EVERYTHING YOU NEED TO KNOW FOR PHYSICS (EXCEPT THE PHYSICS) Part I: Percent Error Percent error is a way of measuring how good your lab data is. It compares the value you have measured with the accepted value—the value you'd see in a textbook or the value you think you ought to be getting. If you have a precise, highly accurate, widely-accepted textbook value (such as the mass of the electron, the universal constant of gravity, etc.), then you assume that the textbook value is correct, and any possible deviation is due to your error. You compare your differences purely against the textbook ("actual") value, %= v actual − v measured × 100 v actual If you do not have a textbook value, if you are working with a calculated, theoretical value (such as the speed of your cart), then you cannot assume that the calculated, theoretical value is 100% accurate. Differences between the measured value and the calculated value are not necessarily signs of an egregious boo-boo in the laboratory: you must address the possibility that your calculated value is somewhat off. You compare your differences not against the calculated value, but against the average of the measured and calculated values. %= v theoretical − v measured × 100 a a= v theoretical + v measured 2 In standard scientific investigations, a percent error of under 5% is acceptable. In a high school laboratory, conditions are not as ideal as we might like them, and we won't quibble with anything under 10%. Anything over 20% is definitely iffy, and anything over 30% is begging for a re-do. (Anyone who beats my personal record of -428.7% error will get a round of applause and should probably seek assistance before re-doing the experiment.) Speaking of which, percent error is an absolute value: it is never negative! No one cares whether you're 67% too high or 67% too low: either way, you're 67% off, your data is lousy, and you have to re-do the lab. Always express percent error as a percent (4.3%) never as a decimal (0.043). At no point in the year should I ever have to answer the question, "What's the equation for percent error?" You will be expected to have the procedures memorized or look them up. -1- Uncertainty Percent error is about accuracy: how truthful your data is. Uncertainty is about precision: how consistent your data is. All data is science comes with some sort of margin of error. The narrower the margin of error, the more precise your data is. The graph above charts the comparative brightness of a certain star as it appears in the sky. The vertical bars represent the uncertainty. As you can see, when the hazy clouds rolled in at 1:30 a.m., the uncertainty became enormous: the computer measured all sorts of numbers not all of which were accurate or relevant. Uncertainty is often graphed in a box-and-whisker plot like this. The mathematical formula for uncertainty is half the difference between the highest and lowest number in your data: vhighest − vlowest u= 2 It is always positive, and it is always expressed with a ±, as shown below. Example: July 2011 daily high temperatures Mean T = 89° Fahrenheit Highest recorded T = high of 99° 99 − 75 u= Lowest recorded T = high of 75° 2 T = 89 ± 12 When we say that this was the 3rd-hottest summer in Chicago history, this is what we mean! -2- 1. The acceleration of gravity is 9.81 m/s2 at sea level. You measure it to be 9.63 m/s2. What is your percent error? Ans. 1.8% How good is your data? Ans. GREAT! 