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UNFORMATTED ATTACHMENT PREVIEW
Physics 1103 Multimedia Project Description These are worth 60 points EACH! That is the same weighting as a midterm! This is a serious assignment and will require serious effort from all group members. You must complete three projects. These are to be completed together as a team/table. If you are unable to sync schedules and wish to complete a project individually, you must email your instructor and get permission. For each project you will design and implement an experiment to test some physical phenomenon/behavior related to this class. For example, you might set up a lever and fulcrum to lift a large weight using a small weight. Or you might film a falling object and use the frame rate to verify the acceleration due to gravity on the surface of Earth. Each project must have the following, and will be graded on these criteria: • • • • • • • • (5 pts) A clear, short statement of the phenomenon you will test (one or two sentences). (10 pts) An explanation of the physics behind this phenomenon (including relevant equations and pictures/diagrams). (5 pts) A clear statement of how you ensured the safety of everyone in your team during this project. DO NOT DO ANYTHING UNSAFE! Do not use electricity irresponsibly or lean off of high platforms to get good data! Your safety is more important than anything in this class! (5 pts) A video or photograph of the experimental setup. (5 pts) A clear statement of each group member’s contribution to the project. Falsifying this to say that students contributed who did not could be considered academic misconduct. (10 pts) A clear explanation of why you set up the experiment the way you did. These are NOT free points! You need to think about potential problems with your setup (friction, forces you forgot about, heat losses, imprecise measurements, etc.). You need to design your experiment to avoid problems and get the best data possible. (10 pts) Data collected (presented in a plot, table, or other suitable format). (10 pts) Interpretation of the data. This means determining whether the data show the behavior you expected. If so, explain how the data show that. If not, give possible explanations why not, cite specific potential sources of error and describe how the experiment could be improved. DO NOT SAY “human error”. You have to identify human errors and correct them in the project. Citing specific sources of error here might mean explaining that results are skewed due to air resistance and proposing that the experiment could be redone in an evacuated chamber (not realistic for this project). Timing and submission: You may present the information you collect in any appropriate format, including video, as a written report, as a PowerPoint presentation, etc. You will post the project in your chosen format and submit a link to the project in Carmen. DO NOT UPLOAD THE PROJECT DIRECTLY TO CARMEN – space limitations prevent us from storing all projects on Carmen; submit only a link to project. Project due dates (due at 11:59 PM): Project 1 – Monday, Oct. 18 Project 2 – Monday, Nov. 1 Project 3 – Monday, Nov. 15 You may complete these early, if you like, and turn them early. But they must be turned in NO LATER than 11:59 PM on the days indicated above, and they may NOT be resubmitted, even if turned in early. Chapter 17: Electric Circuits Goals of Period 17 Section 17.1: Section 17.2: Section 17.3: Section 17.4: To define resistance in series circuits To define resistance in parallel circuits To measure human resistance To illustrate combination circuits 17.1 Resistors in Series As current flows through a resistor, the resistor reduces the electrical potential energy of the circuit by transforming electrical energy into other forms of energy. For example, a toaster’s resistor converts electrical energy into thermal energy, which is then converted into radiant energy. Since a resistor reduces the electrical potential energy per charge, the voltage across the resistor is reduced, producing a voltage drop. Bulbs connected in series When circuit elements are connected so that the same current flows through each element, the elements form a series circuit. Connecting resistors in series gives a larger total resistance. A series connection of resistors makes a single equivalent resistor whose resistance is the sum of the individual resistances. The batteries’ voltage in a series circuit is divided across the circuit elements. Those elements with the greatest resistance receive the most voltage. The sum of the voltage drops across the circuit elements equals the voltage boost of the batteries. For resistors connected in series: RTot = R1 + R2 + R3 + …. (Equation 17.1) where RTot = total resistance of the resistors connected in series (in ohms) R1 = resistance of the first