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Quick question help asap

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
Reply 1
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Zacken
22
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.
Reply 2
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pleasehelp1
OP
1
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?
Reply 3
Avatar for Zacken
Zacken
22
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.
Reply 4
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pleasehelp1
OP
1
great thank v much
Reply 5
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Zacken
22
Original post by pleasehelp1
great thank v much


No problem.
Reply 6
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pleasehelp1
OP
1
T = 2pi √((h^2+k^2)/gh))
Any ideas
How I would put the formula into y=mx+c ??
Reply 7
Avatar for Zacken
Zacken
22
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.
Reply 8
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pleasehelp1
OP
1
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
Reply 9
Avatar for Zacken
Zacken
22
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
ah i see, thanks you have been great help
Reply 11
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Zacken
22
Original post by pleasehelp1
ah i see, thanks you have been great help


No problem.

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