Dr. Who takes you for a ride in his space/time machine, you are still in the pre

Question

Dr. Who takes you for a ride in his space/time machine, you are still in the present day, but as usual, he's not sure where you've landed. He takes a Yo-Yo out of his pocket, unrolls all the string, and lets it tick-tock back and forth several times (it's a simple pendulum!). The Yo-yo is approximately a point mass at the end of a 0.50 m long string, and you notice that it takes 3.95 seconds for one complete period. Where are you? a) Earth("g" = 9.80 m/s^2) b) Venus ("g" = 8.50 m/s^2) c) Mars ("g" = 3.72 m/s^2)d) Mercury ("g" = 2.5 m/s^2) e) Moon ("g" = 1.62 m/s^2) f) Titan ("g" = 1.27 m/s^2)

Explanation / Answer

Solution:

As we know;

The time period of a simple pendulum is given by:

T = 2L/g'

=> 3.95 = 2(0.50/g')

=> (0.50/g') = 3.95 / 2

=> g' = 1.264 m/s2

=> 1.27 m/s2(approx)

f) Titan

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