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TrojanPoem
 one year ago
Verify if this equation is true based on physics dimensions :
T = 2 pi sqrt(l/g)
As T = time, l = length , g = gravity
TrojanPoem
 one year ago
Verify if this equation is true based on physics dimensions : T = 2 pi sqrt(l/g) As T = time, l = length , g = gravity

This Question is Closed

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: we have: \[\sqrt {\frac{l}{g}} = \sqrt {\frac{{{\text{meters}}}}{{{\text{meters/second}}{{\text{s}}^{\text{2}}}}}} \]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I've solved it and got T^1 as the dimension. The problem is , I can't find which answer is correct a) correct b) correct according to dimensions c) wrong d) wrong according to dimensions d or c ? and why ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it is correct! namely option a)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1a good method is to start from the differential equation of the motion of the pendulum

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0m/m/s^2 = sqrt(s^2)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But why b is wrong ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! it is correct, nevertheless your reasoning is not able to justify the presence of the coefficient \(2 \pi\) option b) is wrong since we have the correct formula, and the dimensional analysis can not say anything on possible numerical coefficients which may accompany the various formulas

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0we cannot check if 2 pi is correct therefore b is correct

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in other words, using the dimensional analysis I am able to write this: \[T \propto \sqrt {\frac{l}{g}} \]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, so correct according to dimension ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, since you have this formula: \[T = 2\pi \sqrt {\frac{l}{g}} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think it is option a)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Doesn't a mean "We are fully sure that the equation dimensions are correct, constants are correct" ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1this formula: \[T \propto \sqrt {\frac{l}{g}} \] is correct according to dimensions

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0so if the constant is not written it's correct according to dimensions if it's written , just correct ?
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