Lecture Notes (Prelim to Midterm) PDF

Lesson 1. Physical Quantities and Measurements MEET YOUR TARGETS At the end of this lesson, you will be able to: Solve measurement problems involving conversion of units, expression of measurements in scientific notation. VENTURE Over the years, people have de...

Lesson 1. Physical Quantities and Measurements MEET YOUR TARGETS At the end of this lesson, you will be able to: Solve measurement problems involving conversion of units, expression of measurements in scientific notation. VENTURE Over the years, people have developed units of measurement based on the parts of the human body to facilitate communication. Did you know? In the Philippines, we use nonstandard units of measure such as: Dama (palm) – width of palm Dali (digit) – breadth of a finger Talampakan (foot) – length of a foot Timuro – length of a forefinger Hakbang – a single stride Dakot – a handful, used to measure the amount of grains such as corn and palay Gusi – a jar used to measure volume of liquids such as tuba and vinegar Kaing – a container used to measure the amount of harvested mangoes, tomatoes and salt 1|Page  SHORT HISTORY The use of standard units of measure can be traced back to ancient times. Ancient people used objects such as ropes, stones and sticks to measure other objects. They also create units based on human body parts such as arms, hands and feet. The cubit was described as the distance from man’s elbow to the tip of his middle finger. Another unit of measure was the palm which was the width of a person’s thumb was called uncia. Twelve uncias was equivalent to one foot. The yard is based on the distance from tip of a person’s nose to the tip of his middle finger. Ancient people also used the king’s foot as the standard unit of measure for buying, selling and trading. As long as the king ruled, it would be the standard unit of measure. Body parts, however, vary from one person to another. Hence, people would give different measurements for the same length using the same unit, say, a cubit. To eliminate such confusion, an international convention agreed to use standard units like meter, kilogram and second. A standard is an amount/quantity that everyone agrees on. In 1790, the French National Assembly appointed the committee that created the metric system. The metric system is used for a more convenient and accurate standard unit of measurement. The modernized version of the metric system is called the International System of Measurements, officially abbreviated as SI. CHAT WITH THE EXPERT In this lesson, we will consider three of the fundamental physical quantities – length, mass, and time. Length is the measured distance between two points. The standard unit of measurement for length is the meter. The meter was defined as 1 650 763.73 wavelengths of the orange-red light emitted from a krypton-86 lamp. However, in October 1983, the meter was redefined to be the distance traveled by light in a vacuum at 1/299 792 458 of a second. In effect, this definition establishes that the speed of light in a vacuum is 299 792 458 meters per second. 2|Page  Just as the meter is the unit for measuring length, the kilogram is the unit for measuring mass. The standard for the kilogram is a cylinder of platinum-iridium alloy kept by the International Bureau of Weights and Measures in Sevres, France. The cylinder was nicknamed ‘Le Grand K’; it is the one and only true kilogram. Its mass was originally chosen to be as close as possible to that of pure water at a pressure of 1 atmosphere and a temperature of 4°C, which occupies a volume of 1000 ml or 1 liter. One kilogram (kg) is equal to 1000 grams (g). The second is the SI unit for time. It is operationally defined as 9 192 631 770 times the period of one oscillation of the cesium – 133 atom. Conversion of Units Units in different systems or even different units in the same system can express the same quantity. To avoid confusion, it is, therefore, necessary to convert the units of a quantity from one unit to another. You may need to convert inches into feet, inches into centimeters, or even centimeters into meters. Conversion of units can be done by multiplying the original unit by Did you know? an appropriate ‘conversion factor’.  