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Physics 11 Online
OpenStudy (anonymous):

please someone help What's the gravitational acceleration on a planet where the period of a 2.30 m long pendulum is 2.73 s?

OpenStudy (anonymous):

a=v^2/r v=2pir/t i think its this

OpenStudy (anonymous):

um i dont think that is right wouldnt u have to use the equation g=4pi^2*[ L/T^2]?

OpenStudy (ash2326):

You're right Time period of a pendulum is given by \[T=\frac{1}{2\pi}\sqrt{\frac{L}{g}}\] substitute L= 2.3 m and T=2.73 s then you can find g on the planet

OpenStudy (anonymous):

oh pellet read the problem wrong! LOL ya you would use the one above.

OpenStudy (anonymous):

typically g is always 9.8 but its a different planet. so there is a new g.

OpenStudy (anonymous):

my answer is not coming out right

OpenStudy (anonymous):

lol nevermind... thanks for the help.. i found out my problem i was squaring the L and the T.. I was only suppose square T

OpenStudy (anonymous):

ya :P no problem man :)

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