According to Altman, traditional sexual morality, a person’s sexual life was, to a large extent, not his or her own: the rules imposed a highly confining set of duties on sexual choices and actions. Answers: True False

According to Altman, traditional sexual morality, a person’s sexual life was, to a large extent, not his or her own: the rules imposed a highly confining set of duties on sexual choices and actions. Answers: True False

According to Altman, traditional sexual morality, a person’s sexual life … Read More...
4. In the introduction to the Punctuation Made Simple website, what analogy is used to explain the way punctuation works? What, exactly, about punctuation does this analogy illustrate?

4. In the introduction to the Punctuation Made Simple website, what analogy is used to explain the way punctuation works? What, exactly, about punctuation does this analogy illustrate?

As equivalence, consider of the traffic signs that administer the … Read More...
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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Applications of Regular Expressions

Applications of Regular Expressions

A regular expression is a picture of the patter which … Read More...
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: Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 57 5/9/2014 8:02 PM 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: 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 . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 57 5/9/2014 8:02 PM Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? 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 ? = = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 57 5/9/2014 8:02 PM Hint 1. What is the general quadratic equation? The general quadratic equation is , where , , and are constants. Depending on the value of the discriminant, , the equation may have two real valued 1. solutions if , 2. one real valued solution if , or 3. two complex valued solutions if . 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 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. No; there is no chance he is going to get aboard. Yes; he will get a second chance Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 57 5/9/2014 8:02 PM 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: 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: C and E E and F A and F C and D B and D C and D A and F E and F A and B E and D Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 57 5/9/2014 8:02 PM Correct 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 . The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 2. 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. ANSWER: A and F A and E D and B C and D E and F Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 57 5/9/2014 8:02 PM Correct 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. Express your answers, separated by a comma, in meters to one significant figure. = 5 positive negative Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 57 5/9/2014 8:02 PM 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, , is constant. 1. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of its velocity, , 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 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 corresponds to the moment just after the ball is launched from position and . Its launch velocity, also called the initial velocity, is . 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 with velocity . Its position at this moment is denoted by or since it is at its maximum height. The other point, at time with velocity , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is , also known as ( 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 . Part A , = -2,-5 , Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 57 5/9/2014 8:02 PM How do the speeds , , and (at times ,

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: Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 57 5/9/2014 8:02 PM 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: 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 . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 57 5/9/2014 8:02 PM Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? 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 ? = = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 57 5/9/2014 8:02 PM Hint 1. What is the general quadratic equation? The general quadratic equation is , where , , and are constants. Depending on the value of the discriminant, , the equation may have two real valued 1. solutions if , 2. one real valued solution if , or 3. two complex valued solutions if . 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 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. No; there is no chance he is going to get aboard. Yes; he will get a second chance Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 57 5/9/2014 8:02 PM 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: 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: C and E E and F A and F C and D B and D C and D A and F E and F A and B E and D Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 57 5/9/2014 8:02 PM Correct 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 . The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 2. 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. ANSWER: A and F A and E D and B C and D E and F Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 57 5/9/2014 8:02 PM Correct 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. Express your answers, separated by a comma, in meters to one significant figure. = 5 positive negative Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 57 5/9/2014 8:02 PM 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, , is constant. 1. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of its velocity, , 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 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 corresponds to the moment just after the ball is launched from position and . Its launch velocity, also called the initial velocity, is . 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 with velocity . Its position at this moment is denoted by or since it is at its maximum height. The other point, at time with velocity , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is , also known as ( 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 . Part A , = -2,-5 , Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 57 5/9/2014 8:02 PM How do the speeds , , and (at times ,

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The second task : Tutorial Topic 7 – MIS (to be completed over 2 weeks) Describe a decision support system whose purpose is to help you decide which accommodation would be best for you whilst at college next year. Hints: You will need to decide on the factors that will influence your decision, and decide on any weightings that may apply to these factors. You will also need to supply a formula to derive ‘the best solution’, making sure it can support ‘what if’ features. Factors you might consider – affordable rent values, flat or house, sharing (how many others), location, personal circumstances etc.

The second task : Tutorial Topic 7 – MIS (to be completed over 2 weeks) Describe a decision support system whose purpose is to help you decide which accommodation would be best for you whilst at college next year. Hints: You will need to decide on the factors that will influence your decision, and decide on any weightings that may apply to these factors. You will also need to supply a formula to derive ‘the best solution’, making sure it can support ‘what if’ features. Factors you might consider – affordable rent values, flat or house, sharing (how many others), location, personal circumstances etc.