The diameter of a CD is 0.12m Conversion factors are simply  The diameter of the aorta is 0.018 m. equivalence statements expressed in  The diameter of a proton is 2 x 10 -15 m. the form of a ratio equal to 1.  The mass of a baseball is 0.15 kg. Given: 1 inch = 2.54 cm and  A human heart beats about once every vice versa second.  The human reaction time is 0.1 s. 1 inch = 2.54 cm = 1  The length of a football field is 102 m. 2.54 cm 2.54cm  A normal adult’s arm span or height is almost 2 meters. or  A liter of water has a mass of 1 kilogram.  The time between normal heartbeats is 0.8 s. 1 inch = 1 inch = 1 2.54 cm 1 inch 3|Page  Since conversion factors are equal to 1, they can be multiplied by other factors in equations without changing the validity of the equations. The answer will just be in different units. When using conversion factors, set the problem as follows: Quantity sought = given quantity x conversion factors In converting units, you must take advantage of unit analysis. That is, choosing the appropriate form of conversion factor that will allow cancellation of unwanted units and thus give the answer in the desired unit. Consider the following examples. 1.) How many meters are there in 2.1 miles? 1 mile = 1.6 km 1 km = 1000 m 2.1 mi. x 1.6 km x 1000 m = 3 360 meters 1 mi 1 km 2.) A bus moves at 120 km/h along the highway. What is its speed in m/s? 1 km = 1000 m 1 h = 3 600 s 120 km 1000 m _1 h_ = 33.33 m/s h 1km 3 600 s 3.) Find the distance in (a) meters and (b) miles if the distance between Manila and Bataan is 130 kilometers. Given: d = 130 km Conversion factor: 1 km = 1000 m = 0.62 mi Solution: a.) d = 130 km x 1000 m = 130, 000 m 1km b.) d = 130 km x 0.62 mi = 80. 6 mi 1 km 4.) Starting with the definitions of 1 inch = 2.54 cm, compute the number of kilometers in one mile, to five significant figures. Given: One mile Conversion factors: 1 mi = 5280 ft; 1 ft = 12 in.; 1 in = 2.54 cm 1 m = 100 cm 1 km = 1000 m Solution: a.) d = 1 mi x 5280 ft x 12 in x 2.54cm x _1 m x _1 km_ = 1.609 km 1 mi 1ft 1 in 100cm 1000m 4|Page Practice Time 1. Convert 36 ft/s2 into km/hr. 2. Compute the number of seconds in a day, and in a year. Answer in p. 5 PRACTICE TIME 5|Page  Lesson 2. Accuracy and Precision MEET YOUR TARGETS At the end of this lesson, you will be able to: Differentiate accuracy from precision VENTURE Figure 1. Dartboards showing different accuracy and precision scenarios 6|Page  CHAT WITH THE EXPERT Precision and accuracy are two ways that scientists think about error. Accuracy refers to how close a measurement is to the true or accepted value. Precision refers to how close measurements of the same item are to each other. Precision is independent of accuracy. That means it is possible to be very precise but not very accurate, and it is also possible to be accurate without being precise. The best quality scientific observations are both accurate and precise. If the darts are neither close to the bulls-eye nor close to each other, there is neither accuracy nor precision (Fig. 1.A). If all of the darts land very close together but far from the bulls-eye, there is precision, but not accuracy (Fig. 1.B). If the darts are all about an equal distance from and spaced equally around the bulls-eye, there is mathematical accuracy because the average of the darts is in the bulls-eye. This represents data that is accurate but not precise (Fig. 1.C). However, if you were actually playing darts, this would not count as a bulls-eye! If the darts land close to the bulls-eye and close together, there is both accuracy and precision (Fig. 1.D). 7|Page PRACTICE TIME (1) PRACTICE TIME (2) Situation: A student measures a Situation: A student measures test tube, she reported 15 g in the following temperature mass, but the actual mass of the 40.3°C, 41°C, and 40°C. test was 32 g. What conclusion can you What conclusion can you can you make out from make out from the the situation? situation? QUICK ANALOGY ON ACCURACY AND PRECISION Imagine a basketball player shooting baskets. If the player shoots with accuracy, his aim will always take the ball close to or into the basket. If the player shoots with precision, his aim will always take the ball to the same location which may or may not be close to the basket. A good player will be both accurate and precise by shooting the ball the same way each time and each time making it in the basket. 