Question 99596
{{{8T^2=pi^(2L)}}} we divide both sides by 8
{{{T^2=(pi^(2L)/8)}}} we can extract the root because in this physical example, we are sure that no values are negative (cannot extract negative numbers)
{{{T = sqrt((pi^(2L)/8))}}} after substitution of 10, we get :
{{{T = sqrt((pi^(2*10)/8))}}} = 4.18251339837959879019990694359876215457916259765625 - exactly :-D, by the in real world such precision of computations(in physics) is completly futile, because precision of output can never be higher then precision of input. This means that it is quite useless to compute with higher precision then precision of our measurements.For instance if we measure with unit precision our results can never have higher then unit precision.