2. According to your calculations, the coefficient of friction (traction) for your sneakers is 0.43. You measure it to be 0.51. What is your percent error? Ans. 17% How good is your data? Ans. Not so hot! 3. According to your calculations, the current in your circuit should be 0.38 A. You measure it to be 0.56 A. What is your percent error? What should your next step be? 4. Calculate the percent error. The speed of sound in dry air = 343 m/s (according to the book) Measured frequency (ν) = 241 s-1 Measured wavelength (λ) = 1.31 m Calculated speed = s = λν = 316 m/s What should your next step be? -3- Part II: The Metric System Virtually all of our work will be done in the metric system. (Get used to it.) More helpfully, we’ll be working in fundamental units and other units derived directly from those units. These units, for the most part, are not selected for convenience (minutes, hours, etc.) or for familiarity (pounds, Calories, etc.) but for the way they make the math work out. Metric Units for Physics: Fundamental units Derived units Measures Length Time Mass Temperature Electric current Unit meter second kilogram Kelvin Ampere Symbol m s kg K A Equivalent To Volume Speed Acceleration Force Energy Power Electric charge meter3 meters per second meters per second2 Newton Joule Watt Coulomb m3 m/s m/s2 N J W C m3 m/s m/s2 kg⋅m/s2 kg⋅m2/s2 kg⋅m2/s3 A⋅s Please note that the standard units of mass is NOT the gram, it is the KILOGRAM! This is different than in chemistry, and it is done because a Newton needs to be 1 kg⋅m/s2, not 1 g⋅m/s2. You all have seen metric prefixes before. Here they are, just for old times’ sake: My Table of Metric Prefixes: femto pico nano micro milli centi deci f p n µ m c d 10-15 10-12 10-9 10-6 10-3 10-2 10-1 deka hecto kilo mega giga tera peta da h k M G T P 101 102 103 106 109 1012 1015 -4- Most common Most common You must have these prefixes memorized by no later than Wednesday September 2nd! You will be using them on a weekly basis, especially the ones I have marked as most common. On tests and quizzes, you will be on your own—I will not tell you what kilo stands for! More importantly, you will need to be ready to convert from one unit into another. In general, you do this by replacing the little letter with its corresponding power of 10. Example: Corresponding powers of 10 If… 1 ns = 1 × 10-9 s 1 µs = 1 × 10-6 s 1 ms = 1 × 10-3 s 1 ks = 1 × 103 s 1 Ms = 1 × 106 s Then… 1s=1× 1s=1× 1s=1× 1s=1× 1s=1× 109 ns 106 µs 103 ms 10-3 ks 10-6 Ms Example: Converting 136.8 µA to A To convert a prefix-unit into a base unit, multiply by the associated power of 10. 1 µA = 1 × 10-6 A 136.8 µA = 136.8 (×10-6 A) …see how you replace the µ with ×10-6? 136.8 × 10-6 A = 1.368 × 10-4 A Check: did we want our answer in pure Amperes? Yes! So we’re done. 136.8 µ A = 1.368 × 10-4 A. Example: Converting 324 meters to kilometers To convert a base unit into a prefix-unit, multiply by the opposite power of 10. If 1 km = 1 × 103 m, then 1 m = 1 × 10-3 km. 324 m = 324 (× 10-3 km) 324 × 10-3 km = 0.324 km This is the same as saying: 324 m = 0.324 km 10 3 Example: Converting 901 cg to kg To convert a prefix-unit into another prefix-unit, first convert to a base unit and then proceed. If 1 cg = 1 × 10-2 g, then 1 g = 1 × 102 cg. 901 cg = 901 (× 10-2 g) 901 × 10-2 g = 9.01 g Now you’re ready to worry about the kg! 9.01 g = 9.01 (× 10-3 kg) 9.01 × 10-3 kg = 0.00901 kg… …But an answer of 9.01 × 10-3 kg is much preferred! Fewer zeroes to lose track of, you see. -5- 1. Express the following in meters: 81.1 cm Ans. 0.811 m 603,428 µm 0.39 mm 2. Express the following in kilograms: 34.0 g Ans. 0.0340 kg 652.3 mg Ans. 0.6523 g = 6.523 × 10-4 kg 0.18 Mg 71,300,000 ng 2 Tg -6- Part III: Scientific Notation and Significant Figures Scientific notation you’ve all seen before. It’s when something long and ungainly and full of zeroes gets shortened using powers of 10. Here is a quick review to brush the dust off. To write a BIG number in scientific notation: 306000000 Draw a decimal point where it IS NOW: 306000000. MOVE the decimal RIGHT, counting how many spaces you must move it: 3.06000000 = 3.06 _ 108 To write a small number in scientific notation: 0.00000000098 Draw a decimal point where it SHOULD BE: 0.0000000009.8 MOVE the decimal LEFT, counting how many spaces you must move it: 0.0000000009.8 = 9.8_ 10-10 Whenever possible, leave your answers in scientific notation! Why? Because all numbers written in scientific notation have the proper number of significant figures. Scientific notation is the easiest way to keep track of significant figures—it is the only way to express numbers like 400 with two significant figures.