8|Page Lesson 3. Random Errors and Systematic Errors MEET YOUR TARGETS At the end of this lesson, you will be able to: Differentiate random errors from systematic errors. Estimate errors from multiple measurements of a physical quantity using variance. VENTURE 9|Page  CHAT WITH THE EXPERT Error is the technical term for uncertainty in reading a measurement. Making an error carries with it an implication of a mistake or a blunder. An error in measurement means an uncertainty between the measured value and the standard value. Let’s look at these errors closely and differentiate them. Afterward, we will estimate errors from multiple measurements of a physical quantity using variance. In performing measurements, errors are sometimes committed. They are classified as either random or systematic depending on the situation on how the measurement was obtained. Random errors are defined as variations in the measured data brought by the limitations of the measuring device. Random errors used statistical analysis. Averaging the large numbers of observation can reduce errors. Systematic errors are defined as reproducible inaccurate data that are constantly in the same direction. For example, if a systematic error is identified due to calibration of a measuring instrument based on standard, applying a correction factor to compensate for the effect can reduce the favored measurement. In performing measurements, the plan is to lessen the sources of error and to keep track of the errors that cannot be eliminated. It is important to understand the type of errors that may occur in an experiment to recognize if they arise. Causes of Error in Doing Laboratory Experiments 1.) Inadequate definition (either systematic or random – if two students measure the length of a rope, they will possibly get different results because either one may stretch the rope with a different force. A better way to reduce the error is to determine specific conditions that may affect the measurements. 2.) Unable to include a factor (systematic) – when measuring free fall, air resistance was not considered. A good way to analyze this source of error is to discuss all aspects that could 10 | P a g e probably affect the result before doing the experiment so that considerations can be made before doing the measurements. 3.) Factors due to the environment (either systematic or random) – errors brought by the environment such as vibrations, temperature, noise, or other conditions that may affect the measuring instrument. 4.) Limited scale of the instrument (random) – ammeter stick cannot measure exactly in the smallest scale division. 5.) Unable to calibrate or check zero scale of the instrument (systematic) – if possible, always check the calibration of the instrument before taking measurements. 6.) Variations in the physical measurement (random) – take several measurements over a whole range that is being explored. This will reveal variations in the experiment that might not be noticed. 7.) Parallax (either systematic or random) – whether an experiment’s eye is not aligned with a pointer in a scale, the reading may differ, either too high or low. 8.) Personal errors – are errors that occur from carelessness, poor method, or bias measurement from the experimenter. Estimate errors from multiple measurements of a physical quantity using variance Understanding variance  A large variance indicates that numbers in the set are far from the mean and far from each other.  A small variance indicates that numbers in the set are near from the mean and near from each other.  A variance with zero value indicates that all values within a set of numbers are identical.  Every variance that isn’t zero is a positive number.  A variance cannot be negative because it is mathematically impossible. You can’t have a negative value resulting from a square of a number. Example #1. Using a meter stick, measure the width of a piece of paper. Being careful to keep the meter stick parallel to the edge of the paper (to avoid a systematic error which would cause the measured value to be consistently higher than the correct value), the width of the paper is measured at a number of points on the sheet, and the values obtained are entered in a data table. Note that the last digit is only a rough estimate, since it is difficult to read a meter stick to the nearest tenth of a millimeter (0.01cm). 