On the planet Xenos, an astronaut observes that a 1.00 m long pendulum has a period of 1.50 s. What is the free-fall acceleration on Xenos?

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Answer 1 · 2016-11-09T03:28:56

#sf(g=1.77pi^(2)color(white)(x)"m/s"^(2))#

The time period is given by:

#sf(T=2pisqrt(l/g))#

#:.##sf(sqrt(l/g)=T/(2pi))#

#:.##sf(l/g=T^(2)/(4pi^2))#

#:.##sf(g/l=(4pi^2)/T^2)#

#:.##sf(g=(4pi^2l)/T^2)#

#sf(g=(4pi^(2)xx1.00)/(1.50^(2))=1.77pi^(2)color(white)(x)"m/s"^2)#