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SCIENCE UNIT 3 THE PHYSICS OF MOTION !

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Presentation on theme: "SCIENCE UNIT 3 THE PHYSICS OF MOTION !"— Presentation transcript:

1 SCIENCE 1206 - UNIT 3 THE PHYSICS OF MOTION !
Corner Brook Regional High School SCIENCE UNIT 3 THE PHYSICS OF MOTION !

2 UNIT OUTLINE Measurement and Calculations Vector Quantities
Significant Digits Scientific Notation Converting between Units Accuracy vs. Precision Vector Quantities Scalar Quantities Displacement Calculations Distance Calculations Velocity Calculations Speed Calculations Acceleration Calculations Distance-Time Graph Vector Diagrams Speed Time Graph

3 BOOK SECTIONS: Chapter 9 Chapter 10 Chapter 11
Intro, 9.2, 9.5, 9.6, 9.7, 9.10 Chapter 10 Intro, 10.2, 10.3, 10.4, 10.7 Chapter 11 Intro, 11.1, 11.3, 11.5, 11.7

4 What is PHYSICS? DEFINITION:
The study of motion, matter, energy, and force. Branches include: MECHANICS (motion and forces) WAVES (sound and light) ENERGY (potential and kinetic, thermodynamics) MODERN (quantum physics, nuclear physics)

5 Measurement & Calculations
CERTAINTY Defined as the number of significant digits plus one uncertain (estimated) digit The last digit of any number is always UNCERTAIN, as measurement devices allow you to estimate. EXAMPLE: 2.75 m The “5” is uncertain

6 CERTAINTY TAKE THE FOLLOWING MEASUREMENT and determine the certain digits and the uncertain digit. ANSWER:_____________________________

7 CERTAINTY TAKE THE FOLLOWING MEASUREMENT and determine the certain digits and the uncertain digit. ANSWER:_____________________________ the 5 is uncertain, it was estimated

8 SIGNIFICANT DIGIT RULES
1. EXACT VALUES EXACT VALUES have an INFINITE (∞) NUMBER of SIGNIFICANT DIGITS. TWO TYPES: COUNTED VALUES – directly counted Ex: 20 students, 3 dogs, 5 fingers DEFINED VALUES – always true, constant measures Ex: 60 s/min, 100 cm/m, 1000 m/km

9 SIGNIFICANT DIGITS RULES
2. ZEROS ALL NUMBERS in a value are SIGNIFICANT EXCEPT LEADING ZEROS, and TRAILING ZEROS WITH NO DECIMAL. VALUE NUMBER OF SIG FIGS 600 606 600.0 0.60 0.606 660

10 SIGNIFICANT DIGIT RULES
3. MULTIPLYING and DIVIDING WHEN MULTIPLYING(x) and DIVIDING(/), ANSWER has SMALLEST NUMBER of SIGNIFICANT DIGITS. EXAMPLE: 6.15 x 8.0 = ÷ 2 =

11 SIGNIFICANT DIGIT RULES
3. MULTIPLYING and DIVIDING WHEN MULTIPLYING(x) and DIVIDING(/), ANSWER has SMALLEST NUMBER of SIGNIFICANT DIGITS. EXAMPLE: 6.15 x 8.0 = ÷ 2 =

12 SIGNIFICANT DIGITS RULES
4. ADDING AND SUBTRACTING WHEN ADDING(+) and SUBTRACTING(-), ANSWER has SMALLEST NUMBER of DECIMAL PLACES. EXAMPLE: =

13 SIGNIFICANT DIGITS RULES
4. ADDING AND SUBTRACTING WHEN ADDING(+) and SUBTRACTING(-), ANSWER has SMALLEST NUMBER of DECIMAL PLACES. EXAMPLE: = 116

14 SIGNIFICANT DIGITS RULES
5. ROUNDING When ROUNDING, if the number is 5 or GREATER, ROUND UP. Remember, round only once! VALUE ROUND to 2 SIG FIGS 61.3 s 12.70 m/s 36.5 km 99.0 m/s2 46.4 min

15 SIGNIFICANT DIGITS RULES
5. ROUNDING When ROUNDING, if the number is 5 or GREATER, ROUND UP. Remember, round only once! VALUE ROUND to 2 SIG FIGS 61.3 s 12.70 m/s 36.5 km 99.0 m/s2 46.4 min

16 SCIENTIFIC NOTATION A convenient way of expressing very large and small numbers. Expressed as a number between 1 and 10 and multiplied by 10x (x = exponent). LARGE numbers exponent is # of spaces to the LEFT SMALL numbers NEGATIVE exponent is # of spaces to the RIGHT

