Book review The Shareholder Value Myth: How Putting Shareholders First Harms Investors, Corporations, and the Public by Lynn Stout Provide 1) a 900 word review of this book (word range 900-1,200) and 2) a 350 word reflection where you force yourself to relate the message of the book . As per the format of the review, I like the ones done by the folks of the WSJ. This is an example: http://forums.delphiforums.com/diversecity/messages?msg=17531.1264 or http://www.wsj.com/articles/book-review-how-adam-smith-can-change-your-life-by-russ-roberts-1413846808?KEYWORDS=book+reviews

Book review The Shareholder Value Myth: How Putting Shareholders First Harms Investors, Corporations, and the Public by Lynn Stout Provide 1) a 900 word review of this book (word range 900-1,200) and 2) a 350 word reflection where you force yourself to relate the message of the book . As per the format of the review, I like the ones done by the folks of the WSJ. This is an example: http://forums.delphiforums.com/diversecity/messages?msg=17531.1264 or http://www.wsj.com/articles/book-review-how-adam-smith-can-change-your-life-by-russ-roberts-1413846808?KEYWORDS=book+reviews

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. An experiment is performed by passing light through one or two slits and the light pattern to the right is produced. Determine the best answer and explain how you can tell: the pattern produced was a result of a double slit interference experiment the pattern produced was a result of a single slit diffraction experiment More information is needed, i.e. the wavelength of the source, the distance it is from the slit(s) and how far the slit is to the viewing screen, before it can be determined which type of experiment produced the pattern

. An experiment is performed by passing light through one or two slits and the light pattern to the right is produced. Determine the best answer and explain how you can tell: the pattern produced was a result of a double slit interference experiment the pattern produced was a result of a single slit diffraction experiment More information is needed, i.e. the wavelength of the source, the distance it is from the slit(s) and how far the slit is to the viewing screen, before it can be determined which type of experiment produced the pattern

Single slit (Choice B) Reason: The central maximum is two … Read More...
ELEC153 Circuit Theory II M2A3 Lab: AC Series Circuits Introduction Previously you worked with two simple AC series circuits, R-C and R-L circuits. We continue that work in this experiment. Procedure 1. Setup the following circuit in MultiSim.The voltage source is 10 volts peak at 1000 Hz. Figure 1: Circuit for analysis using MultiSim 2. Change R1 to 1 k and C1 to 0.1 uF. Connect the oscilloscope to measure both the source voltage and the voltage across the resistor.You should have the following arrangement. Figure 2: Circuit of figure 1 connected to oscilloscope To color the wires, right click the desired wire and select “Color Segment…” and follow the instructions. Start the simulation and open the oscilloscope. You should get the following plot: Figure 3: Source voltage (red) and the voltage (blue) across the resistor The red signal is the voltage of the source and the blue is the voltage across the resistor. The colors correspond to the colors of the wires from the oscilloscope. 3. From the resulting analysis plotdetermine the peak current. To determine the peak current measure the peak voltage across the resistor and divide by the value of the resistor (1000 Ohms). Record it here. Measured Peak Current 4. Determine the peak current by calculation. Record it here. Does it match the measured peak current? Explain. Calculated Peak Current 5 Determine the phase shift between the current in the circuit and the source voltage. We look at the time between zero crossings to determine the phase shift between two waveforms. In our plot, the blue waveform (representing the circuit current or the voltage across the resistor) crosses zero before the red waveform (the circuit voltage). So, current is leading voltage in this circuit. This is exactly what should happen when we have a capacitive circuit. 6. To determine the phase shift, we first have to measure the time between zero crossings on the red and blue waveforms. This is done by moving the oscillator probes to the two zero crossing as is shown in the following figure Figure 4: Determining the phase shift between the two voltage waveforms We can see from the figure that the zero crossing difference (T2 – T1) is approximately 134 us. The ratio of the zero-crossing time difference to the period of the waveform determines the phase shift, as follows: Using our time values, we have: How do we know if this phase shift is correct? In step 4 when you did your manual calculations to find the peak current, you had to find the total impedance of the circuit, which was: Now, the current will be: Here, the positive angle on the current indicates it is leading the circuit voltage. 7. Change the frequency of the voltage source to 5000 Hz. Estimulate and perform a Transient Analysis to find the new circuit current and phase angle. Measure them and record them here: Measured Current Measured Phase Shift 8. Perform the manual calculations needed to find the circuit current and phase shift. Record the calculated values here. Do they match the measured values within reason? What has happened to the circuit with an increase in frequency? Calculated Current Calculated Phase Shift Writeup and Submission In general, for each lab you do, you will be asked to setup certain circuits, simulate them, record the results, verify the results are correct by hand, and then discuss the solution. Your lab write-up should contain a one page, single spaced discussion of the lab experiment, what went right for you, what you had difficulty with, what you learned from the experiment, how it applies to our coursework, and any other comment you can think of. In addition, you should include screen shots from the MultiSim software and any other figure, table, or diagram as necessary.