* And significant figures in physics can make the difference between an answer that looks correct and an answer that looks incorrect. So let’s talk significant figures. RULE 1: ALL COUNTING NUMBERS 1-9 ARE SIGNIFICANT. This means that 3.1416 has five significant figures and 6.11 has three significant figures. Count them! RULE 2: ALL PLACE-HOLDING ZEROES ARE NOT SIGNIFICANT. • Any trailing zero that is NOT followed by a decimal point is defined as placeholding. Examples: The zeroes in 300 and 21,000 and 7,500 and 8,890 are all place-holders. This means that 300 has one significant figure, 21,000 has two, and 8,890 has three. The place-holders don’t count! Counterexamples: The zeroes in 300. and 21,004 and 7,500.0 and 8,890.0000 are all significant. The presence of a decimal point or a final integer makes all the difference! This means that 300. has three significant figures, 21,004 has five, and 8,890.0000 has eight. A Theory: any zero after the decimal point is significant. This is true EXCEPT IN CASE OF… How, you ask? 400 with two significant figures is 4.0 × 102. This is markedly different from 4 × 102 and 4.00 × 102. * -7- • Any zero in a pure decimal that comes in a long starting line of zeroes is defined as place-holding. Examples: The zeroes in 0.03 and 0.00096 and 0.00000131 are all place-holders. This means that 0.03 has one significant figure, 0.00096 has two, and 0.00000131 has three. The place-holders don’t count! Counterexamples: The trailing zeroes in 0.030 and 0.000960 and 0.00130100 are all significant. Any zero that comes after a decimal point AND after an integer counts! This means that 0.030 has two sig figs, 0.000960 has three, and 0.00130100 has six. When taking data in the lab, be sure to take at least two significant digits, if not three. Always specify whether your zeroes are trailing or not—write 50. rather than 50, if you can say for sure that your number is more precise than “fiftyish.” I will always try to give you at least three significant figures. There are different rules for significant figures math depending on whether you are adding/subtracting or multiplying/dividing. Add/subtract math is designed for precision and uniformity in data. Multiply/divide math is designed for scaling up or scaling down. Learn and/or memorize these steps! To Add or Subtract: • Perform the addition or subtraction. • Take the number with the fewest decimal places, and truncate your answer to that number of decimal places. Example: 30.6 kg + 29.9991 kg 30.6 kg: one decimal place 29.9991 kg: two decimal places 30.6 kg + 29.9991 kg = 60.5991 kg ≈ 60.6 kg Final answer: 60.6 kg with one decimal place To Multiply Or Divide: • Perform the multiplication or division. • Take the number with the fewest significant figures, and truncate your answer to that number of significant figures. Example: 5.412 N _ 30. m 30. m: two significant figures 5.412 N: four significant figures (5.412 N)(30. m) = 162.36 N•m ≈ 160 N•m Final answer: 160 Newton-meters with only two significant figures Moral of the story: hope you’re given more than two significant figures! -8- 1. How many significant figures in… 3,100 N Ans. 2 501.0 N 930.0 N Ans. 4 930 N 0.00375 N 0.000375 N 0.0003750 N Ans. 4 2. A tank is 13.0 m long, 0.14 m deep, and 3.7776 m wide. Calculate the volume of fluid in the tank. Ans. 6.9 m3 3. A torque is a twisting force (as opposed to a pulled force). If one balance beam experiences four separate torques… τ1 = 17.23 N×m τ2 = -13.966 N×m τ3 = -13.3400 N×m τ4 = 10 N×m …what is the total torque on the balance beam? (Hint: “total” means add.) Extra