17 SCIENTIFIC NOTATION EXAMPLES
ROUND THE FOLLOWING to 2 SIGNIFICANT DIGITS. VALUE SCIENTIFIC NOTATION 100 m 3500 s 926,000,000,000 h m s 0.1 m/s2

18 SCIENTIFIC NOTATION EXAMPLES
ROUND THE FOLLOWING to 2 SIGNIFICANT DIGITS. VALUE SCIENTIFIC NOTATION 100 m 3500 s 926,000,000,000 h m s 0.1 m/s2

19 HOMEWORK DO Worksheets 1,2, 3 (p. 4-6)in your handout for homework.

20

21

22

23 From the Physics Worksheets Booklet

24

25 0.0300

26 WHEN ADDING(+) and SUBTRACTING(-), ANSWER has SMALLEST NUMBER of DECIMAL PLACES.
WHEN MULTIPLYING(x) and DIVIDING(/), ANSWER has SMALLEST NUMBER of SIGNIFICANT DIGITS.

27

28 Scientific Notation

29 5 x 10-3 2.5 x 10-1 5.05 x 103 2.5 x 10-2 8 x 10-4 2.5 x 10-3 1 x 103 5 x 102 5 x 103 1 x 106 1500 0.335 0.0015 10,000 0.0375 0.1 375 4 220,000

30 CONVERTING BETWEEN UNITS
BASE UNIT A unit from which other units may be derived, including units for the following: Length  metres, m Mass  kilogram, kg Time  second, s Temperature  kelvin, K In science, we use SI BASE UNITS, from the INTERNATIONAL SYSTEM OF UNITS. DERIVED UNIT A unit which is derived from base units. Ex: m/s

31 Some Examples of Base Units

32 CONVERTING BETWEEN UNITS
METRIC PREFIXES Values placed in front of the base units. PREFIX SYMBOL FACTOR giga G 109 mega M 106 kilo k 103 hecta h 102 deca da 101 SI BASE UNITS deci d 10-1 centi c 10-2 milli m 10-3 micro μ 10-6 nano n 10-9

33

34 CONVERTING BETWEEN UNITS
To convert, using the following system: TO THE RIGHT  multiply by 10 TO THE LEFT  divide by 10 G M k h da SI BASE UNITS d c m μ n DIVIDE BY 10 MULTIPLY BY 10

35 CONVERTING BETWEEN UNITS
EXAMPLES: 1.6 m = ____________ μm 340 N = ____________ hN

36 CONVERTING BETWEEN UNITS
EXAMPLES: 1.6 m = ____________ μm (micro-meter) 340 N = ____________ hN (hecta-newton) 1.6 x 106 = 1,600,000 340 /100 = 3.4

37 CONVERTING BETWEEN UNITS
EXAMPLES: 1250 cm = ____________ km 4.7 Gg = ____________ ng

38 CONVERTING BETWEEN UNITS
EXAMPLES: 1250 cm = ____________ km 4.7 Gg = ____________ ng 1250/ = .0125 (1 km = 1000m and 1 m=100 cm) 4.7 x 1018 Do the metric hops!

39 CONVERTING BETWEEN UNITS
In addition to using metric prefixes, we also convert between SI UNITS and other accepted systems of measurement. Here are some helpful CONVERSION FACTORS you should know when studying MOTION: 1 km = 1000 m 1 h = 3600 s 1 m/s = 3.6 km/h

40 So again…. 1 meter/second = 3.6 km/hr
..and where did that number come from…. So again…. 1 meter/second = 3.6 km/hr

41 CONVERTING BETWEEN UNITS
CONVERT THE FOLLOWING: 23 min = ____________ h 0.47 h= ____________ s

42 CONVERTING BETWEEN UNITS
CONVERT THE FOLLOWING: 23 min = ____________ h 0.47 h= ____________ s 23/60 = = .38 1700 WHEN MULTIPLYING(x) and DIVIDING(/), ANSWER has SMALLEST NUMBER of SIGNIFICANT DIGITS.

43 CONVERTING BETWEEN UNITS
CONVERT THE FOLLOWING: 4.5 km/h = ____________ m/s 30.2 m/s = ____________ km/h

44 CONVERTING BETWEEN UNITS
CONVERT THE FOLLOWING: 4.5 km/h = ____________ m/s 30.2 m/s = ____________ km/h WHEN MULTIPLYING(x) and DIVIDING(/), ANSWER has SMALLEST NUMBER of SIGNIFICANT DIGITS

45 HOMEWORK DO WORKSHEETS 4 and 5 (p in your handout) for homework.

46

47

48 Pages 4 and 5 of Work Book Divide Multiply

49 That’s 5 hops to the right, so multiply by 105
Let’s convert 1000 km to cms. That’s 5 hops to the right, so multiply by 105 1000 x 105 = 100,000,000 cms in 1000 km 1000 km x 1000 m x 100 cm = 100,000,000 cm 1 km m

50 #2. 250 x which is the same as 2.5 x 10-7 (2.5 x 10-4)

51 4 spaces to the left 9 spaces to the right

52

53 WHEN MULTIPLYING(x) and DIVIDING(/), ANSWER has SMALLEST NUMBER of SIGNIFICANT DIGITS.

54

55 WHEN ADDING(+) and SUBTRACTING(-), ANSWER has SMALLEST NUMBER of DECIMAL PLACES.
WHEN MULTIPLYING(x) and DIVIDING(/), ANSWER has SMALLEST NUMBER of SIGNIFICANT DIGITS.