11 | P a g e  Observation Width (cm) #1 31.33 #2 31.15 #3 31.26 #4 31.02 #5 31.20 Therefore, the average or the mean of the measures of the width of a piece of paper is equal to 31.192 cm. Step 2. Compute for the variance of the data. Formula of variance (s2) s2= the value of the variance xi = the value of each data x = the mean value of the data n = number of data Observation Width (cm) (xi – x) (xi – x)2 (xi – x)2 #1 31.33 31.33 – 31.192 = 0.138 (0.138)2 0.019044 #2 31.15 31.15 - 31.192 = -0.042 (-0.042)2 0.001764 #3 31.26 31.26 - 31.192 = 0.068 (0.068)2 0.004624 #4 31.02 31.02 - 31.192 = -0.172 (-0.172)2 0.029584 #5 31.20 31.20 - 31.192 = 0.008 (0.008)2.000064 Σ(xi – x)2 = 0.05508 s2 = 0.05508 5–1 Conclusion: Therefore, the value of the variance of the set of data is s2 = 0.05508 0.01377. Since the value of the variance is small, the 4 values of the measurements are near from the mean and near from each other. s2 = 0.01377 Example #2. Assume a gold ring with a certain mass to be sold. By observing the ring’s mass, the estimated mass is between 10 and 20 grams. However, this is not the best estimate. A device, such as a weight balance or digital weight balance, will show an exact mass of 17.43 grams. 15 | P a g e Step 1. Find the average/ mean of the data. Reading Mass (g) Reading #1 17.46 g Average/mean (x) = 17.46 + 17.42 + 17.44 + 17.45 Reading #2 17.42 g 4 Reading #3 17.44 g x = 69.77 Reading #4 17.45 g 4 x = 17. 44 Therefore, the average/ mean of the measurements of the mass of the gold ring is 17.44 Step 2. Calculate the variance of the given data. Reading mass(g) (xi – x) (xi – x)2 (xi – x)2 #1 17.46 g 17.46 – 17.44 = 0.02 (0.02)2 0.0004 #2 17.42 g 17.42 - 17.44 = -0.02 (-0.02)2 0.0004 #3 17.44 g 17.44 - 17.44 = 0 (0)2 0 #4 17.45 g 17.45 - 17.44 = 0.01 (0.01)2 0.0001 Σ(xi – x)2 = 0.0018 s2 = 0.0018 4–1 Conclusion: Therefore, the value of the variance of the set of data is s2 = 0.0018 0.0006. Since the value of the variance is very small, the 3 values of the measurements are very near from the mean and very near from each other. s2 = 0.0006 Estimating Uncertainty for Multiple Measurements, Relative Uncertainty, and Uncertainty Percentage Example #3. The following measurement values are estimation of a falling object’s acceleration. Calculate the best estimate mean. Solve for relative uncertainty. 16 | P a g e  Observation Acceleration (m/s2) #1 9.69 m/s2 #2 9.76 m/s2 #3 9.85 m/s2 #4 9.91 m/s2 #5 9.83 m/s2 #6 9.78 m/s2 #7 9.72 m/s2 #8 9.74 m/s2 #9 9.91 m/s2 #10 9.71 m/s2 Step #1. Solve for the best estimate mean. Average/ mean (x) = 9.69 + 9.76 + 9.85 + 9.91 + 9.83 + 9.78 + 9.72 + 9.74 + 9.91 + 9.71 10 x = 9.7910 m/s2 is the best estimate mean of the given data Step #2. Solve for the variance of the given data Observation Acceleration (m/s2) (xi – x) (xi – x)2 (xi – x)2 #1 9.69 m/s2 9.69 – 9.7910 (-0.101)2 0.010201 #2 9.76 m/s2 9.76 – 9.7910 (-0.031)2 0.000961 #3 9.85 m/s2 9.85 – 9.7910 (0.059)2 0.003481 #4 9.91 m/s2 9.91 – 9.7910 (0.119)2 0.014161 #5 9.83 m/s2 9.83 – 9.7910 (0.039)2 0.001521 #6 9.78 m/s2 9.78 – 9.7910 (-0.011)2 0.000121 #7 9.72 m/s2 9.72 – 9.7910 (-0.071)2 0.005041 #8 9.74 m/s2 9.74 – 9.7910 (-0.051)2 0.002601 #9 9.91 m/s2 9.91 – 9.7910 (-0.101)2 0.010201 #10 9.71 m/s2 9.71 – 9.7910 (-0.081)2 0.006561 Σ(xi – x)2 = 0.05485 s2 = 0.05485 10 -1 s2 = 0.05485 9 s2 = 0.00609 is the value of the variance of the given data. 17 | P a g e  Example #4. Using a timer, a pendulum has a period (T) of oscillation: T = 0.44 seconds. If the measurement is repeated a number of times, the best way to analyze the different measured values is by averaging or obtaining the mean. Assume the following five measurements of an oscillating pendulum in seconds: Period Time (second) 1 0.46 2 0.44 3 0.45 4 0.44 5 0.41 Step #1. Solve for the mean of the data. Average/mean (x) = 0.46 + 0.44 + 0.45 + 0.44 +0.41 5 x = 2.2 5 x = 0.44 best estimate mean of the given data Step #2. Solve for the variance. Period Time (second) (xi – x) (xi – x)2 (xi – x)2 1 0.46 0.46 - 0.44 (0.02)2 0.0004 2 0.44 0.44 - 0.44 (0)2 0 3 0.45 0.45 - 0.44 (0.01)2 0.0001 4 0.44 0.44 - 0.44 (0)2 0 5 0.41 0.41 - 0.44 (-0.03)2 0.0009 Σ(xi – x)2 = 0.0014 s2 = 0.0014 5 -1 s2 = 0.0014 4 s2 = 0.00035 is the value of the variance of the given 18 | P a g e 19 | P a g e