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Standards of Measurements

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Presentation on theme: "Standards of Measurements"— Presentation transcript:

1 Standards of Measurements

2 Accuracy and Precision
Accuracy – how close a measured value is to the actual value Precision – how close the measured values are to each other

3 Significant Figures All nonzero digits are significant.
1, 2, 3, 4, 5, 6, 7, 8, 9 Zeros within a number are always significant. Both 4308 and have four significant figures.

4 Significant Figures Zeros that set the decimal point are not significant. 470,000 has two significant figures. has two significant figures. Trailing zeros that aren't needed to hold the decimal point are significant. 4.00 has three significant figures.

5 Significant Figures If the least precise measurement in a calculation has three significant figures, then the calculated answer can have at most three significant figures. Mass = grams Volume = 4.42 cubic centimeters. Rounding to three significant figures, the density is 7.86 grams per cubic centimeter.

6 Scientific Notation For large numbers, moving the decimal to the left will result in a positive number = 3.46 x 105 For small numbers, moving the decimal to the right will result in a negative number = 1.45 x 10-4 For numbers less than 1 that are written in scientific notation, the exponent is negative.

7 Scientific Notation Before numbers in scientific notation can be added or subtracted, the exponents must be equal. 5.32 x x 104 5.32 x x 105 x 105 6.24 x 105

8 Scientific Notation When numbers in scientific notation are multiplied, only the number is multiplied. The exponents are added. (3.33 x 102) (2.71 x 104) (3.33) (2.71) x 102+4 9.02 x 106

9 Scientific Notation When numbers in scientific notation are divided, only the number is divided. The exponents are subtracted. 4.01 x 109 1.09 x 102 4.01 x 109-2 1.09 3.67 x 107

10 Scientific Notation A rectangular parking lot has a length of 1.1 × 103 meters and a width of 2.4 × 103 meters. What is the area of the parking lot? (1.1 x 103 m) (2.4 x 103 m) (1.1 x 2.4) (10 3+3) (m x m) 2.6 x 106 m2

11 SI Units Kilo- (k) 1000 Hecto- (h) 100 Deka- (da) 10 Base Unit m, L, g
Deci- (d) 0.1 Centi- (c) 0.01 Milli- (m) 0.001 Mnemonic device: King Henry Died By Drinking Chocolate Milk

12 Metric System Meter (m) – The basic unit of length in the metric system Length – the distance from one point to another A meter is slightly longer than a yard

13 Metric System Liter (L) – the basic unit of volume in the metric system A liter is almost equal to a quart

14 Metric System Gram (g) – The basic unit of mass

15 D M V D = M V Derived Units Combination of base units
Volume – length  width  height 1 cm3 = 1 mL Density – mass per unit volume (g/cm3) D M V D = M V

16 D M V Density V = 825 cm3 M = DV D = 13.6 g/cm3 M = 13.6 g x 825 cm3
1) An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass. GIVEN: V = 825 cm3 D = 13.6 g/cm3 M = ? WORK: M = DV M = 13.6 g x cm3 cm M = 11,220 g D M V

17 D M V Density D = 0.87 g/mL V = M V = ? M = 25 g V = 25 g 0.87 g/mL
2) A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid? GIVEN: D = 0.87 g/mL V = ? M = 25 g WORK: V = M D D M V V = g 0.87 g/mL V = 28.7 mL

18 D M V Density M = 620 g D = M V = 753 cm3 D = ? D = 620 g 753 cm3
3) You have a sample with a mass of 620 g & a volume of 753 cm3. Find the density. GIVEN: M = 620 g V = 753 cm3 D = ? WORK: D = M V D M V D = g 753 cm3 D = 0.82 g/cm3

19 Dimensional Analysis / Unit Factors
Dimensional analysis – a problem-solving method that use any number and can be multiplied by one without changing its value

20 Dimensional Analysis / Unit Factors
How many hours are there in a year? There’s 8,760 hours in a year. 24 hr 1 day 365 days 1 year 8760 hr 1 year x =

21 Dimensional Analysis / Unit Factors
The distance from Dixons Mills to Thomasville is 9 miles. How many feet is that? There’s 47,520 feet in 9 miles. 9 mi 1 5280 ft 1 mi 47520 ft 1 x =

22 Dimensional Analysis / Unit Factors
Convert 36 cm/s to mi/hr 3600 sec 1 hr 1 in 2.54 cm 1 ft 12 in 36 cm sec x x x 1 mi 5280 ft mi hr = = 0.805 mi/hr

23 Temperature A degree Celsius is almost twice as large as a degree Fahrenheit. You can convert from one scale to the other by using one of the following formulas:

24 Temperature Convert 90 degrees Fahrenheit to Celsius
oC = 5/9 (oF - 32) oC = 5/9 ( ) oC = (58) oC = 32.2

25 Temperature Convert 50 degrees Celsius to Fahrenheit
oF = 9/5 (oC ) + 32 oF = 9/5 (50 ) + 32 oF = 1.8 (50) + 32 oF = oF = 122

26 Temperature The SI base unit for temperature is the kelvin (K).
A temperature of 0 K, or 0 kelvin, refers to the lowest possible temperature that can be reached. In degrees Celsius, this temperature is –273.15°C. To convert between kelvins and degrees Celsius, use the formula:

27 Temperature


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