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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online