SOLUTION: Use the model for the period of a pendulum, T, such that T = 2π sqrt(L/g), where the length of the pendulum is L and the acceleration due to gravity is g. If the gravit

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Use the model for the period of a pendulum, T, such that T = 2π sqrt(L/g), where the length of the pendulum is L and the acceleration due to gravity is g. If the gravit      Log On


   

Question 1124528: Use the model for the period of a pendulum, T, such that
T = 2π sqrt(L/g), where the length of the pendulum is L and the acceleration due to gravity is g.
If the gravity is 32 ft/s^2 and the period equals 9 s, find the length to the nearest in.
(12 in. = 1 ft).
Round your answer to the nearest in.
*The answer of 787 inches was incorrect so I am unsure where I am going wrong.*
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!



T+=+2pi%2A+sqrt%28L%2Fg%29...i g=32+%28ft%2Fs%5E2+%29 and T=9s, we have
9s+=+2pi%2A+sqrt%28L%2F32+%28ft%2Fs%5E2+%29%29...convert 32ft to in

32ft=32%2A12in=384in


9s%2F%282%2Api%29+=+sqrt%28L%2F+%28384in%2Fs%5E2+%29%29.........square both sides
%289s%2F%282%2Api%29%29%5E2+=+%28sqrt%28L%2F+%28384in%2Fs%5E2+%29%29%29%5E2
%2881s%5E2%2F%284%2A%28pi%29%5E2%29+%29=+L%2F+%28384in%2Fs%5E2+%29
L=%2881cross%28s%5E2%29%2F%284%2A%28pi%29%5E2%29%29+%2A+%28384in%2Fcross%28s%5E2+%29%29
L=%2881%2F%284%2A%28pi%29%5E2%29+%29%29%2A+384in
L=2.051753968757339871+%2A+384in
L=787.873524002818510464in round it
L=788in