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momr1958
Group Title
I need help with this problem.
Table
L (feet) 0.5, 1.0, 1.5
T (seconds) 0.78, 1.11, 1.36
T=2 pie L/G squared
Solve the formula for G. Not sure how to do this can someone help me please.
 2 years ago
 2 years ago
momr1958 Group Title
I need help with this problem. Table L (feet) 0.5, 1.0, 1.5 T (seconds) 0.78, 1.11, 1.36 T=2 pie L/G squared Solve the formula for G. Not sure how to do this can someone help me please.
 2 years ago
 2 years ago

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wasiqss Group TitleBest ResponseYou've already chosen the best response.0
yes dw:1322993285116:dw
 2 years ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
now gm can b easily made subject
 2 years ago

momr1958 Group TitleBest ResponseYou've already chosen the best response.0
no it is not t2 it is T=2*pie L/G squared my computer does not have the squared sign on it.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
notationwise x squared is x^2
 2 years ago

momr1958 Group TitleBest ResponseYou've already chosen the best response.0
then i am saying it wrong it is the square root of L/G
 2 years ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
i know the formula so thats y my answer is correctdw:1322993751354:dw
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
sqrt(L/G) is good :) \[T=2pi\sqrt{\frac{L}{G}}\]
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
to under the sqrt part; we have to square; they undo each other
 2 years ago

momr1958 Group TitleBest ResponseYou've already chosen the best response.0
Yes that is the way my problem is written can you help me solve for G using the table I put in on my first post.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
using the table? prolly. It might mean that you want the primary root.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[T=2pi\sqrt{\frac{L}{G}}\] \[\left(T=2pi\sqrt{\frac{L}{G}}\right)^2\] \[T^2=4pi^2\frac{L}{G}\] \[GT^2=4pi^2L\] \[G=\frac{4pi^2L}{T^2}\] \[G=\left(\frac{2pi}{T}\right)^2L\] gotta see if that matches up
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
or rather, use the values of T and L to find values for G
 2 years ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
i did it hours b4 amistre
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
good, then you can tell me if im on the right track :)
 2 years ago

momr1958 Group TitleBest ResponseYou've already chosen the best response.0
All I want is to solve for G using L(ft) 0.5, 1.0, and 1.5 and T(sec) 0.78, 1.11, and 1.36 can this be done and give me a step by step on how to do it.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
It can be done. and I just posted the steps for it.
 2 years ago

momr1958 Group TitleBest ResponseYou've already chosen the best response.0
But you are not sure if you are on the right track. If you are then thank you and I will see if it works out. If not on the right track then I will check some where else for the help. Sorry not trying to be rude but this is not easy for me to understand.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
momr, if you have confusion with the steps, I can try to explain them. im pretty confident that I have done it correctly
 2 years ago

momr1958 Group TitleBest ResponseYou've already chosen the best response.0
Thank you I will work it out with your steps replacing the numbers for the letters. Thank you again for being patient with me.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
good luck with it :) and dont feel uncomfortable in posting any question you may have.
 2 years ago

dometryapple Group TitleBest ResponseYou've already chosen the best response.0
dw:1323023125528:dw
 2 years ago
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