resistor (in ohms) R2 = resistance of the second resistor (in ohms) R3 = resistance of the third resistor (in ohms) (Example 17.1) Four resistors with resistances of 1.0 ohm, 0.5 ohm, 1.5 ohms, and 3.0 ohms are connected in series in a circuit. What is the total resistance of these resistors when they are connected in series? RTot = R1 + R2 + R3 + R4 = 1.0 + 0.5 + 1.5 +3.0 = 6.0 © The Ohio State University, 2014 139 8/20/14 This principle follows from Ohm’s law: voltage = current x resistance. In a series circuit, the same amount of current flows through each element. The load device with the greatest resistance requires the largest voltage to push the current through it. (Example 17.2) A 0.1 amp current flows through two resistors in series. Resistor 1 has a resistance of 10 ohms and Resistor 2 has 20 ohms. What is the voltage drop across Resistor 1? (Assume that the connecting wires have no resistance.) V = I R = (0.1 amps) x (10 ohms) = 1 volt Concept Check 17.1 a) What is the voltage drop across Resistor 2 in example 17.2? ______________ b) What is the voltage boost given by the battery in example 17.2? Assume that the only load devices in the circuit are Resistors 1 and 2 and that the connecting wires have no resistance. ____________________ 17.2 Resistors in Parallel When circuit elements are connected so that the current has multiple paths to flow through, a parallel circuit is formed. Current flows more easily through a parallel circuit because the current has multiple paths through the resisting material. A parallel connection of resistors makes a single equivalent resistor that has less total resistance than a circuit with just one resistor. The more parallel paths for current to follow, the lower the resistance. The total resistance of a parallel circuit Bulbs connected in parallel is always less than the resistance of the smallest resistor in the circuit. Figure 17.1 illustrates the comparative total resistance of four circuits consisting of identical resistors connected to identical batteries. In parallel circuits, sum of the voltage drops in each independent parallel branch (path) equals the voltage boost of the batteries. © The Ohio State University, 2014 140 8/20/14 Fig. 17.1 Comparison of Total Resistance of Circuits with Identical Resistors Most resistance Less resistance Even less resistance Adding resistors in series increases the total circuit resistance. Least resistance Adding resistors in parallel decreases the total circuit resistance. In series circuits, each added resistor increases the circuit resistance, and the total circuit resistance is the sum of the individual resistances. In parallel circuits, each added resistor decreases the circuit resistance, and the total circuit resistance is found by summing the inverses of each individual resistance as shown in Equation 17.2. 1 RTot 1 R1 1 R2 1 R3 (Equation 17.2) …. where RTot = total resistance of the resistors connected in parallel (in ohms) R1 = resistance of the first resistor (in ohms) R2 = resistance of the second resistor (in ohms) R3 = resistance of the third resistor (in ohms) (Example 17.3) Three resistors with resistances of 3 ohms, 4 ohms, and 6 ohms are connected in parallel in a circuit. What is the total resistance of this parallel circuit? 1 RTot 1 R1 1 R2 1 R3 . 1 3 1 1 4 6 4 3 2 12 12 12 9 12 Summing fractions involves finding a denominator common to each fraction. In this example, 12 is evenly divisible by the denominators of the fractions. Summing the fractions gives 1/RTot. To find RTot divide the numerator by the denominator. 1 RTot 9 12 RTot 12 9 1.33 As expected, the total resistance of the three parallel resistors is less than the resistance of the smallest resistor. © The Ohio State University, 2014 141 8/20/14 Skills and Strategies #13: Summing Fractions When adding fractions, each term must have the same denominator. The common denominator must be divisible by the denominator of each of the fractions. In some cases, a common denominator can be found by inspection. For example, when adding the fractions 1/2, 1/3, and 1/5, we find that 30 is divisible by each denominator. 1/2 = 15/30; 1/3 = 10/30; and 1/5 = 6/30 The total of the fractions is found by summing their numerators and dividing their sum by the common denominator. 15 10 6 30 30 30 31 30 1.03 If a common denominator cannot easily be found by inspection, one can be found by multiplying together the denominators of each fraction. To add 1/3, 1/4, and 1/7, multiply 3 x 4 x 7 = 84. The numerator of each fraction is the number of times the denominator can be divided into 84. 