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Is 2pi*sqrt(l/g) reducable to a dimension of time? Socratic
WebAug 3, 2015 · 2nd) Square both sides: (T√ (g/L))2 = π2 This gets rid of the square root, so we have: T2g/L = π2 3rd) Multiply both sides by L, then divide by T2: (T2g/L)L = π2L This cancels out the L on the left side: T2g = n2L Now divide by T2 : T2g/T2 = n2L/T2 The T 2 on the left side cancels out and we're left with: g = n2L/T2 WebHow do I transform the equation T=2π√l/g so that l is the subject of the formula? 15 John Privett Software Developer (2010–present) Author has 4.5K answers and 5.6M answer … songs about being different for kids
The period of oscillation of a simple pendulum is T = 2pi√(L/g).
WebExpert Answer. 100% (3 ratings) T = 2π √ (L/g). Therefore, T2 = …. View the full answer. Transcribed image text: T = 2 pi square root of L/g Rearrange to solve for g. Previous … WebFeb 26, 2016 · This formula we've been using recently, I don't understand where it comes from. Homework Equations T = 2pi * sqrt (m/k) The Attempt at a Solution Hooke's law: F = -kx E (total) = .5mv^2 + .5kx^2 Circumference of a Circle: C = 2 * pi * r I figure these piece together, but I don't understand how. WebJun 14, 2013 · the period of oscillation of a simple pendulum is T = 2pi root(L/g) . L is about 10 cm amd is known to 1 mm accuracy. the period of oscillation is about 0.5 s . the time of 100 oscillations is measured with a wrist watch of 1s resolution what is the accuracy in the determination of g smalley chase