As any DSS system, my DSS system will have the … Read More...
COMM 1311: Written Communication Assignment 5 Argumentation Essay (Chapter 10, pp. 218-232, Arlov) Purpose of Assignment • The purpose of this assignment is to enable the student to write an essay with a compelling argumentation that shows critical thinking. A persuasive essay is a writer’s attempt to convince readers of the validity of a particular opinion on a controversial issue. Objectives • The student will be able to correctly structure an essay and bring forward a compelling thesis and argument. • The student will understand the creativity of the writing process and use his own ideas. • The student will be able to craft a compelling essay and show critical thinking. • The student will show that he is able to argue both sides of a topic and is willing to acknowledge a different opinion. Instructions 1. Establish a subject Choose a topic that interests you. An argument does not have to be a burning issue, but it must be a debatable topic. It can be anything you feel strongly about but it has to be approved by the instructor. 2. Present a clear thesis and identify the controversy Your thesis should inform readers of your purpose and how you will proceed in your argumentation. 3. Follow an organizational pattern and provide support The body paragraphs of the essay should provide specific support. These supports may include personal experience, statistics, facts, or experts’ opinions. They may be garnered from scientific journals, magazines, books, newspapers, textbooks, studies, or interviews. Select only the facts that are relevant. 4. Consider differing opinions A persuasive essay may be strengthened by acknowledging conflict viewpoints and discussing them. 4. Draw a conclusion Restate your position in different words from the introduction. Do not introduce new material in the conclusion. You may want to conclude by encouraging some specific call to action. Requirements The essay topic must meet the approval of the instructor: • Have a complete cover page • have at least 500 words • use full sentences (and no bullet points) • must have page numbers • must have a reference page Example writing (not a complete essay): Boxing: Countdown to Injury A left hook smashes into the fighter’s jaw. A following right slams his head the opposite direction. An uppercut to the jaw snaps his head back, momentarily stopping the blood flow to his brain. The boxer drops, hitting the mat with a thud. His brain bounces off his skull for the second time in a matter of seconds. Is this what we should call a sport? Because of injuries, neurological damage, and ring deaths, the rules of professional boxing should be changed. Boxing has always been a brutal sport. The ancient Greeks used gloves studded with metal spikes, which slashed the face and body and split skulls. Although gloves are no longer spiked, boxers today sustain injuries ranging from cuts and bruises to broken bones. It is not uncommon to see a boxer leave the ring with a cut on his face, an eye swollen shut, and a nose enlarged and bloody. Often, healing in is incomplete because these areas receive the same blows again and again in other matches. In fact, repeated blows almost cost Sugar Ray Leonard his sight when his retina detached in his left eye. Besides superficial injuries, boxers suffer short-term neurological damage as a result of staggering blows to the head. A knockout punch, for example, is often delivered with such force that the brain smashes against the skull, tearing nerve fibers and blood vessels, resulting in a concussion. Even a blow to the neck can close the carotid artery, the main artery to the brain, whereby oxygen and blood to the brain are disrupted, resulting in dizziness and confusion. Later, the boxers often have no memory of the moments before or after a knockout blow. Submission Criteria Due Date: Sunday, December 6, 2015. Late assignments will receive an automatic ZERO grade. Where to deliver hard copies: In class Assessment Criteria CRITERIA Assessment Rubric Argumentation Essay SCORES Introduction Introduces the issue and its importance, says what your essay will cover 2 Organization The sound structure of the essay 1 Expression Sentences, phrases, metaphors, verbs etc. The strength of the language used 4 Conclusion Restate the issue, summarizes the strength of the arguments in the essays, gives your opinion about which essay is the strongest with supporting reasons 1 Mechanics Followed guidelines, professional format, punctuation, spelling, and capitalization are correct, use of headings, no bullet points 2 TOTAL 10% Plagiarism, copying from the internet or any other sources without citation will result in an automatic ZERO grade and a procedure of Academic Misconduct will filed against you. The complete essay has to be created and written by you alone. Prior assignments CAN NOT be used.

