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Physics n00b

  1. Feb 3, 2005 #1
    I know nothing bout this stuff, lil help? :P

    I have no idea what this means

    What are the dimensions of viscosity η if the equation for the force on a sphere moving through liquid is F=6πηrv where r is the radius of the sphere, v is the speed and F is the force?

    or this...

    Show that the equation for the final velocity of an object undergoing uniform acceleration is dimensionally consistent.
    v² = u²+2as

    could someone help me out and explain what I need to do for these questions :smile:
     
  2. jcsd
  3. Feb 3, 2005 #2

    arildno

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    Do you know what "dimension" means in this context?
     
  4. Feb 3, 2005 #3
    no I dont :(
     
  5. Feb 3, 2005 #4

    arildno

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    If I ask you what "units" does force have, does that help?

    In this context, "dimension" means "units"
     
  6. Feb 3, 2005 #5
    force is newtons or something right?
     
  7. Feb 3, 2005 #6

    arildno

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    Absolutely right!

    In this case, try to write "newtons" in terms of the base type of units "kilograms" "meter" "second"

    Also set up what units velocity has in these base units.
    Post your work.
     
  8. Feb 3, 2005 #7
    velocity is metres/second/second isnt it? ms^-2 ???
     
  9. Feb 3, 2005 #8

    arildno

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    Nope, that's acceleration".

    Velocity measures how far (a length!) you travel over a period (that's a time!).

    So, the units of "v" is [tex][v]=\frac{m}{s}[/tex]
    Now, the units of force is newtons, which also can be written as:
    [tex][F]=N=kg*\frac{m}{s^{2}}[/tex] (mass times acceleration!)
    (I use the bracket notation, say [F], to mean I'm interested in the UNITS of F, not its particular measured value. Okay?)

    So, your first equation was:
    [tex]F=6\pi\eta{rv}[/tex]
    You need to end up with the same type of units on both sides of the equation, so you must have:
    [tex][F]=[\eta][r][v][/tex] ([tex]6\pi[/tex] is just a number without dimebsion)
    Now, let's fill in:
    [tex]kg*\frac{m}{s^{2}}=[\eta]\frac{m^{2}}{s}[/tex]

    What units is therefore [tex][\eta][/tex] ?
     
  10. Feb 3, 2005 #9
    η= kg (ms)/(s^2*m^2)

    is that right?
     
  11. Feb 3, 2005 #10

    arildno

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    Yep!
    But try simplifying that expression a bit!
     
  12. Feb 3, 2005 #11

    arildno

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    Oops!
    Just to make it absolutely clear, what you've found, are the DIMENSIONS of [tex]\eta[/tex]
    that is [tex][\eta][/tex], not [tex]\eta[/tex] itself..
     
  13. Feb 3, 2005 #12
    but thats what I needed to find in the question anyways :P

    η= kg*1/ms? is that the simplified one?

    what about that other question, how do I do that?
     
  14. Feb 3, 2005 #13

    arildno

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    You've got the simplified one correct: [tex][\eta]=\frac{kg}{ms}[/tex]
    You have only been asked to find the UNITS, not the numerical value of the viscosity.
    (That's just the same as saying that all lengths can be measured in "meters", but that doesn't mean the lengths can't have different values (say, one length is 2meters, another 3meters)

    As for your second:
    Each term in your equation must have the same dimensions.
    You have three "terms:
    [tex]v^{2},u^{2},2as[/tex]
    I'll leave it to yourself to verify that these three terms have, indeed, the same dimensions.
     
  15. Feb 3, 2005 #14
    so that answer for the 1st would be correct considerin what the question is asking?

    Im still not sure what I need to do for that second one

    v^2 = final velocity
    v^2 = (m/s)^2

    u^2 = initial velocity
    u^2 = (m/s)^2 also

    2as = 2*acceleration*distance
    2as = 2((m/s^2)*m) = 2(m^2/s^2)

    umm, am I way off? or kinda on the right track?
     
  16. Feb 3, 2005 #15
    oh wait

    its just the terms right
    so

    as=(m^2/s^2) = (m/s)^2
     
  17. Feb 3, 2005 #16

    Integral

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    Looks like you have it figured out. Good job.
     
  18. Feb 3, 2005 #17
    yay! thankyou
    I got a few other questions, I think I know them, but I dont want to learn how to do something wrong :P

    What are the dimensions of
    a)momentum (mass x velocity)
    b) work (force x displacement)
    c) impulse (force x time)

    a) p=mv
    b) J=Fm
    c) I=Ft

    is that right?
     
  19. Feb 3, 2005 #18
    I'd set out the first problem like this
    [itex]
    \begin{align*}
    \left[v^2\right] &= \rm m^2\, s^{-2} \\
    \left[u^2\right] & = \rm m^2\, s^{-2} \\
    \left[2as\right] & = \left(\rm m\, s^{-2}\right)(\mathrm{m}) \\
    & = \rm m^2\, s^{-2} \\
    \intertext{therefore}
    \left[u^2 + 2as\right] & = \rm m^2\, s^{-2} \\
    & = \left[v^2\right]
    \end{align*}
    [/itex]

    The units of momentum are [itex][p] = \mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1}[/itex],the units of work are [itex][W] = \mathrm{N}\,\mathrm{m} = (\mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-2})(\mathrm{m})=\mathrm{kg}\,\mathrm{m}^{2}\,\mathrm{s}^{-2}[/itex], and the units of impulse are [itex][J] = (\mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-2})(\mathrm{s}) = \mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1} = [p] [/itex].
     
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