ELEC153 Circuit Theory II M2A3 Lab: AC Series Circuits Introduction Previously you worked with two simple AC series circuits, R-C and R-L circuits. We continue that work in this experiment. Procedure 1. Setup the following circuit in MultiSim.The voltage source is 10 volts peak at 1000 Hz. Figure 1: Circuit for analysis using MultiSim 2. Change R1 to 1 k and C1 to 0.1 uF. Connect the oscilloscope to measure both the source voltage and the voltage across the resistor.You should have the following arrangement. Figure 2: Circuit of figure 1 connected to oscilloscope To color the wires, right click the desired wire and select “Color Segment…” and follow the instructions. Start the simulation and open the oscilloscope. You should get the following plot: Figure 3: Source voltage (red) and the voltage (blue) across the resistor The red signal is the voltage of the source and the blue is the voltage across the resistor. The colors correspond to the colors of the wires from the oscilloscope. 3. From the resulting analysis plotdetermine the peak current. To determine the peak current measure the peak voltage across the resistor and divide by the value of the resistor (1000 Ohms). Record it here. Measured Peak Current 4. Determine the peak current by calculation. Record it here. Does it match the measured peak current? Explain. Calculated Peak Current 5 Determine the phase shift between the current in the circuit and the source voltage. We look at the time between zero crossings to determine the phase shift between two waveforms. In our plot, the blue waveform (representing the circuit current or the voltage across the resistor) crosses zero before the red waveform (the circuit voltage). So, current is leading voltage in this circuit. This is exactly what should happen when we have a capacitive circuit. 6. To determine the phase shift, we first have to measure the time between zero crossings on the red and blue waveforms. This is done by moving the oscillator probes to the two zero crossing as is shown in the following figure Figure 4: Determining the phase shift between the two voltage waveforms We can see from the figure that the zero crossing difference (T2 – T1) is approximately 134 us. The ratio of the zero-crossing time difference to the period of the waveform determines the phase shift, as follows: Using our time values, we have: How do we know if this phase shift is correct? In step 4 when you did your manual calculations to find the peak current, you had to find the total impedance of the circuit, which was: Now, the current will be: Here, the positive angle on the current indicates it is leading the circuit voltage. 7. Change the frequency of the voltage source to 5000 Hz. Estimulate and perform a Transient Analysis to find the new circuit current and phase angle. Measure them and record them here: Measured Current Measured Phase Shift 8. Perform the manual calculations needed to find the circuit current and phase shift. Record the calculated values here. Do they match the measured values within reason? What has happened to the circuit with an increase in frequency? Calculated Current Calculated Phase Shift Writeup and Submission In general, for each lab you do, you will be asked to setup certain circuits, simulate them, record the results, verify the results are correct by hand, and then discuss the solution. Your lab write-up should contain a one page, single spaced discussion of the lab experiment, what went right for you, what you had difficulty with, what you learned from the experiment, how it applies to our coursework, and any other comment you can think of. In addition, you should include screen shots from the MultiSim software and any other figure, table, or diagram as necessary.

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Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

