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    I am doing some revision for a physics test here, keep getting confused with this simple bit of math
    T = 2pi √((h^2+k^2)/gh))
    What is the unit of K?
    h= m
    g= ms^-2
    t= s
    2pi = no unit

    im thinking k is maybe m^2
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    (Original post by pleasehelp1)
    im thinking k is maybe m^2
    Not quite; you want h and k to have the same units to add their squares together.
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    yeah thats whats I was thinking

    k= m
    h= m
    g= ms^-2
    t= s
    then the formula would look this in terms of its units
    S = √((m^2+m^2) / ms^-2 m))
    so S = √((m^2) / ms^-2 m))
    then cancel out m so that S = √(1 / s^-2 ))
    s = s
    Does this look correct?
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    (Original post by pleasehelp1)
    yeah thats whats I was thinking

    k= m
    h= m
    g= ms^-2
    t= s
    then the formula would look this in terms of its units
    S = √((m^2+m^2) / ms^-2 m))
    so S = √((m^2) / ms^-2 m))
    then cancel out m so that S = √(1 / s^-2 ))
    s = s
    Does this look correct?
    Yup.
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    great thank v much
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    (Original post by pleasehelp1)
    great thank v much
    No problem.
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    T = 2pi √((h^2+k^2)/gh))
    Any ideas
    How I would put the formula into y=mx+c ??
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    (Original post by pleasehelp1)
    T = 2pi √((h^2+k^2)/gh))
    Any ideas
    How I would put the formula into y=mx+c ??
    Square both sides.
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    Im plotting a graph ht^2 on y-axis and h^2 on x

    so is T = 2pi √((h^2+k^2)/gh))
    put into y=mx+c
    by sqauring both T = 4pi^2 ((h^2+k^2)/gh))
    then bring h over for the y axis
    ht^2=4pi^2((h^2+k^2)/g))

    is that it completely put into y=mx+c
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    (Original post by pleasehelp1)
    Im plotting a graph ht^2 on y-axis and h^2 on x

    so is T = 2pi √((h^2+k^2)/gh))
    put into y=mx+c
    by sqauring both T = 4pi^2 ((h^2+k^2)/gh))
    then bring h over for the y axis
    ht^2=4pi^2((h^2+k^2)/g))

    is that it completely put into y=mx+c
    Seperate it like this:

    ht^2 = (4pi^2/g) h^2 +4pi^2 k^2/g

    So that it's in y = mx + c
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    ah i see, thanks you have been great help
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    (Original post by pleasehelp1)
    ah i see, thanks you have been great help
    No problem.
 
 
 
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