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    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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    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.
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    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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