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**Unformatted text preview: **Practice on significant figures and error analysis 1 Problems Exercise 1 37.76 + 3.907 + 226.4 = . . . Exercise 2 319.15 - 32.614 = . . . Exercise 3 104.630 + 27.08362 + 0.61 = . . . Exercise 4 125 - 0.23 + 4.109 = . . . Exercise 5 2.02 2.5 = . . . Exercise 6
600.0 5.2302 = ... Exercise 7 0.0032 273 = . . . Exercise 8 (5.5)3 = . . . Exercise 9 0.556 (40 - 32.5) = . . . Exercise 10 45 3.00 = . . . Exercise 11 What is the percent uncertainty in the measurement 2.26 m 0.25 m? Exercise 12 A useful and easy-to-remember approximate value for the number of seconds in a year is 107 . Determine the percent error in this approximate value. (There are 365.24 days in one year.) Exercise 13 With a wooden ruler you measure the length of a rectangular piece of sheet metal to be 13 mm. You use a micrometer caliper to measure the width of the rectangle and obtain the value 4.98 mm. Give your answers to the following questions with the correct number of significant figures. (a) What is the area of the rectangle? (b) What is the ration of the rectangle's width to its length? (c) What is the perimeter of the rectangle? Exercise 14 A rectangular piece of aluminum is 3.70 0.01 cm long and 2.30 0.01 cm wide. (a) Find the area of the rectangle and the uncertainty in the area. (b) Verify that the fractional uncertainty in the area is equal to the sum of the fractional uncertainties in the length and in the width. 1 2 Solutions 1 268.1 2 286.54 3 132.32 4 129 5 5.0 6 114.7 7 0.87 8 1.7 102 9 4 10 1.4 102 11 11% 12 0.45%
mm 13 (a)(13 mm) (4.98 mm) = 65 mm2 (two significant figures) (b) 4.98 mm = 13 0.38 (also two significant figures) (c)36 mm (to the nearest millimeter) 14 The area is 8.51 0.06 cm2 , where the extreme values in the piece's length and width are used to find the uncertainty in the area. The fractional 0.06 cm2 uncertainty in the area is 8.51 cm2 = 0.71%, and the fractional uncertainties cm in the length and width are 0.01 cm = 0.27% and 0.01 cm = 0.44%. 3.70 cm 2.3 2 ...

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