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How do you solve for L in T = 2 π √ L g ?

Answer 1

Amelia Adams

#L=(gT^2)/(4pi^2)#

As #T=2pisqrt(L/g)#

#T^2=4pi^2xxL/g#

or #T^2xxg/(4pi^2)=4pi^2xxL/gxxg/(4pi^2)#

or #(gT^2)/(4pi^2)=L#

Hence #L=(gT^2)/(4pi^2)#

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Answer 2

Parker Abbott

To solve for ( L ) in the equation ( T = 2\pi \sqrt{L/g} ), where ( T ) represents the period of oscillation, and ( g ) is the acceleration due to gravity, follow these steps:

Square both sides of the equation to isolate the square root term: ( T^2 = 4\pi^2 \frac{L}{g} ).
Multiply both sides of the equation by ( g ) to move ( L ) out of the denominator: ( T^2 g = 4\pi^2 L ).
Divide both sides of the equation by ( 4\pi^2 ) to isolate ( L ): ( L = \frac{T^2 g}{4\pi^2} ).
This expression gives the value of ( L ) in terms of ( T ) and ( g ).

Therefore, ( L = \frac{T^2 g}{4\pi^2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.