1/3 = 28/84; 1/4 = 21/84; and 1/7 = 12/84 The total of the fractions 1/3, 1/4, and 1/7, is 28 84 21 12 84 84 61 84 0.73 Concept Check 17.2 a) One electric circuit consists of a 1.5 ohm resistor connected to a battery. A second circuit consists of two resistors (1.5 ohms and 4.0 ohms) connected in parallel and attached to a battery. Which circuit has less total resistance? Why? ______________________________________________________________ b) What is the total resistance of a circuit connected in parallel that consists of a 2 ohm, a 3 ohm, and a 4 ohm resistor? ______________ © The Ohio State University, 2014 142 8/20/14 17.3 Resistance of Humans In class, we will measure the resistance of your body when you and your classmates are connected together in series and in parallel. The resistance of a human body from one hand to another is typically 5,000 to 40,000 ohms. Your resistance depends on how wet or sweaty your hands are and on how close your blood is to the surface of your skin. Electrically, your body is like a sack of salty water. The fluids in your body are good conductors of electric current, but the wall of the sack, your skin, has a high resistance when it is dry. Sweaty or wet hands lower the resistance of your skin, so that a better connection is made to the salty fluids inside. If you rub your hands together briskly, blood is brought closer to the surface of your skin, lowering your skin resistance. Some lie detectors work on the principle that people sweat under the stress of lying. The lie detector measures changes in the resistance of the human skin. 17.4 Combination Series and Parallel Circuits In class we will consider a circuit consisting of two bulbs in parallel connected in series to a third bulb. When all three bulbs are lit, current flows from the batteries through the single bulb (bulb 1). The current then splits, with some current flowing through the parallel bulb on the left (bulb 2) and the remainder of the current flowing through the bulb on the right (bulb 3). If the resistances of the two parallel bulbs are equal, the current splits evenly with one-half flowing through each of the parallel bulbs. 1 2 3 The voltage boost of the batteries is divided across bulbs 1, 2, and 3. Since bulbs in parallel have less resistance than a single bulb, bulbs 2 and 3 in parallel have less resistance and, thus, less voltage than bulb 1. If bulbs 2 and 3 have equal resistance, they will have equal voltages. As in any circuit, the sum of the voltage drops across the elements in each path connected to the batteries is equal to the total voltage boost from the batteries. (Example 17.4) In the combination circuit shown above, the bulbs have equal resistances and the connecting wires have no resistance. The batteries have a total voltage boost of 5.5 volts and bulb 2 has a voltage drop of 1.3 volts. What is the voltage drop across bulb 3? What is the voltage drop across bulb 1? Since bulbs 2 and 3 are connected in parallel and have equal resistances, they have equal voltages. Therefore, bulb 3 has a voltage drop of 1.3 V. The parallel path from the batteries, through bulb 1 and through bulb 2 has a total voltage drop equal to the voltage boost from the batteries, 5.5 volts. Since bulb 2 has a voltage drop of 1.3 V, bulb 1 must have 5.5 V – 1.3 V = 4.2 V voltage drop. © The Ohio State University, 2014 143 8/20/14 Period 17 Summary 17.1: The resistances of resistors in series add. RTot = R1 + R2 + R3 + …. The voltage drops across the resistors equals the voltage boosts across the circuit’s energy sources. The larger the resistance, the greater the voltage drop across the resistor. 17.2: Resistors in parallel provide additional paths for current flow and reduce the total resistance of the circuit. The more resistors in parallel, the lower the resistance of the circuit. The total circuit resistance is less than the resistance of the smallest resistor. For two resistors connected in parallel, the total circuit resistance is 1 RTot 1 R1 1 R2 1 R3 … 17.3: The resistance of the human body depends on factors such as moisture on the skin. These differences in resistance have been used in some lie detectors. 17.4: Circuits that combine series and parallel elements follow the same principles as described for simple series and simple parallel circuits. 1) The circuit elements with the greatest resistance have the largest voltage drop. 2) The sum of the total voltage drops across the circuit elements in each circuit path connected to the batteries equals the total voltage boost from the batteries. Solutions to Chapter 17 Concept Checks 17.1 a) b) V = I R = (0.1 amps) x (20 ohms) = 2 volts The battery’s voltage boost equals the sum of the voltage drops across the circuit elements. 1 volt + 2 volts = 3 volts © The Ohio State University, 2014 144 8/20/14 17.2 a) The circuit with 2 parallel resistors has less total resistance than the circuit with one resistor. Adding resistors in parallel provides more pathways for current to flow. Each additional parallel pathway reduces the total circuit resistance. b) 1 1 RTot R1 1 R2 1 . R3 1 1 1 2 3 4 6 4 3 12 12 12 13 12 To sum the fractions, find a denominator common to each fraction. In this example, 12 is divisible by each of the denominators of the fractions. Summing the fractions gives 1/RTot. To find RTot divide the numerator by the denominator. 