SOLUTION: Hello: Here is my problem that I need assistance with. Thank you. The period T (in seconds) of a simple pendulum is a function of its length L (in feet), given by T(L) = 2 p

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hello: Here is my problem that I need assistance with. Thank you. The period T (in seconds) of a simple pendulum is a function of its length L (in feet), given by T(L) = 2 p      Log On


   

Question 81536This question is from textbook Coll. Algebra
: Hello:
Here is my problem that I need assistance with. Thank you.
The period T (in seconds) of a simple pendulum is a function of its length L (in feet), given by T(L) = 2 pi square root of L/G where G = 32.2 feet per second. Per second is the acceleration of gravity. Express length L as a function of the period T. This question is from textbook Coll. Algebra

Answer by tutor_paul(519) About Me  (Show Source):
You can put this solution on YOUR website!
T=2%28pi%29sqrt%28L%2F32.2%29
The problem is asking you to solve for L in terms of T.
First get the square root term alone on one side of the equation:
T%2F%282pi%29=sqrt%28L%2F32.2%29
Square both sides to get rid of that square root:
T%5E2%2F%284pi%5E2%29=L%2F32.2
Solve for L:
highlight%28L%28t%29=%28%2832.2%29T%5E2%29%2F4pi%5E2%29
Good Luck,
tutor_paul@yahoo.com