please just answer part C! Suppose the perimeter P 2 W+2*L and the area A L*W of

Question

please just answer part C!

Suppose the perimeter P 2 W+2*L and the area A L*W of a rectangle are to be determined from the following measurements of the lengths of the width and length W 107.4 t 0.4 m L 9.02 t 0.17 m (a) Calculate the perimeter of the rectangle and report your results to correct significant figures P 232.8 (b) Calculate the area of the rectangle and report your results to correct significant figures 40D 969 m2 (c) Use the rules of error propagation to determine the uncertainty in the recta perimeter and area. For the perimeter calculation, determine the overa uncertaint calculated by (1) considering that the length and width were each measured only once (P 2 W.+2 L), and by (2) considering what if the length and width were each measured twice (P W+W +L+L). In both cases, follow the rules to add the absolute errors in quadrature dA 402 dP X m two measurements X m four measurements (d) For the perimeter calculations, compare the overall uncertainty calculated by (1) considering that the length and width were each measured only once (P 2 W.+2 L), and by (2) Considering what if the length and width were each measured twice (P W+W +L+L). Which consideration yields the smaller uncertainty? Four measurements. O Two measurements Cannot determine

Explanation / Answer

C.) A = L*W

A = 9.02*107.4 = 968.748 m^2

dA/A = (dW/W) + (dL/L) = (0.4/107.4) + (0.17/9.02) = 0.02257

so dA = 0.02257*968.748 = 21.9 m^2

P of two = 2W + 2L

P = 2(107.4) + 2(9.02) = 232.84 m

(dPtwo/P) = d(W+L)/(W+L) = (0.17+0.4)/(107.4 + 9.02)

So dP two = 2(0.17 + 0.4) = 1.14 m

P of four =

(dPfour/P) = 2d(W+L)/2(W+L) = d(W+L)/(W+L) = (0.17+0.4)/(107.4 + 9.02)

So dP four = 2(0.17 + 0.4) = 1.14 m

ANSWERS:

dA = 21.9 m^2

dP two = 1.14 m

dP four = 1.14 m

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