1. Develop a thought experiment that attempts to uncover hidden assumptions about human freedom. 2. Find a paragraph from a book, magazine, ect. First, tell whether there are claims in the paragraph. If there are, identify the types of claims (descriptive, normative, a priori, a posteriori) in the paragraph

1. Develop a thought experiment that attempts to uncover hidden assumptions about human freedom. 2. Find a paragraph from a book, magazine, ect. First, tell whether there are claims in the paragraph. If there are, identify the types of claims (descriptive, normative, a priori, a posteriori) in the paragraph

Let us think of a thought experiment that wants to … 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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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...
Chapter 4 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Advice for the Quarterback A quarterback is set up to throw the football to a receiver who is running with a constant velocity directly away from the quarterback and is now a distance away from the quarterback. The quarterback figures that the ball must be thrown at an angle to the horizontal and he estimates that the receiver must catch the ball a time interval after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Express in terms of and . Hint 1. Equation of motion in y direction What is the expression for , the height of the ball as a function of time? Answer in terms of , , and . v r D  tc y = 0 x = 0 g v0y v0y tc g y(t) t g v0y ANSWER: Incorrect; Try Again Hint 2. Height at which the ball is caught, Remember that after time the ball was caught at the same height as it had been released. That is, . ANSWER: Answer Requested Part B Find , the initial horizontal component of velocity of the ball. Express your answer for in terms of , , and . Hint 1. Receiver’s position Find , the receiver’s position before he catches the ball. Answer in terms of , , and . ANSWER: Football’s position y(t) = v0yt− g 1 2 t2 y(tc) tc y(tc) = y0 = 0 v0y = gtc 2 v0x v0x D tc vr xr D vr tc xr = D + vrtc Typesetting math: 100% Find , the horizontal distance that the ball travels before reaching the receiver. Answer in terms of and . ANSWER: ANSWER: Answer Requested Part C Find the speed with which the quarterback must throw the ball. Answer in terms of , , , and . Hint 1. How to approach the problem Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. ANSWER: Answer Requested Part D xc v0x tc xc = v0xtc v0x = + D tc vr v0 D tc vr g v0 = ( + ) + D tc vr 2 ( ) gtc 2 2 −−−−−−−−−−−−−−−−−−−  Typesetting math: 100% Assuming that the quarterback throws the ball with speed , find the angle above the horizontal at which he should throw it. Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . Hint 1. Find angle from and Think of velocity as a vector with Cartesian coordinates and . Find the angle that this vector would make with the x axis using the results of Parts A and B. ANSWER: Answer Requested Direction of Velocity at Various Times in Flight for Projectile Motion Conceptual Question For each of the motions described below, determine the algebraic sign (positive, negative, or zero) of the x component and y component of velocity of the object at the time specified. For all of the motions, the positive x axis points to the right and the positive y axis points upward. Alex, a mountaineer, must leap across a wide crevasse. The other side of the crevasse is below the point from which he leaps, as shown in the figure. Alex leaps horizontally and successfully makes the jump. v0  v0x v0y v0  v0x v0y v0xx^ v0yy^   = atan( ) v0y v0x Typesetting math: 100% Part A Determine the algebraic sign of Alex’s x velocity and y velocity at the instant he leaves the ground at the beginning of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Algebraic sign of velocity The algebraic sign of the velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, to the right is defined to be the positive x direction and upward the positive y direction. Therefore, any object moving to the right, whether speeding up, slowing down, or even simultaneously moving upward or downward, has a positive x velocity. Similarly, if the object is moving downward, regardless of any other aspect of its motion, its y velocity is negative. Hint 2. Sketch Alex’s initial velocity On the diagram below, sketch the vector representing Alex’s velocity the instant after he leaves the ground at the beginning of the jump. ANSWER: ANSWER: Typesetting math: 100% Answer Requested Part B Determine the algebraic signs of Alex’s x velocity and y velocity the instant before he lands at the end of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Sketch Alex’s final velocity On the diagram below, sketch the vector representing Alex’s velocity the instant before he safely lands on the other side of the crevasse. ANSWER: Answer Requested ANSWER: Answer Requested Typesetting math: 100% At the buzzer, a basketball player shoots a desperation shot. The ball goes in! Part C Determine the algebraic signs of the ball’s x velocity and y velocity the instant after it leaves the player’s hands. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s initial velocity On the diagram below, sketch the vector representing the velocity of the basketball the instant after it leaves the player’s hands. ANSWER: Typesetting math: 100% ANSWER: Correct Part D Determine the algebraic signs of the ball’s x velocity and y velocity at the ball’s maximum height. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s velocity at maximum height Typesetting math: 100% On the diagram below, sketch the vector representing the velocity of the basketball the instant it reaches its maximum height. ANSWER: ANSWER: Answer Requested PSS 4.1 Projectile Motion Problems Learning Goal: Typesetting math: 100% To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 9.00 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 17.