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?
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- Thread Starter
- 22-09-2008 18:58
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- 22-09-2008 22:30
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)
ie Weight = mass x gravity
anyone feel free to correct me if im wrong.
- 23-09-2008 13:56
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