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# Physics watch

1. I'm stuck with some physics homework. Teacher's given us questions on a topic she hasn't even taught us yet.

Explain how the following expression for upthrust is derived.
4/3nr^3pg where n is pi, r is radius of a falling particle through a fluid, and p is the density of the fluid. What is the question asking me to do?
2. What syllabus are you on? Because I havent covered it (im on edexcel)

Im assuming the question is asking you to find the base units.

so basically 4/3.Pi are constants, so they are not derived from anything

r^3 is measured in metres, and would therefore be m^3

p = density
therefore using the equation
p = m/v (mass over volume) which is kg/m^3 or kgm^-3 (over volume would make it become to -3)

g im assuming is gravitation strength and would be ms^-2

Using all of the above:

m^3 . kg m^-3 . ms^-2

(m^3 would cancel out with m^-3)

leaving

kg ms^-2

ie Weight = mass x gravity

anyone feel free to correct me if im wrong.
3. upthrust is the weight of displaced water.

so use the equation W=mg

W is upthrust so upthrust=mg

mass is density into volume so

upthrust = density (p) x volume (v) x g.

assuming that the particle is a sphere,
volume of sphere = 4/3nr^3 (n is pi)

so now the equation looks like this

upthrust= volume xdensity(p) x g
= 4/3nr^3 x p x g
= 4/3 nr^3pg.

hope i am correct

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Updated: September 23, 2008
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