Physics for Scientists and Engineers, Chapter 1 Lecture Notes PDF

Storyline Chapter 1: Physics and Measurement Physics for Scientists and Engineers, 10e Raymond A. Serway 1 John W. Jewett, Jr.  Standards of Length, Mass, and Time Standards in measurement must: be readily ac...

Storyline Chapter 1: Physics and Measurement Physics for Scientists and Engineers, 10e Raymond A. Serway 1 John W. Jewett, Jr.  Standards of Length, Mass, and Time Standards in measurement must: be readily accessible possess some property that can be measured reliably yield same result not change with time 2  Length Length: distance between two points in space Meter: distance traveled by light in vacuum during a time interval of 3 1/299 792 458 second Mass Kilogram: mass of a specific platinum– iridium alloy cylinder kept at the International Bureau of Weights4 and Measures at Sèvres, France  Time Second: 9 192 631 770 times the period of vibration of radiation from the cesium-133 atom 5 Powers of 10 6  Density m ρ≡ V Styrofoam Lead 7  Quick Quiz 1.1 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? (a) The aluminum cam is larger. (b) The iron cam is larger. (c) Both cams have the same size. 8  Quick Quiz 1.1 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? (a) The aluminum cam is larger. (b) The iron cam is larger. (c) Both cams have the same size. 9  Modeling and Alternative Representations A model is a simplified substitute for the real problem that allows us to solve the problem in a relatively simple way. Two primary conditions for using particle model: 1. Size of the actual object of no consequence in analysis of its motion. 2. Any internal processes occurring in object of no consequence in analysis of its motion. First category: geometric model 10  Example 1.1: Finding the Height of a Tree You wish to find the height of a tree but cannot measure it directly. You stand 50.0 m from the tree and determine that a line of sight from the ground to the top of the tree makes an angle of 25.0 ° with the ground. How tall is the tree? opposite side h tan θ = adjacent side 50.0 m h = ( 50.0 m ) tan θ ( 50.0 m ) tan 25.0° = 23.3 m 11 Modeling and Alternative Representations Second category: simplification model Third category: analysis model Fourth category: structural model 12  Modeling and Alternative Representations A representation is a method of viewing or presenting the information related to the problem 13  Dimensions speed: [ v ] L/T = = area: [ A] L 2 14  Dimensional Analysis 1 2 L 2 x = at L= ⋅T =L 2 T2 n m x∝a t  a n t m = L= L1T 0 ( L/T ) 2 n L1T 0 → ( Ln T m − 2 n ) = Tm = L1T 0 2 x ∝ at 15  Quick Quiz 1.2 True or False: Dimensional analysis can give you the numerical value of constants of proportionality that may appear in an algebraic expression. 16  Quick Quiz 1.2 True or False: Dimensional analysis can give you the numerical value of constants of proportionality that may appear in an algebraic expression. False 17  Example 1.2: Analysis of an Equation Show that the expression v = at, where v represents speed, a acceleration, and t an instant of time, is dimensionally correct. L [v] = T L L [ at ] = T = T2 T 18  Example 1.3: Analysis of a Power Law Suppose we are told that the acceleration a of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say rn, and some power of v, say vm. Determine the values of n and m and write the simplest form of an equation for the acceleration. n m a = kr v m n+m L n L L = 2 L=   T T T m + m 1 and n= = m 2 → n =−1 2 −1 2 v =a kr = v k r 19  Conversion of Units =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 ( exactly ) 20  Conversion of Units Convert 15.0 in. to centimeters. 2.54 cm 1 in. 2.54 cm → = 1 in.  2.54 cm  ( ) 15.0 in. = 15.0 in.   = 38.1 cm  1 in.  21  Quick Quiz 1.3 The distance between two cities is 100 mi. What is the number of kilometers between the two cities? (a) smaller than 100 (b) larger than 100 (c) equal to 100 22  Quick Quiz 1.3 The distance between two cities is 100 mi. What is the number of kilometers between the two cities? (a) smaller than 100 (b)larger than 100 (c) equal to 100 23  Example 1.4: 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 the driver exceeding the speed limit of 75.0 mi/h?  1 mi   60 s   60 min  ( ) 38.0 m / s     = 85.0 mi/h  1609 m   1 min   1h  Yes, he is speeding! 24  Example 1.4: Is He Speeding? What if the driver were from outside the United States and is familiar with speeds measured in kilometers per hour? What is the speed of the car in km/h?  mi   1.609 km   85.0    h   1 mi  = 137 km/h 25 Estimates and Order-of-Magnitude Calculations −2 0.0086 m  10 m 0.0021 m  10−3 m 720 m  103 m 26  Example 1.5: Breaths in a Lifetime Estimate the number of breaths taken during an average human lifetime.  400 days   25 h   60 min  5 1 yr    = 6 × 10 min  1 yr   1 day   1h     number of= minutes ( 70 yr ) ( 6 ×10 5 min/yr ) = 4 × 107 min number of breaths = (10 breaths/min )( 4 × 10 7 min ) 8 = 4 × 10 breaths 27  Example 1.5: Breaths in a Lifetime What if the average lifetime were estimated as 80 years instead of 70? Would that change our final estimate? ( 80 yr )( 6 × 10 5 min/yr ) 5 = × 10 7 min (10 breaths/min )( 5 × 10 7 min ) 5 = × 108 breaths 9 on the order of 10 breaths 28 Significant Figures ( 6.0 ± 0.1) cm 29  Significant Figures 1500 g → ? significant figures 1.500 × 103 g → 4 significant figures 3 1.50 × 10 g → 3 significant figures 2 1.5 × 10 g → 2 significant figures −4 2.3 × 10 → 2 significant figures → 0.0023 2.30 × 10−4 → 3 significant figures → 0.000 230 30  Significant Figures When multiplying several quantities, the number of significant figures in the final answer is the same as the number of significant figures in the quantity having the smallest number of significant figures. The same rule applies to division. 31  Significant Figures A = π r= π ( 6.0 cm )= 1.1× 10 cm 2 2 2 2 calculator answer: 113.097 3355 113 cm 32  Significant Figures When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum or difference. 23.2 + 5.174 = 28.374 → 23.2 has one decimal place → sum = 28.4 1.0001 + 0.0003 = 1.000 4 1.002 − 0.998 = 0.004 33  Significant Figures In this book, most of the numerical examples and end-of- chapter problems will yield answers having three significant figures. When carrying out estimation calculations, we shall typically work with a single significant figure. last digit dropped > 5: increase last retained digit by 1: 1.356 → 1.35 last digit dropped = 5: increase last retained rounded to nearest even number: 1.345 → 1.34 last digit dropped < 5: leave last retained as is: 1.343 → 1.34 34  Example 1.6: Installing a Carpet A carpet is to be installed in a rectangular room whose length is measured to be 12.71 m and whose width is measured to be 3.46 m. Find the area of the room. 43.976 6 m 2 12.71 m × 3.46 m = 44.0 m 2 3.46 → 3 sig figs → A = 35