SOLUTION: A Pendulum Is L Meters Long. The Time, R, In Seconds That It Takes To Swing Back And Forth Once Is Given By T=2 Square Root Of L . Suppose The Pendulum Takes 5.42 Seconds To Swing

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: A Pendulum Is L Meters Long. The Time, R, In Seconds That It Takes To Swing Back And Forth Once Is Given By T=2 Square Root Of L . Suppose The Pendulum Takes 5.42 Seconds To Swing       Log On


   



Question 1201771: A Pendulum Is L Meters Long. The Time, R, In Seconds That It Takes To Swing Back And Forth Once Is Given By T=2 Square Root Of L . Suppose The Pendulum Takes 5.42 Seconds To Swing Back And Forth Once. What Is Its Length? Carry Your Intermediate Computations To At Least Four Decimal Places, And Round Your Answer To The Nearest Tenth. Meters ×
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!

+T+=+2%2Asqrt%28L%29
Here T = 5.42
4.52 = 2*sqrt(L)
4.52/2 = sqrt(L)
2.26 sqrt(L)
L = (2.26)^2
L=5.1076 m

Answer by ikleyn(53567) About Me  (Show Source):
You can put this solution on YOUR website!
.
A Pendulum Is L Meters Long. The Time, R, In Seconds That It Takes To Swing Back And Forth Once Is Given
By T=2 Square Root Of L . Suppose The Pendulum Takes 5.42 Seconds To Swing Back And Forth Once.
What Is Its Length? Carry Your Intermediate Computations To At Least Four Decimal Places,
And Round Your Answer To The Nearest Tenth. Meters ×
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        Calculations in the post by @mananth are incorrect, leading to a wrong answer.
        I came to bring a correct solution.



T = 2%2Asqrt%28L%29


5.42 = 2%2Asqrt%28L%29


5.42/2 = sqrt%28L%29


2.71 = sqrt%28L%29


L = 2.71%5E2 = 7.4341 meters  (with four decimal places).


ANSWER.  The length is 7.4 meters (rounded to the nearest tenth of a meter).

Solved correctly.