COMM 1311: Written Communication Assignment 5 Argumentation Essay (Chapter 10, pp. 218-232, Arlov) Purpose of Assignment • The purpose of this assignment is to enable the student to write an essay with a compelling argumentation that shows critical thinking. A persuasive essay is a writer’s attempt to convince readers of the validity of a particular opinion on a controversial issue. Objectives • The student will be able to correctly structure an essay and bring forward a compelling thesis and argument. • The student will understand the creativity of the writing process and use his own ideas. • The student will be able to craft a compelling essay and show critical thinking. • The student will show that he is able to argue both sides of a topic and is willing to acknowledge a different opinion. Instructions 1. Establish a subject Choose a topic that interests you. An argument does not have to be a burning issue, but it must be a debatable topic. It can be anything you feel strongly about but it has to be approved by the instructor. 2. Present a clear thesis and identify the controversy Your thesis should inform readers of your purpose and how you will proceed in your argumentation. 3. Follow an organizational pattern and provide support The body paragraphs of the essay should provide specific support. These supports may include personal experience, statistics, facts, or experts’ opinions. They may be garnered from scientific journals, magazines, books, newspapers, textbooks, studies, or interviews. Select only the facts that are relevant. 4. Consider differing opinions A persuasive essay may be strengthened by acknowledging conflict viewpoints and discussing them. 4. Draw a conclusion Restate your position in different words from the introduction. Do not introduce new material in the conclusion. You may want to conclude by encouraging some specific call to action. Requirements The essay topic must meet the approval of the instructor: • Have a complete cover page • have at least 500 words • use full sentences (and no bullet points) • must have page numbers • must have a reference page Example writing (not a complete essay): Boxing: Countdown to Injury A left hook smashes into the fighter’s jaw. A following right slams his head the opposite direction. An uppercut to the jaw snaps his head back, momentarily stopping the blood flow to his brain. The boxer drops, hitting the mat with a thud. His brain bounces off his skull for the second time in a matter of seconds. Is this what we should call a sport? Because of injuries, neurological damage, and ring deaths, the rules of professional boxing should be changed. Boxing has always been a brutal sport. The ancient Greeks used gloves studded with metal spikes, which slashed the face and body and split skulls. Although gloves are no longer spiked, boxers today sustain injuries ranging from cuts and bruises to broken bones. It is not uncommon to see a boxer leave the ring with a cut on his face, an eye swollen shut, and a nose enlarged and bloody. Often, healing in is incomplete because these areas receive the same blows again and again in other matches. In fact, repeated blows almost cost Sugar Ray Leonard his sight when his retina detached in his left eye. Besides superficial injuries, boxers suffer short-term neurological damage as a result of staggering blows to the head. A knockout punch, for example, is often delivered with such force that the brain smashes against the skull, tearing nerve fibers and blood vessels, resulting in a concussion. Even a blow to the neck can close the carotid artery, the main artery to the brain, whereby oxygen and blood to the brain are disrupted, resulting in dizziness and confusion. Later, the boxers often have no memory of the moments before or after a knockout blow. Submission Criteria Due Date: Sunday, December 6, 2015. Late assignments will receive an automatic ZERO grade. Where to deliver hard copies: In class Assessment Criteria CRITERIA Assessment Rubric Argumentation Essay SCORES Introduction Introduces the issue and its importance, says what your essay will cover 2 Organization The sound structure of the essay 1 Expression Sentences, phrases, metaphors, verbs etc. The strength of the language used 4 Conclusion Restate the issue, summarizes the strength of the arguments in the essays, gives your opinion about which essay is the strongest with supporting reasons 1 Mechanics Followed guidelines, professional format, punctuation, spelling, and capitalization are correct, use of headings, no bullet points 2 TOTAL 10% Plagiarism, copying from the internet or any other sources without citation will result in an automatic ZERO grade and a procedure of Academic Misconduct will filed against you. The complete essay has to be created and written by you alone. Prior assignments CAN NOT be used.

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Give an example of when an analogy can be a really good advance organizer and an example of when an analogy can be a very poor advance organizer.

Give an example of when an analogy can be a really good advance organizer and an example of when an analogy can be a very poor advance organizer.

