MEASUREMENT AND CALCULATIONS

1. MEASUREMENT AND CALCULATIONS Chapter 9.2

2. Why is it important to be accurate? • Drug design • Construction • Sports

3. Example • At the 2008 Olympics, Usain Bolt and Asafa Powell were very close. • They both registered times of 9.7 seconds. • How are significant digits important here? • Usain’s time was 9.69s, Asafa’s was 9.72s

4. Significant Digits • The international agreement about the correct way to record measurements. • The number of significant digits in your answer is important because depending on your method of measurement, you can only be certain to a specific place value. • Ex. You time a foot race with an analog clock. Can you be certain of the winner?

5. Significant Digits Significant digits allow us to be certain of our calculations. This is very important to Science because we must always show how certain we are in our measurements. The greater the number of significant digits, the greater the certainty of measurement.

6. SIGNIFICANT DIGITS All whole numbers and zeroes between whole numbers are significant. Zeroes at the beginning or the end of a number may or may not be significant. Does the zero have to do with the accuracy of the value? Yes? – SIGNIFICANT Is the zero simply showing how big or small the value is? Called a “placeholder” – NOT SIGNIFICANT • Examples: • 30.95 – 4 sig figs • 4.03 – 3 sig figs • 0.04 – 1 sig fig ( leading zeros don’t count) • 0.5060 – 4 sig figs • 120 – 2 sig figs • 120. – 3 sig figs

7. SIGNIFICANT DIGITS • Placeholder zeroes. 0.000009has only one significant digit. • Any zeros added on the right side ARE significant because they indicate the accuracy of the measurement. 0.00000900 contains three significant digits.

8. PRACTICESignificant Digits • 1) 1.02 Km = _______ significant Digits • 2) 0.32 cm = _______ significant Digits • 3) 3600 kg = _______ significant Digits • 4) 20.060 L = ______ significant Digits • 5) 0.0030 g = ______ significant Digits

9. ROUNDING NUMBERS • If the digit after the digit to be rounded is 5 or larger, round up. If not round down. • Example: • 9.147 cm rounded to three Sig. Figs. Digits is 9.15 cm. • 7.23 g rounded to two Sig. Figs. Digits is 7.2 g.

10. TRY THESEROUNDING QUESTIONS • 0.0327 rounded to one Sig. Fig. Digit • 15.430 rounded to three Sig. Fig. Digits • We now can apply these two concepts to basic mathematical calculations.

11. Multiplying or Dividing SIGNIFICANT DIGITS • When multiplying or dividing significant digits, you round to the value with the least total number of sig. figs. • Example: • 4.62 x 0.035 = 0.1617 = 0.16 • 107.45 ÷ 6.40 = 16.7890 = 16.8

12. ADDING OR SUBTRACTINGSIGNIFICANT DIGITS • When adding or subtracting, you round to the value with the least number of Sig. Figs. after the decimal. • EXAMPLE: • 1.2 + 3.08 + 7.60 = 11.88 = 11.9 • 10.013 – 1.07 = 8.943 = 8.94

13. PRACTICE • 1) (2.4)(6.16) = ______ = _____ • 2) 16.1 – 2.4 = ______ = _____ • 3) 4.1 ÷ 8.6 = ______ = _____ • 4) 6.105 + 0.12 = ____ = _____

14. ORDER OF OPERATIONSSignificant Digits • You will come across problems involving both x / ÷ and + / - . This is done by using the following rules: • Do multiplication/division BEFORE • addition/subtractions. • Follow the rules of significant digits at • each step.

15. ORDER OF OPERATIONSSignificant Digits • EXAMPLE: 4.3 ÷ 1.2 + 6.12 = 3.58333 + 6.12 =3.6 + 6.12 =9.72 =9.7 Division is performed first. This number needs to be rounded to the value with the least total number of sig. digs. This number needs to rounded to the value with the least number of Sig. Digs. after the decimal.

16. PRACTICE 1) 42 – (2.2)(1.3) 2) (6.2)(4.3) – 12 6.1

17. Converting Units, Scientific Notation • You must understand the metric system to effectively convert. Examples: 1 gram (g) = 0.001 kilogram 1 gram (g) = 100 milligrams Examples: 1 metre (m) = 100 cm 1 metre = 0.001 kilometres

18. Scientific Notation 1.903 x 0.000001 = 0.000001903 = 2 x 10-6 • Examples: 1903 x 0.1 = 190.3 = 1.9 x 103 Unit conversion 5.1g x 1.2 kg =5.1g x 1200g =6120g =6.1 x 103 g

19. You will also have to use a conversion method that does not involve the metric system or has more than one unit. Example: 1) How many hours is 20.5 minutes? 20.5 min x 1 hour = 0.34166 = 0.342 h 60 min 2) How many m/s is 5km/h? 5 km x 1 h x 1000 m = 5000 m=1.388 1m/s h 3600s 1 km 3600 s

20. P. 349: #6 abcd, #9abcd

21. REARRANGING FORMULAS • You must isolate the variable you are trying to solve for. • To accomplish this you need to use the opposite operation that is indicated. • EXAMPLE: d = vt ( rearrange for v ) Divide by t because vt is multiplication. d = v t

22. There is an easy way to rearrange three part equations using the pie method. • EXAMPLE: • This does not work for equations such as: • a = vf – vi OR c = 2πr • T v = d / t t = d / v d = vt D V T

23. PRACTICE • 1) c = m / v ( rearrange for m ) • 2) a = ½ bh ( rearrange for h) • ANSWER: • 1) m = cv • 2) h = 2a/b

24. STEPS FOR SOLVING WORD PROBLEMS 1) List all the known and the unknown from the problem. 2) Select the best formula which uses the known and unknown. (be careful of extraneous info.) 3) Substitute the information into the equation. 4) calculate 5) round with appropriate significant digits. 6) Write a sentence answer.

25. QUESTIONS • Text Page 349 • #1,3,4,6,7,8,9