Chapter 1: Physics and Measurement 1.1. International System of Units (SI) Fundamental Basic Quantities: Derived Quantities: Length = meter = m Force = Newton = N = kg  (m/s2) Mass = kilogram = kg Energy = Joule = J = N  m Time = second = s Power = Watt = W = (J/s) Other derived units: For example: 1 millimeter = 1 mm = 10-3 m 1 kilometer = 1 km = 1000 m = 103 m 1 kilogram = 1 kg = 1000 grams = 103 g

(The speed of light = c = 3 x 108 m/s). Base Unites The (SI) Base Unite: m (meter): The distance traveled by light in vacuum during a time of 1/299792458 second. (The speed of light = c = 3 x 108 m/s). kg (kilogram):The mass of a specific platinum – iridium cylinder. S (second): The time taken by 9192631770 light oscillations of a particular wavelength emitted by a cesium-133 atom.

The U.S. customary system The U.S. customary system, is used in the United States. In this system, the units of: Length = foot, mile, yard, inch, Mass = slug, Time = second. ----------------------------------------------------------------------------------- feet = ft mile = mi inch = in 1 ft = 12 in. 1 yard = 3 ft = 36 in. 1 mi = 5280 ft = 1760 yards 1 mi = 1 609 m = 1.609 km 1 ft = 0.304 8 m = 30.48 cm 1 m = 39.37 in = 3.281 ft 1 in. = 0.025 4 m = 2.54 cm

1.2 Matter and Model Building

(aluminum) = 2.7 g/cm3 - (Lead) = 11.3 g/cm3 The density () of any substance is defined as its mass (m) per unit volume (V): (aluminum) = 2.7 g/cm3 - (Lead) = 11.3 g/cm3 Mass (atom, nucleus, proton, neutron, electrons) is measured by: atomic mass unit (u) [ 1 u = 1.6605387 X 10-27 kg] Quick Quiz 1.1 (Page 10): In a machine shop, two cams are produced, one of aluminum and one of iron. Both cams have the same mass. Which cam is larger? The aluminum cam The iron cam Both cams have the same size Density of aluminum is lower than the density of iron (Answer a)

1.4 Dimensional Analysis The symbols to specify the dimensions of: length ( L ), mass ( M ), and time ( T ). For example: The dimension of speed ( V ) is: ( V ) = ( L / T ). The dimension of area (A) is: (A) = ( L2 ).

Example 1.2 (Page 12): Analysis of an Equation Show that the expression (v = at) is dimensionally correct, where (v) represents speed, (a) acceleration, and (t) an instant of time. Solution: For the speed term, v = [L / T] So, the dimension of [at ] is: Therefore the expression (v = at) is dimensionally correct. If the expression is (v = at2) it would be dimensionally incorrect. (Try)

1.5 Conversion of Units Example: Covert (15 inch) to (cm) Some units: 1 kg = 2.2 pound = 2.2 lb 1 mile = 1609 m = 1.609 km 1 ft = 0.3048 m = 30.48 cm 1 m = 39.37 inch = 3.281 ft 1 inch = 0.0254 m = 2.54 cm Example: change (200 lb) into (kg) (200) lb = (200) (1/2.2) = (90.9) kg Example: Covert (15 inch) to (cm) 15 (inch) = 15 (2.54 cm) = 38.1 cm Quick Quiz 1.3: The distance between two cities 100 mile. The number of kilometers between the two cities is: Smaller than 100 (b) Larger than 100 (c) Equal to 100

Example 1.4 (Page 13): Is He Speeding? On an interstate highway in a rural region of Wyoming, a car is traveling at a speed of 38.0 m/s. Is this car exceeding the speed limit of 75.0 mi/h? Solution: (38 m/s) = (38) (60/1609) = 85 mi/h (The car exceeding 75 mi/h) What If? What if the driver is from outside the U.S. and is familiar with speeds measured in km/h? What is the speed of the car in km/h? (85 mi/h) = (85) (1.609) = 137 km/h Example: change (200 lb) into (kg) (200) lb = (200) (1/2.2) = (90.9) kg Example 1.5: How many (milligrams) are in a (kilogram)? Solution: (1 kg) = (1) (1000 g) (1000 mg) = 1000000 mg = 106 mg

You have (1.5) pounds of gold. Find its volume (V) Example 1.6: You have (1.5) pounds of gold. Find its volume (V) in cm3 if the density () of gold is (19.3) g/cm3. Note: (1) kg = (2.2) lb → lb = (1/2.2) kg = (1000/2.2) g solution mass = ρ V → V = (mass ∕ ρ) Then: V = (1.5) (1000/2.2) / (19.3) = 35.3 cm3

It is often useful to compute an approximate answer to a given physical problem even when little information is available. 0.0086 = 0.86  10-2  1  10-2  10-2 0.0021 = 2.1  10-3  10-3 720 = 0.720  103  1  103  103

Example 1.5 (Page 14): Breaths in a Lifetime Estimate the number of breaths taken during an average life span. Solution: The number of minutes (min) in a year (yr) is approximately: (1 yr) = (1) (400 days) (25 h) (60 min) ≈ 6 x 105 min Thus, in 70 years there will be: (70 yr) = (70) (6 x 105 min) ≈ 4 x 107 min At a rate of (10 breaths/min), an individual would take = (10) (4 x 107 min) = 4 x 108 ≈ 109 breaths What If? What if the average life span were estimated as 80 years instead of 70? Would this change our final estimate? (80 yr) = (80) (6 x 105 min) ≈ 5 x 107 min At a rate of (10 breaths/min), an individual would take = (10) (5 x 107 min) = 5 x 108 ≈ 109 breaths No effect of the increased life span on the number of breaths!

Significant Figures All non-zero digits are significant Zeros are significant when between other non-zero digits after the decimal point and another significant figure can be clarified by using scientific notation 3 significant figures 5 significant figures 6 significant figures

Operations with Significant Figures Accuracy -- number of significant figures When multiplying or dividing, round the result to the same accuracy as the least accurate measurement When adding or subtracting, round the result to the smallest number of decimal places of any term in the sum. Example: meter stick: 2 significant figures Example: rectangular plate: 4.5 cm by 7.3 cm area: 32.85 cm2 33 cm2 Example: 135 m + 6.213 m = 141 m

Solved Problems The standard kilogram (m = 1 kg) is a platinum – iridium cylinder (39.0 mm) in height (h) and (39 mm) in diameter (2 r). What is the density () of the material? Solution h = 39 mm = 39 x 10-3 m r = (39/2) = (19.5 mm) = 19.5 x 10-3 m The Volume of the Cylinder (V) = (π r2) (h) = (3.14) (19.5 x 10-3)2 (39 x 10-3) = 4.66 x 10-5 m3 The density () of the cylinder = (m/V) = (1/ 4.66 x 10-5) = 2.15 x 104 kg/m3

Area of each wall = (8 ft x 12 ft) = (96) ft2 18. A worker is to paint the walls of a square room (8 ft) high and (12 ft) along each side. What surface area in square meters must she cover? solution Area of each wall = (8 ft x 12 ft) = (96) ft2 Note: The room has 4 surface, Then: Total area of the room = 4 (96) ft2 = 4 (96) = 384 ft2 = (384) (0.3048 m) 2 = 35.7 m2