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Rearranging formula

I need some help with rearranging formula.
the first one is. Make r the subject of the formula:
H = w^2 / 2g (R^2 - r^2)
and the second one is. Make a the subject of the formula:
T = (Sqrt pbh / 4 + a^2)

Please give some working along with the answer
Thanks in advance :wink:
(edited 7 years ago)
Original post by kszyni1
I need some help with rearranging formula.
the first one is. Make r the subject of the formula:
H = w^2 / 2g (R^2 - r^2)
and the second one is. Make a the subject of the formula:
T = (Sqrt pbh / 4 + a^2)

Please give some working along with the answer
Thanks in advance :wink:


Are you still stuck? What have you tried?
Reply 2
Original post by SeanFM
Are you still stuck? What have you tried?


I've rearranged the second one but I'm stuck on the first one I'm not sure how to get rid of the brackets and all of the powers :/
Original post by kszyni1
I've rearranged the second one but I'm stuck on the first one I'm not sure how to get rid of the brackets and all of the powers :/


As you've written it it is ambiguous.

Is it H=w2(2g)×(R2r2) H = \frac{w^2}{(2g) \times (R^2 - r^2)}?
Reply 4
yes
Everytime I see this topic, I fully get disgusted
God just needs to help I do well tbh:frown:

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Reply 6
Original post by kszyni1
yes


If you multiply both sides of equation by R2r2 R^2-r^2 and divide both sides by H H you get
R2r2=w22gH \displaystyle R^2-r^2=\frac{w^2}{2gH} .
Can you do anything from here?
Reply 7
Original post by B_9710
If you multiply both sides of equation by R2r2 R^2-r^2 and divide both sides by H H you get
R2r2=w22gH \displaystyle R^2-r^2=\frac{w^2}{2gH} .
Can you do anything from here?


I'm fine form there thanks a lot

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