An ‘advance organizer is a cognitive instructional policy used to … Read More...
Instructions Multiple Attempts Not allowed. This test can only be taken once. Force Completion This test can be saved and resumed later. Question 1 1. According to Matthews, Egyptian culture is thought to be ______. more pessimistic in outlook that Mesopotamian culture. Egyptians were more fun and had more parties than Mesopotamians. more optimistic in outlook that Mesopotamian culture. Ancient Sumerians could really throw a party. Question 2 1. The Rosetta Stone, found in the Rosetta section of the Nile by Napoleon, allowed historians to translate Egyptian hieroglyphics because it had each of the following languages EXCEPT ______. Hieratic Hieroglyphics Greek Coptic Question 3 1. Select the legacies of Near Eastern civilizations. Choose all the correct answers. Lunar calendar that guides planting Number system based on 60 Newspapers Algebra Pschology Writing Question 4 1. Each of the following would be considered a quality of Egyptian art EXCEPT _____. people were portrayed with serene looks on their faces. sculptures, paintings and reliefs portray figures in profile and from the front simultaneously. images of people housed their ka or spirit figures were often portrayed in a contorted, twisting and dramatic fashion. Question 5 1. Hatshepsut became pharaoh because _______. she claimed title of pharaoh because of her visions of Aten. she pulled a sword out of a pyramid-shaped stone. she inherited the title from her mother, Nefertiti. she was the chief queen to Thutmose II and seized power when he died. Question 6 1. Place the civilizations in the correct chronological order. Persian Akkadian Assyrian Sumerian Babylonian Question 7 1. Match the description with the phrase. A time when a more free and fluid style of artwork was created. Read Answer Items for Question 7 A set of rules or a set of established works of art Read Answer Items for Question 7 Tokens or emblems of royal authority Read Answer Items for Question 7 The northern more fertile area of the Nile’s delta Read Answer Items for Question 7 Answer A. Regalia B. Lower Egypt C. The Amarna period under Akhenaten D. Canon Question 8 1. After reading the ancient Sumerian text from 1700 BCE in which a father “Lectures His Son,” what conclusions might we draw about parallels between contemporary and ancient fathers? Select all the correct answers. Both modern and ancient fathers often want their sons to be successful. Both modern and ancient fathers often encourage their sons not to be idle. Both modern and ancient fathers often think education is important. Both modern and ancient fathers think the purchase of shares in reeds and canes is a wise investment. Both modern and ancient fathers often watch over their sons and are concerned about their future. Both modern and ancient fathers enjoy golf. Question 9 1. The oldest surviving medical textbooks from ancient Egypt might include each of the following EXCEPT _____. health insurance forms advice about getting rid of vermin magical spells to heal illnesses diagnostic information about maladies of organs Question 10 1. Circle each of the developments of the Neolithic age. Circle all that apply. plows microchip indoor plumbing kiln-fired bricks wheel hammers boats flying buttress internal combustion engine knives Save and Submit Click Save and Submit to save and submit. Click Save All Answers to save all answers. Click Close Window to close window.

Instructions Multiple Attempts Not allowed. This test can only be taken once. Force Completion This test can be saved and resumed later. Question 1 1. According to Matthews, Egyptian culture is thought to be ______. more pessimistic in outlook that Mesopotamian culture. Egyptians were more fun and had more parties than Mesopotamians. more optimistic in outlook that Mesopotamian culture. Ancient Sumerians could really throw a party. Question 2 1. The Rosetta Stone, found in the Rosetta section of the Nile by Napoleon, allowed historians to translate Egyptian hieroglyphics because it had each of the following languages EXCEPT ______. Hieratic Hieroglyphics Greek Coptic Question 3 1. Select the legacies of Near Eastern civilizations. Choose all the correct answers. Lunar calendar that guides planting Number system based on 60 Newspapers Algebra Pschology Writing Question 4 1. Each of the following would be considered a quality of Egyptian art EXCEPT _____. people were portrayed with serene looks on their faces. sculptures, paintings and reliefs portray figures in profile and from the front simultaneously. images of people housed their ka or spirit figures were often portrayed in a contorted, twisting and dramatic fashion. Question 5 1. Hatshepsut became pharaoh because _______. she claimed title of pharaoh because of her visions of Aten. she pulled a sword out of a pyramid-shaped stone. she inherited the title from her mother, Nefertiti. she was the chief queen to Thutmose II and seized power when he died. Question 6 1. Place the civilizations in the correct chronological order. Persian Akkadian Assyrian Sumerian Babylonian Question 7 1. Match the description with the phrase. A time when a more free and fluid style of artwork was created. Read Answer Items for Question 7 A set of rules or a set of established works of art Read Answer Items for Question 7 Tokens or emblems of royal authority Read Answer Items for Question 7 The northern more fertile area of the Nile’s delta Read Answer Items for Question 7 Answer A. Regalia B. Lower Egypt C. The Amarna period under Akhenaten D. Canon Question 8 1. After reading the ancient Sumerian text from 1700 BCE in which a father “Lectures His Son,” what conclusions might we draw about parallels between contemporary and ancient fathers? Select all the correct answers. Both modern and ancient fathers often want their sons to be successful. Both modern and ancient fathers often encourage their sons not to be idle. Both modern and ancient fathers often think education is important. Both modern and ancient fathers think the purchase of shares in reeds and canes is a wise investment. Both modern and ancient fathers often watch over their sons and are concerned about their future. Both modern and ancient fathers enjoy golf. Question 9 1. The oldest surviving medical textbooks from ancient Egypt might include each of the following EXCEPT _____. health insurance forms advice about getting rid of vermin magical spells to heal illnesses diagnostic information about maladies of organs Question 10 1. Circle each of the developments of the Neolithic age. Circle all that apply. plows microchip indoor plumbing kiln-fired bricks wheel hammers boats flying buttress internal combustion engine knives Save and Submit Click Save and Submit to save and submit. Click Save All Answers to save all answers. Click Close Window to close window.

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