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ITM 220 – Excel Assignment, ver 5.1 Individual (not group) Effort Data Analysis with Spreadsheet Software (Excel) OVERVIEW – The purpose of this assignment is to give you a chance to demonstrate your understanding of data analysis tools available in today’s spreadsheets such as Microsoft Excel. You’ll be working with data extracted from an ERPsim game performed by ITM 220 students a few months ago. This is not data from our class, so when considering the performance of “your team” (use data for Team “H” as your team data) don’t be surprised that it is significantly different from what you may recall for our ERPsim game. APPROACH – 1. Begin with the data found in this Excel Workbook–Worksheet named “ERPsimRawData” 2. Use tools and techniques as we discussed in class to study the data: a. Autofilter (and Advanced Filter) b. Conditional Formatting c. Lookup Table (VLOOKUP) d. Pivot Table e. Charts (smart ones, please) 3. Create multiple worksheets to document your answers to the following questions: Q1. What were average Muesli prices (per box) in the 2nd round? Organize by product. Label your answer worksheet “Q1”. COMMENT: It is necessary that you use a pivot table and be sure to use a chart, as well. Q2. How does your company (Team H – Leipzig) compare to your chief competitors–Munich and Stuttgart–on product prices? Use the lookup function (VLOOKUP) to add city name to your raw data and then use city name in your answer. Label your answer worksheet “Q2”. COMMENT: A pivot table is expected–chart, too. Q3. Which team has the most total sales revenue for each round? Label your answer worksheet “Q3”. COMMENT: Pivot table is expected–chart, too. Q4. Fill in the missing data in the worksheet labeled “FinancialForecast” –similarly to what we did in class. Use Scenario Manager to build 3 scenarios for projected company financial performance: Best Case, Expected case, and Worst Case scenarios (as we did in class) by modifying the appropriate blue highlighted cells in the Assumptions area. (Only 4 of these such highlighted cells will be used. Right?) Create an appropriate (your choice) chart illustrating the Summary Table results. Label your answer worksheet “Q4”. COMMENT: Must use Scenario Manager and you should be able to answer using a single worksheet for summary table and chart (as we did in class). Q5. Duplicate (NOT copy/paste) the “FinancialForecast” worksheet on which you have entered your assumptions data for the Expected Case. Label this new worksheet “Q5.” Use Goal seek to determine the average daily Grocery Chains sales needed in order to achieve Earnings Before Taxes of 500,000 Euros in Round 1. Discuss the plausibility of attaining such daily sales (probably based on the result of your answer to question 4). (And notice that growth rates are irrelevant to answer this question since we are talking FIRST round.) COMMENT: This is very easy using Goal Seek. However, a manual solution using trial and error is possible. Please describe the technique you used. Note: If you need multiple worksheets to answer a question label them as “Q1a”, “Q1b”, etc. DELIVERABLE – When you’re done you’ll have one Excel file to submit using the Assignments link on Blackboard. See the syllabus for due date. Below are a couple of helpful reference tables (use for lookup function?).

ITM 220 – Excel Assignment, ver 5.1 Individual (not group) Effort Data Analysis with Spreadsheet Software (Excel) OVERVIEW – The purpose of this assignment is to give you a chance to demonstrate your understanding of data analysis tools available in today’s spreadsheets such as Microsoft Excel. You’ll be working with data extracted from an ERPsim game performed by ITM 220 students a few months ago. This is not data from our class, so when considering the performance of “your team” (use data for Team “H” as your team data) don’t be surprised that it is significantly different from what you may recall for our ERPsim game. APPROACH – 1. Begin with the data found in this Excel Workbook–Worksheet named “ERPsimRawData” 2. Use tools and techniques as we discussed in class to study the data: a. Autofilter (and Advanced Filter) b. Conditional Formatting c. Lookup Table (VLOOKUP) d. Pivot Table e. Charts (smart ones, please) 3. Create multiple worksheets to document your answers to the following questions: Q1. What were average Muesli prices (per box) in the 2nd round? Organize by product. Label your answer worksheet “Q1”. COMMENT: It is necessary that you use a pivot table and be sure to use a chart, as well. Q2. How does your company (Team H – Leipzig) compare to your chief competitors–Munich and Stuttgart–on product prices? Use the lookup function (VLOOKUP) to add city name to your raw data and then use city name in your answer. Label your answer worksheet “Q2”. COMMENT: A pivot table is expected–chart, too. Q3. Which team has the most total sales revenue for each round? Label your answer worksheet “Q3”. COMMENT: Pivot table is expected–chart, too. Q4. Fill in the missing data in the worksheet labeled “FinancialForecast” –similarly to what we did in class. Use Scenario Manager to build 3 scenarios for projected company financial performance: Best Case, Expected case, and Worst Case scenarios (as we did in class) by modifying the appropriate blue highlighted cells in the Assumptions area. (Only 4 of these such highlighted cells will be used. Right?) Create an appropriate (your choice) chart illustrating the Summary Table results. Label your answer worksheet “Q4”. COMMENT: Must use Scenario Manager and you should be able to answer using a single worksheet for summary table and chart (as we did in class). Q5. Duplicate (NOT copy/paste) the “FinancialForecast” worksheet on which you have entered your assumptions data for the Expected Case. Label this new worksheet “Q5.” Use Goal seek to determine the average daily Grocery Chains sales needed in order to achieve Earnings Before Taxes of 500,000 Euros in Round 1. Discuss the plausibility of attaining such daily sales (probably based on the result of your answer to question 4). (And notice that growth rates are irrelevant to answer this question since we are talking FIRST round.) COMMENT: This is very easy using Goal Seek. However, a manual solution using trial and error is possible. Please describe the technique you used. Note: If you need multiple worksheets to answer a question label them as “Q1a”, “Q1b”, etc. DELIVERABLE – When you’re done you’ll have one Excel file to submit using the Assignments link on Blackboard. See the syllabus for due date. Below are a couple of helpful reference tables (use for lookup function?).

