# Lab1 - Sheet1 Table 1 a(mm Part A triangle#1 Part A...

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Sheet1 Page 1 Table 1 a (mm) Δa (mm) b (mm) Δb (mm) c (mm) Δc (mm) 61 0.29 103 0.29 120 0.29 42 0.29 103 0.29 111 0.29 61.3 0.01 103.6 0.01 120.28 0.01 42.48 0.01 103.4 0.01 111.7 0.01 Table 2 Part A Part B <r> (mm) 0.03 -0.09 Δ<r> (mm) 0.29 0.01 0.37 0.01 0 0 2 2 0.84 252.42 0.66 0 Final Questions Part A, triangle #1 Part A, triangle #2 Part B, triangle #1 Part B, triangle #2 σ r (mm) for theory r = 0 v =degrees of freedom Χ 2/v = reduced X 2 P(X 2 ,v) 1. ∆d=[(∂d/∂a)∆a 2 + (∂d/∂b)∆b2] .5 => ∆d = [(a∆a) 2 + (b∆b) 2 ] .5 /d => ∆d = [∆a 2 (a 2 +b 2 )] .5 /d since d/d =1 ∆d=∆a Since ∂r/∂c=1 and ∂r/∂d= -1 ∆r=[1 2 ∆c 2 + -(1) 2 ∆d 2 )] .5 2. The data that we gathered does confirm the Pythagorean Theorem for Part A beca because the ruler isn't percise and thus the standard deviation is larger. The data we gathered does not confirm the Pythagorean Theorem for Part B becaus because the Vernier Caliper is more percise making the standard deviation smaller. 3. Although the readings obtained from the ruler will be off, the conclusion about the

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Lab1 - Sheet1 Table 1 a(mm Part A triangle#1 Part A...

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