1 RTot 13 12 © The Ohio State University, 2014 RTot 12 13 145 0.92 8/20/14 © The Ohio State University, 2014 146 8/20/14 Chapter 18: Electricity Use and Safety Goals of Period 18 Section 18.1: Section 18.2: Section 18.3: Section 18.4: Section 18.5: To define linear and exponential growth rates To explore the growth of electricity use in the U.S. To learn how to prevent electric fires To learn how to prevent electric shocks To examine the consequences of electric shocks 18.1 Linear and Exponential Growth A growth rate is the ratio of the amount of increase to the time elapsed. We consider two common models for growth rates – linear and exponential growth – using sample data for the increase in the number of oil wells and hydroelectric dams in a hypothetical country. Figure 18.1 presents the data for dams. Fig. 18.1: Linear Growth of Dams 10 Time Periods Dams 1998 0 5 1999 1 6 2000 2 7 2001 3 8 9 N (Number of Dams) Years 8 7 6 2002 4 9 5 1998 1999 2000 2001 2002 t (Time in Years) The growth rate of dams producing hydroelectric power is called linear because the number of dams increases by a constant amount during each time period (1 dam/year). When graphed, linear data form a straight line as shown in Figure 18.1. The equation describing linear growth uses the symbol N for the number of dams at a given time. This number depends on B, the initial number of dams (B = 5); t, the number of one-year time periods elapsed; and A, the amount of increase per time period. The value of A determines how steeply the straight graph line rises or falls and is known as the slope of the line. The relationship between A, B, N, and t is © The Ohio State University, 2013 145 8/18/13 N=Axt+B Applying the equation to the data in Fig. 18.1 verifies that the equation is correct. (A) (t) (B) In 2008, N = In 2009, N = 1 x 1 + 5 = 6 In 2010, N = 1 x 2 + 5 = 7 In 2011, N = 1 x 3 + 5 = 8 In 2012, N = 1 x 4 + 5 = 9 1 x 0 + 5 = 5 Next we consider a second type of growth rate called exponential growth. Figure 18.2 illustrates exponential growth using sample data for the increase in the number of oil wells in a hypothetical country. Fig. 18.2: Exponential Growth of Oil Wells Oil Wells Number of Wells as exponentials 1998 0 1 1 = 20 1999 1 2 2 = 21 2000 2 4 4 = 22 2001 3 8 8 = 23 2002 4 16 16 N (Number of Oil Wells) Time Periods Years 4 16 = 2 12 8 4 0 1998 1999 2000 2001 2002 t (Time in Years) The graph of exponential growth is not a straight line because the amount of the increase per time period is not constant. Exponential growth is characterized by a doubling of the amount of the quantity during a fixed time period. Since the amount increases by a factor of two, exponential growth is described by base 2 raised to an exponential power that is equal to the number of time periods elapsed. The last column of Figure 18.2 illustrates this basis of exponential growth. © The Ohio State University, 2013 146 8/18/13 The equation for the exponential growth uses B for the initial amount of the quantity (B = 1 oil well) and t for the number of one-year time periods elapsed. (We assume that there is only one doubling per time period.) N is the number of wells. In 2008, In 2009, In 2010, In 2011, In 2012, N = Bx 2 t N N N N N 0 = 1×2 = 1×2 = 1×2 = 1×2 = 1×2 1 2 3 4 0 = 2 = 1 = 2 = 2 = 2 = 4 = 2 = 8 = 2 = 16 1 2 3 4 Figure 18.3 uses the sample data on dams and oil well to compare the rates of linear and exponential growth. Figure 18.3 Increase in the Number of Dams and Oil Wells When studying energy, we may be interested in the amount of a quantity at a particular time, such as the amount of electrical energy used in 2010. More often we are interested in the © The Ohio State University, 2013 147 8/18/13 increase or decrease in a quantity over a period of time, such as the increase in energy use over the past 50 years. Table 18.1 presents sample data of the population of a town and energy used by that town from 1940 to 2000. Figure 18.4 graphs these data. Table 18.1 Sample Data on Population and Megajoules (MJ) of Energy Use Year Population Energy Use (MJ) Year Population Energy Use (MJ) 1940 500 50 1980 1,500 800 1950 750 100 1990 1,750 1,600 1960 1,000 200 2000 2,000 3,200 1970 1,250 400 2010 2,250 6,400 Population or Energy Use (in mega joules) Fig. 18.4 Sample Data on Population and Energy Use 6750 6500 6250 6000 5750 5500 5250 5000 4750 4500 4250 4000 3750 3500 3250 3000 2750 2500 2250 2000 1750 1500 1250 1000 750 500 250 0 1940 Energy Use Population 1950 © The Ohio State University, 2013 1960 1970 1980 Time (in years) 1…
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