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Units Help

  1. Jun 10, 2006 #1
    Hey all,

    this is my first post here, I have just discovered the site, and it looks great :smile:

    I need some help on units, regarding square roots.

    When you find the reciprocal of a quantity, like say, mass, the new unit is kg-1.

    I'm not sure about when you find the square root of a quantity, what unit would it be?

    Regards,
    Physique.
     
  2. jcsd
  3. Jun 11, 2006 #2

    Tide

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    When you take the square root of a physical quantity you get the square root of the units as well.
     
  4. Jun 11, 2006 #3
    Thanks Tide!

    I'm investigating the relationship between time and mass, and I think I've got it, time is proportional to the square root of mass, but what on earth does the gradient mean?

    time is on the y-axis, square root of mass is on the x-axis
     
  5. Jun 11, 2006 #4

    Tide

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    "Gradient" generally refers to (rate of) change with respect to spatial position. If you could provide more detail on the problem you are trying to solve then we could offer more clarification.
     
  6. Jun 12, 2006 #5

    HallsofIvy

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    How did you arrive at the idea that "time is proportional to the square root of mass"? I imagine that you have some formula in which something, say a raindrop, is accumulating mass as it falls. In that case, I would be more inclined to say "mass is proportional to the square of time". In order for that to make any sense, your constant of proportionality would have to "fix" the units:
    For example, if we have m= ct2, mass is measured in kg and time in seconds, then we have kg= c*sec2 so c must have units of kg/sec2 in order to "cancel" the sec2 and introduce kg into the formula.
    Yes, you could also write [itex]t= c\sqrt{m}[/itex] in which case c must have units of [tex]\frac{sec}{\sqrt{kg}}[/tex]. While mathematics does not deal in "cause and effect", physics does. As soon as you introduce "mass" and "time", it makes more sense to me to think of time as the independent variable and mass as the dependent variable!

    Notice that just because you can do a mathematical operation doesn't mean it will make physical sense! If A= x2, where A is the area of a square of side length x, then [itex]x= \sqrt{A}[/itex]. Now, there, of course, A would have to be measured in something like "square inches" or "square meters" and, of course, [itex]x= \sqrt{A}[/itex] would me measured in "inches" or "meters". That's why, in formulas, we typically write "square meters" as "m2" or "square inches" as in2.

    On the other hand, I would not expect to have any physical quantity with units of [itex]\sqrt{m}[/itex] or [itex]\sqrt{kg}[/itex]. In the examples above, your constants (which may not be "physical" quantities) had to change the units appropriately
     
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