56 REMEMBER TO READ ALL INSTRUCTIONS!
ROUND FIRST!!!! Ex: 

57 1 m/s = 3.6 km/h 1 h = 3600 s 1 km = 1000 m 65 min = 0.045 days
1 d = min (60 x 24) WHEN MULTIPLYING(x) and DIVIDING(/), ANSWER has SMALLEST NUMBER of SIGNIFICANT DIGITS.

58

59 ASSESSMENT TIME! UNIT 3 QUIZ 1
Significant Digits, Rounding, Scientific Notation, Converting Between Units Closed Book Quiz Similar to Worksheet 5 CLASS QUIZ DATE:____________________

60 ACCURACY Accuracy measures how close a measurement is to an ACCEPTED or TRUE VALUE. It is expressed as a PERCENT VALUE (%). Often, poor accuracy is a result of flaws in equipment or procedure. EXAMPLE: Accepted Value  ag = 9.80 m/s2 Experimental Value  ag = 9.50 m/s2 Accuracy = 96.9 %

61 PRECISION Precision measures the reliability, repeatability, or consistency of a measurement. It is expressed as the accepted value ± a discrepancy. Often, poor precision is a result of flaws in techniques by the experimenter. EXAMPLE: Accepted Value  ag = m/s2 Experimental Value  ag = 9.50 m/s2 Precision = 9.80 m/s2 ± .306

62 ACCURACY vs. PRECISION EXAMPLE:
Describe the ACCURACY and PRECISION of each of the following results.

63

64 HOW DO WE DESCRIBE WHAT WE OBSERVE IN SCIENCE?
QUALITITATIVE DESCRIPTIONS Describing with words. These descriptions are made using the 5 senses. Example: colour of a solution odor of a chemical product sound of thunder QUANTITATIVE DESCRIPTIONS Describing with numbers (i.e., quantities). These descriptions are made by counting and measuring. height of a building speed of an airplane

65 REARRANGING EQUATIONS
POINTS to REMEMBER: Whatever you do to ONE SIDE of an EQUATION, you must do to the OTHER SIDE. Do not move the item you are trying to isolate. Move EVERYTHING ELSE!!! “Do the opposite” to move a variable. For example, to move a variable that is multiplied, divide by it.

66 REARRANGING EQUATIONS
Rearrange the following equations to solve for the variable indicated: a = v Solve for v. t

67 REARRANGING EQUATIONS
Rearrange the following equations to solve for the variable indicated: a =v Solve for v t Multiply both sides by t: t x a = v x t = t x a =v t Cross multiply: v = a x t

68 REARRANGING EQUATIONS
y = mx + b Solve for m.

69 REARRANGING EQUATIONS
y =mx + b Solve for m. Subtract b on both sides: y – b = mx + b – b y – b = mx Divide both sides by X: y – b = mx so m = y – b x x x

70 HOMEWORK DO WORKSHEET 6 (page 14 in your handout) for homework.

71

72 a = v solve for t: t at = v and t = v a
Rearrange the following equations and solve for the variable indicated. a) a = v solve for t: t at = v and t = v a

73 Solve for n: b) c = n (cross multiply) v cv = n

74 c) Solve for M n = m M nM = m M = m n

75 Solve for R d) PV = nRT nT nT PV = R nT

76 Solve for X e) y = mx + b y – b = mx +b – b y – b = mx y – b = x m

77 Solve for r f) C = 2∏r C = r 2∏

78 Solve for a: g) d = 1aT2 2 2d = 1aT22 2d = aT2 T T2

79 Solve for T h) d = 1aT2 2 2d = aT2 2d = T2 a √2d = T

80 Solve for ΔT i) V2 = V1 + aΔT V2 – V1 = a ΔT V2 – V1 = ΔT a

81 Solve for Vf d = (Vi + Vf)T 2 2d = (Vi + Vf)T 2d = (Vi + Vf)T x 1 T T
j) d = (Vi + Vf)T 2 2d = (Vi + Vf)T 2d = (Vi + Vf)T x 1 T T 2d=Vi + Vf T 2d – Vi = Vf


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