Scan_Doc0023 - 1.6 Dimensional Analysis 17 Solution a. The...

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-~ number must be rounded to two ~tfigures. Table 1.4 English-Metric Equivalents zth 1 m = 1.094 yd 2.54 em = 1 in 1 kg = 2.205 Ib 453.6 g = lIb 1 L = 1.06 qt 1 ft 3 = 28.32 L e 1.6 Dimensional Analysis 17 Solution a. The result is 1.71 X 10- 4 , which has three significant figures because the term with the least precision (1.05 X 10- 3 ) has three significant figures. b. The result is 7 with no decimal point because the number with the least number of decimal places (21) has none. c. R = PV = (2.560)(8.8) T 275.15 The correct procedure for obtaining the final result can be represented as follows: (2.560)(8.8) 22.528 275.15 = 275.15 = 0.0818753 = 0.082 = 8.2 X 10~2 = R The final result must be rounded to two significant figures because 8.8 (the least pre- cise measurement) has two significant figures. To show the effects of rounding at in- termediate steps, we will carry out the calculation as follows: (2.560)(8.8) 275.15 Rounded to two significant figures ! 23
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