credit: what does this mean about the balance beam? 4. You calculate the temperature of Hell to be 104.1º C. Your measured value is 1.0 × 102 º C. Calculate your percent error. Ans. 4% … not bad! -9- Part IV: Unit Analysis Unit analysis is the science of units. It has two primary uses: I) The answer to the question, "Mrs. Eliaser, what units is the answer supposed to be in?" II) A foolproof method of checking your equations without an answer key To try some unit analysis, take the equation you're working on, and replace every letter with its associated unit. Then look and see if the statement you have written is true! If it is true, then you have a correct equation; if it is false, then you have some checkin' to do! Unit analysis only makes sense when you write your units as FRACTIONS! Get into the habit, and remember that PER means DIVIDE! mph Example: F = ma m m/h h m h m = 14 kg a = 2.0 m/s2 F = ma = (14)(2.0) = 28 … But 28 whatses? Well, if you'd have put the units next to the numbers, you'd see 28 whatses! Watch! ma = (14 kg)(2.0 m/s2) …now group your units together like this… ma = (14)(2.0)(kg)(m/s2) …or better still, like this… ma = (14)(2.0)(kg)( sm ) 2 Then just multiply the units together and see! F = 28 kg ⋅ sm 2 The unit of F—the kilogram-meter-per-second-squared—is named after Sir Isaac Newton, such 2 that 1 Newton = 1 kg ⋅ m/s . Example: F = Gm1m2 d2 You know that F is measured in Newtons (N), m is measured in kilograms (kg), and d is measured in meters (m). What are the units of G? Simply replace each letter with its component unit, then rearrange. Don’t forget little things like squares! F= Gm1 m2 d2 N= G(kg)(kg) N= (m) 2 Now multiply both sides by m2: N ⋅ m 2 = G ⋅ kg 2 And divide by kg2: N ⋅ m2 =G kg 2 -10- G ⋅ kg 2 m2 2 The units for G are N ⋅ m kg 2 . Example: distance = speed × time d = st s = 35 mph t = 15 seconds d = st = (35 mph)(15 sec) = 525 miles After 15 seconds, you have gone 525 miles. Right? Um? d = distance = miles s = speed = miles per hour t = time = seconds distance = (speed)(time) miles = (mph)(seconds) ⎛ mi ⎞ (sec ) mi = ⎝ hr ⎠ Now, think. Is this a true statement? Of course not! An example of a true statement would be: mi = ⎛ mi ⎞ (hr ) ⎝ hr ⎠ Conclusion: to use the equation d = st with the data above, you must first convert t in to hours. Otherwise, the equation will be wrong! A Note Regarding Fundamental Units: We already mentioned fundamental units in the context of the metric system. They are defined by standard weights and measures in France: there is a standard “kilogram” of a platinumiridium alloy, etc. Other units, including kph, Newtons, degrees Celsius, etc. are derived relative to these fundamental units. In unit analysis, it is usually preferable to use the fundamental units, since more things will cancel out that way. More importantly, all of your calculations must be done in fundamental units or in ways that directly relate to them. (That is to say, metersper-second instead of miles-per-hour.) Otherwise, your answers will be inaccurate! -11- 1. Using the equation p = mv, where m is measured in kilograms (kg) and v is measured in meters per second (m/s), find the units of p. Ans. p is measured in kg·m/s Extra credit: what’s p in physics? 2. Using the equation F = kQ1Q2 , where F is measured in Newtons (N), Q is measured in d2 Coulombs (C), and d is measured in meters (m), find the units of k. Extra credit: what’s the name of this equation and what topic is it describing? 3. Area = length × width. Find the basic metric unit for area using unit analysis. (Careful! This is not the unit you are used to seeing in math!) Ans. A is measured in m2 force to find the basic metric units of pressure. (Careful! This area is not the unit you are used to seeing in chemistry! Need the metric unit for force? I recommend page 3.) 4. Use the equation pressure = -12-

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