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Significant Figures Uncertainty of Measurement All measurements are assumed to be approximate with the last digit estimated. 0 1 2 The length in “cm” here is written as: 1.45 cm The last digit “5” is estimated as 1/2 of the interval between 4 and 5. Estimated Measurements (Cont.) Length = 1.45 cm 0 1 2 The last digit is estimated, but is significant. It tells us the actual length is between 1.40 cm and 1.50. It would not be possible to estimate yet another digit, such as 1.453. This measurement of length can be given in three significant figures - the last is estimated. Significant Figures When writing numbers, zeros used ONLY to help in locating the decimal point are NOT significant—others are. See examples. 0.0062 cm 4.0500 cm 0.1061 cm 2 significant figures 5 significant figures 4 significant figures 50.0 cm 3 significant figures 50,600 cm 3 significant figures Pacific-Atlantic Rule 0.0020 = (2) 70020 = (4) 15.800 = (5) 385 = (3) 95.0 = (3) 6,000 = (1) Rule 1. When approximate numbers are multiplied or divided, the number of significant digits in the final answer is the same as the number of significant digits in the least accurate of the factors. 45 N 6.97015 N/m2 Example 1: P (3.22 m)(2.005 m) Least significant factor (45) has only two (2) digits so only two are justified in the answer. The appropriate way to write the answer is: P = 7.0 N/m2 Rule 2. When approximate numbers are added or subtracted, the number of significant digits should equal the smallest number of decimal places of any term in the sum or difference. Ex 2: 9.65 cm + 8.4 cm – 2.89 cm = 15.16 cm Note that the least precise measure is 8.4 cm. Thus, answer must be to nearest tenth of cm even though it requires 3 significant digits. The appropriate way to write the answer is: 15.2 cm Example 3. Find the area of a metal plate that is 8.71 cm by 3.2 cm. A = LW = (8.71 cm)(3.2 cm) = 27.872 cm2 Only 2 digits justified: A = 28 cm2 Example 4. Find the perimeter of a metal plate that is 8.71 cm long and 3.2 cm wide. P = 8.71 cm + 3.2 cm + 8.71 cm + 3.2 cm Ans. to tenth of cm: P = 23.8 cm Working with Numbers Classroom work and laboratory work should be treated differently. In class, the uncertainties are shown in the numbers given. Round to the smallest number of significant figures. In lab, we know the limitations of the measurements. The estimated digit is either 0 or 1/2 of the smallest unit of the device . Classroom Example: A car traveling initially at 46.0 m/s undergoes constant acceleration of 2.0 m/s2 for a time of 4.30 s. Find total displacement, given formula. 1 2 d vi t at 2 1 d (46 m / s)( 4.3s) (2.0m / s 2 )( 4.3s) 2 2 197.8m 18.49m 216.29m For class work, we assume all given info is accurate to 2 significant figures. d = 220 m Lab Example 1: The length of a sheet of metal is measured as 233.5 mm and the width is 9.0 mm. Find the area. (Multiplication Rule) Note that the precision of each measure is to the nearest tenth of a millimeter. However, the length has four significant digits and the width has only two. How many significant digits are in the product of length and width (area)? Two (9.0 has least significant digits). Lab Example 1 (Cont.): The length of a sheet of metal is measured as 233.5 mm and the width is 9.0 mm. Find area. Area = LW = (233.5 mm)(9.0 mm) Area = 2101.5 mm2 W = 9.0 mm But we are entitled to only two significant digits. Therefore, the answer becomes: Area = 2100 mm2 L = 233.5 mm Lab Example 2: Find the perimeter of sheet of metal, with measured L = 233.5 mm and W = 9.0 mm. (Addition Rule) P = 233.5 mm + 9.0 mm + 233.5 mm + 9.0 mm P = 485 mm Note: is Note: The The answer result has determined by the more significant digits least precise measure. than the width factor (the tenth of a mm) in this case. W = 9.0 mm L = 233.5 mm Perimeter = 485.0 mm