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4meisu
 2 years ago
t = 2π √l/g Rearrange to make g the subject.
4meisu
 2 years ago
t = 2π √l/g Rearrange to make g the subject.

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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large t=2\pi \frac{\sqrt \ell}{g}\]Is that formatted correctly? With the way it's written, it's a little hard to tell whether or not you have the g placed in the square root.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1If this is correct, here are the steps we would take: ~Multiply both sides by g,\[\large g t=2\pi \frac{\sqrt \ell}{\cancel g}\cancel g\]The g's will cancel out on the right, ~Divide both sides by t,\[\large \frac{g \cancel t}{\cancel t}=2\pi \frac{\sqrt \ell}{t}\]The t's will cancel on the left, giving us,\[\large \color{brown}{g=2\pi \frac{\sqrt \ell}{t}}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Alternatively, if you meant to write,\[\large t=2\pi \sqrt{\frac{\ell}{g}}\]We would start by writing it like this,\[\large t=2\pi \dfrac{\sqrt \ell}{\sqrt g}\]Then we would apply similar steps to the last method, ~Multiply both sides by \(\sqrt g\), \[\large t \sqrt g=2\pi \frac{\sqrt \ell}{\cancel{\sqrt g}}\cancel{\sqrt g} \qquad \rightarrow \qquad t \sqrt g=2\pi \sqrt \ell\]~Then divide both sides by t, \[\large \frac{\cancel t \sqrt g}{\cancel t}=2\pi\frac{\sqrt \ell}{t} \qquad \rightarrow \qquad \sqrt g=2\pi \frac{\sqrt \ell}{t}\] ~Then we'll finish up by squaring both sides,\[\large \left(\sqrt g\right)^2=\left(2\pi \frac{\sqrt \ell}{t}\right)^2 \qquad \rightarrow \qquad g=2^2 \pi^2\frac{(\sqrt \ell)^2}{t^2}\]When we square a term, we square `every part of it` as I have done above. Which gives us,\[\large \color{brown}{g=4\pi^2\frac{\ell}{t^2}}\]

4meisu
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you, that was very helpful and easy to understand :)
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