ITM 220 – Excel Assignment, ver 5.1  Individual (not group) … Read More...
Lab Assignment-09 Note: Create and save m-files for each problem individually. Copy all the m-files into a ‘single’ folder and upload the folder to D2L. Read chapters 2 and chapter 3.1-3.3 of the textbook (Introduction to MATLAB 7 for Engineers), solve the following problems in MATLAB. Given A= [■(3&-2&1@6&8&-5@7&9&10)] ; B= [■(6&9&-4@7&5&3@-8&2&1)] ; C= [■(-7&-5&2@10&6&1@3&-9&8)] ; Find the following A+B+C Verify the associative law (A+B)+C=A+ (B+C) D=Transpose(AB) E=A4 + B2 – C3 Find F, given that F = E-1 * D-1 – (AT) -1 Use MATLAB to solve the following set of equations 5x+7y + 9z = 12 7x- 4y + 8z = 86 15x- 9y – 6z = -57 Write a function that accepts temperature in degrees F and computes the corresponding value in degree C. The relation between the two is Aluminum alloys are made by adding other elements to aluminum to improve its properties, such as hardness or tensile strength. The following table shows the composition of five commonly used alloys, which are known by their alloy numbers ( 2024, 6061, and so on) [Kutz, 1999]. Obtain a matrix algorithm to compute the amounts of raw materials needed to produce a given amount of each alloy. Use MATLAB to determine how much raw material each type is needed to produce 1000tons of each alloy. Composition of aluminum alloys Alloy % Cu % Mg % Mn % Si % Zn 2024 4.4 1.5 0.6 0 0 6061 0 1 0 0.6 0 7005 0 1.4 0 0 4.5 7075 1.6 2.5 0 0 5.6 356.0 0 0.3 0 7 0

Lab Assignment-09 Note: Create and save m-files for each problem individually. Copy all the m-files into a ‘single’ folder and upload the folder to D2L. Read chapters 2 and chapter 3.1-3.3 of the textbook (Introduction to MATLAB 7 for Engineers), solve the following problems in MATLAB. Given A= [■(3&-2&1@6&8&-5@7&9&10)] ; B= [■(6&9&-4@7&5&3@-8&2&1)] ; C= [■(-7&-5&2@10&6&1@3&-9&8)] ; Find the following A+B+C Verify the associative law (A+B)+C=A+ (B+C) D=Transpose(AB) E=A4 + B2 – C3 Find F, given that F = E-1 * D-1 – (AT) -1 Use MATLAB to solve the following set of equations 5x+7y + 9z = 12 7x- 4y + 8z = 86 15x- 9y – 6z = -57 Write a function that accepts temperature in degrees F and computes the corresponding value in degree C. The relation between the two is Aluminum alloys are made by adding other elements to aluminum to improve its properties, such as hardness or tensile strength. The following table shows the composition of five commonly used alloys, which are known by their alloy numbers ( 2024, 6061, and so on) [Kutz, 1999]. Obtain a matrix algorithm to compute the amounts of raw materials needed to produce a given amount of each alloy. Use MATLAB to determine how much raw material each type is needed to produce 1000tons of each alloy. Composition of aluminum alloys Alloy % Cu % Mg % Mn % Si % Zn 2024 4.4 1.5 0.6 0 0 6061 0 1 0 0.6 0 7005 0 1.4 0 0 4.5 7075 1.6 2.5 0 0 5.6 356.0 0 0.3 0 7 0

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Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F  Fx Fy  Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER:  Fx Fy F  Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F  /2 cos() sin() F  sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos()  /2 F  |F| cos() |F| sin() sin() cos()  N  N N  Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the   cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle.     x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N  Nx, Ny = −Nsin(),Ncos() T  T  Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a  A x Ax |Ax| OL = OMcos( ) A  = 0.985,0.174 B b   Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c  =  C  C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C  = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B  = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C  = -5,4,0 −A + B − C  = -6,4,3 3A − 2C  = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A −2A + 3B − C  = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F  Fx Fy  Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER:  Fx Fy F  Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F  /2 cos() sin() F  sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos()  /2 F  |F| cos() |F| sin() sin() cos()  N  N N  Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the   cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle.     x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N  Nx, Ny = −Nsin(),Ncos() T  T  Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a  A x Ax |Ax| OL = OMcos( ) A  = 0.985,0.174 B b   Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c  =  C  C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C  = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B  = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C  = -5,4,0 −A + B − C  = -6,4,3 3A − 2C  = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A −2A + 3B − C  = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

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