
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I'm pretty sure it is that simple. First semester was just Calc I review and a few new concepts, but second semester isn't too much fun. I'll get through it, though.

Ask me more chem questions; I've been tutoring some sophmores and juniors taking chem, so it has been a real refresher of s[b][/b]hit I did two years ago.

but physics is moar fun
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  • 3 weeks later...

Name: Wolfy Eddy (AKA Eddy)
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*Math up to College AB Calculus (still studying BC part but I can still do a few questions on this section)
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High School Level:
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Told ya I was studying....so 2012 will probably hit ;)

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  • 2 weeks later...
* Name: Byakuya Kuchiki
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