How do you solve for #g# in #T=2pisqrt(L/g)#?

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Kayla Share
Aug 10, 2016

Answer:

#g=(4pi^2L)/T^2#

Explanation:

#color(red)("First,")# what you want to do is square both sides of the equation to get rid of the square root, since squaring a number is the inverse of taking the square root of a number:

#T^2 = (2pi)^2sqrt((L/g)^2)#

#T^2=(2pi)^2 (L/g)#

#color(blue)("Second,")# we simplify the #(2pi)^2# part so it's easier to read. Remember, when you square something in parenthesis, you are squaring every single term inside:

#T^2=4pi^2 (L/g)#

#color(purple)("third,")# multiply both sides by #g#:
#gxxT^2=4pi^2 (L/cancel"g")xxcancelg#

#gxxT^2=4pi^2L#

#color(maroon)("finally,")# divide both sides by #T^2# to get #g# by itself:

#(gxxcancelT^2)/cancelT^2=(4pi^2L)/T^2#

Thus, #g# is equal to:

#g=(4pi^2L)/T^2#

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