Question 1026263: The formula T = 2π√(l/g) relates a pendulum's period, T, in seconds (the time it takes to swing back and forth) to its length, l, in centimeters using g, the gravitational acceleration of 981 cm/sec^2.
How long would the pendulum have to be (to the nearest tenth of a centimeter) to make a pendulum with the given period?
QUESTION:
1 sec
NOTE:
√(l/g) means square root of l/g
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The formula T = 2π√(l/g) relates a pendulum's period, T, in seconds
(the time it takes to swing back and forth) to its length, l,
in centimeters using g, the gravitational acceleration of 981 cm/sec^2.
How long would the pendulum have to be (to the nearest tenth of a centimeter) to make a pendulum with the given period?
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sqrt(L/g) = T/(2pi)
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L/981 = [T/(2pi)]^2
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Length = 981[T/(2pi)]^2
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Cheers,
Stan H.
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