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: and . Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are , . is the same for the horizontal and vertical components of the motion. Find from one component, and then use that value for the other component. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly. Visualize Part A Which diagram represents an accurate sketch of the rock’s trajectory? Hint 1. The launch angle In a projectile’s motion, the angle of the initial velocity above the horizontal is called the launch angle. ANSWER: m/s  d m g m/s2 ax = 0 ay = −g xf = xi +vixt, yf = yi +viyt− g(t 1 2 )2 vfx = vix = constant, and vfy = viy − gt t t v i Typesetting math: 100% Typesetting math: 100% Correct Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile’s acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn’t change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? ANSWER: ay At ground level below the point where the rock is launched At the point where the rock strikes the ground At the peak of the trajectory At the point where the rock is released At ground level below the peak of the trajectory Typesetting math: 100% Correct It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below: Solve Part C Find the height from which the rock was launched. Express your answer in meters to three significant figures. yi yi Typesetting math: 100% Hint 1. How to approach the problem The time needed to move horizontally to the final position = 17.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for . Hint 2. Find the time spent in the air How long ( ) is the rock in the air? Express your answer in seconds to three significant figures. Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time the rock spent in the air? ANSWER: Hint 2. Find the x component of the initial velocity What is the x component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: t xf = d m yi t yi t t xf = xi + vixt yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt vix = 7.79 m/s Typesetting math: 100% Hint 3. Find the y component of the initial velocity What is the y component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: Answer Requested Assess Part D A second rock is thrown straight upward with a speed 4.500 . If this rock takes 2.181 to fall to the ground, from what height was it released? Express your answer in meters to three significant figures. Hint 1. Identify the known variables What are the values of , , , and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: t = 2.18 s viy = 4.50 m/s yi = 13.5 m m/s s H yf viy t a Typesetting math: 100% Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, . ANSWER: ANSWER: Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 4.500 , and fell the same vertical distance of 13.5 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly. ± Arrow Hits Apple An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at the same height above the ground as it was shot. Use for the magnitude of the acceleration due to gravity. Part A , , , = 0,4.500,2.181,-yf viy t a 9.810 m, m/s, s, m/s2 H yi = H yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt = − 2g( − ) v2f y v2i y yf yi H = 13.5 m viy = m/s m  = 45 D = 220 m g = 9.8 m/s2 Typesetting math: 100% Find , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables and for the initial components of velocity. Then use kinematics to relate them and solve for . What is the vertical component of the initial velocity? Express your answer symbolically in terms of and . Hint 1. Find Find the horizontal component of the initial velocity. Express your answer symbolically in terms of and given symbolic quantities. ANSWER: Hint 2. Find What is the vertical component of the initial velocity? Express your answer symbolically in terms of . ANSWER: ANSWER: ta vy0 vx0 ta vy0 ta D vx0 vx0 ta vx0 = D ta vy0 vy0 vx0 vy0 = vx0 vy0 = D ta Typesetting math: 100% Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find in terms of and . Give your answer in terms of and . Hint 1. Find When applied to the y-component of velocity, in this problem the formula for with constant acceleration is What is , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of only. ANSWER: ANSWER: Hint 3. Put the algebra together to find symbolically. If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial component to find an expression for Express your answer symbolically in terms of given variables. ANSWER: ta vy0 g vy0 g vy(ta) v(t) −g vy(t) = vy0 − g t vy(ta ) vy0 vy(ta) = −vy0 ta = 2vy0 g ta D ta g ta ta2 t2 = a 2D g Typesetting math: 100% ANSWER: Answer Requested Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: Answer Requested ANSWER: ta = 6.7 s tf = 1.1 s td = 5.6 s Typesetting math: 100% Answer Requested Video Tutor: Ball Fired Upward from Accelerating Cart 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 Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? Hint 1. Determine how long the ball is in the air How will doubling the initial upward speed of the ball change the time the ball spends in the air? A kinematic equation may be helpful here. The time in the air will ANSWER: be cut in half. stay the same. double. quadruple. Typesetting math: 100% Hint 2. Determine the appropriate kinematic expression Which of the following kinematic equations correctly describes the horizontal distance between the ball and the cart at the moment the ball lands? The cart’s initial horizontal velocity is , its horizontal acceleration is , and is the time elapsed between launch and impact. ANSWER: ANSWER: Correct The ball will spend twice as much time in the air ( , where is the ball’s initial upward velocity), so it will land four times farther behind the cart: (where is the cart’s horizontal acceleration). Video Tutor: Ball Fired Upward from Moving Cart 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. d v0x ax t d = v0x t d = 1 2 axv0x t2 d = v0x t+ 1 2 axt2 d = 1 2 axt2 the same distance twice as far half as far four times as far by a factor not listed above t = 2v0y/g v0y d = 1 2 axt2 ax Typesetting math: 100% Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance. Hint 1. How to approach the problem While the crate is on the plane, it shares the plane’s velocity. What is the crate’s velocity immediately after it is released? Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate’s horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that’s not very realistic in this situation.) ANSWER: Correct At the moment it is released, the crate shares the plane’s horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell. Score Summary: Your score on this assignment is 0%. Before the plane is directly over the target After the plane has flown over the target When the plane is directly over the target Typesetting math: 100% You received 0 out of a possible total of 0 points. Typesetting math: 100%

Chapter 4 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Advice for the Quarterback A quarterback is set up to throw the football to a receiver who is running with a constant velocity directly away from the quarterback and is now a distance away from the quarterback. The quarterback figures that the ball must be thrown at an angle to the horizontal and he estimates that the receiver must catch the ball a time interval after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Express in terms of and . Hint 1. Equation of motion in y direction What is the expression for , the height of the ball as a function of time? Answer in terms of , , and . v r D  tc y = 0 x = 0 g v0y v0y tc g y(t) t g v0y ANSWER: Incorrect; Try Again Hint 2. Height at which the ball is caught, Remember that after time the ball was caught at the same height as it had been released. That is, . ANSWER: Answer Requested Part B Find , the initial horizontal component of velocity of the ball. Express your answer for in terms of , , and . Hint 1. Receiver’s position Find , the receiver’s position before he catches the ball. Answer in terms of , , and . ANSWER: Football’s position y(t) = v0yt− g 1 2 t2 y(tc) tc y(tc) = y0 = 0 v0y = gtc 2 v0x v0x D tc vr xr D vr tc xr = D + vrtc Typesetting math: 100% Find , the horizontal distance that the ball travels before reaching the receiver. Answer in terms of and . ANSWER: ANSWER: Answer Requested Part C Find the speed with which the quarterback must throw the ball. Answer in terms of , , , and . Hint 1. How to approach the problem Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. ANSWER: Answer Requested Part D xc v0x tc xc = v0xtc v0x = + D tc vr v0 D tc vr g v0 = ( + ) + D tc vr 2 ( ) gtc 2 2 −−−−−−−−−−−−−−−−−−−  Typesetting math: 100% Assuming that the quarterback throws the ball with speed , find the angle above the horizontal at which he should throw it. Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . Hint 1. Find angle from and Think of velocity as a vector with Cartesian coordinates and . Find the angle that this vector would make with the x axis using the results of Parts A and B. ANSWER: Answer Requested Direction of Velocity at Various Times in Flight for Projectile Motion Conceptual Question For each of the motions described below, determine the algebraic sign (positive, negative, or zero) of the x component and y component of velocity of the object at the time specified. For all of the motions, the positive x axis points to the right and the positive y axis points upward. Alex, a mountaineer, must leap across a wide crevasse. The other side of the crevasse is below the point from which he leaps, as shown in the figure. Alex leaps horizontally and successfully makes the jump. v0  v0x v0y v0  v0x v0y v0xx^ v0yy^   = atan( ) v0y v0x Typesetting math: 100% Part A Determine the algebraic sign of Alex’s x velocity and y velocity at the instant he leaves the ground at the beginning of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Algebraic sign of velocity The algebraic sign of the velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, to the right is defined to be the positive x direction and upward the positive y direction. Therefore, any object moving to the right, whether speeding up, slowing down, or even simultaneously moving upward or downward, has a positive x velocity. Similarly, if the object is moving downward, regardless of any other aspect of its motion, its y velocity is negative. Hint 2. Sketch Alex’s initial velocity On the diagram below, sketch the vector representing Alex’s velocity the instant after he leaves the ground at the beginning of the jump. ANSWER: ANSWER: Typesetting math: 100% Answer Requested Part B Determine the algebraic signs of Alex’s x velocity and y velocity the instant before he lands at the end of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Sketch Alex’s final velocity On the diagram below, sketch the vector representing Alex’s velocity the instant before he safely lands on the other side of the crevasse. ANSWER: Answer Requested ANSWER: Answer Requested Typesetting math: 100% At the buzzer, a basketball player shoots a desperation shot. The ball goes in! Part C Determine the algebraic signs of the ball’s x velocity and y velocity the instant after it leaves the player’s hands. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s initial velocity On the diagram below, sketch the vector representing the velocity of the basketball the instant after it leaves the player’s hands. ANSWER: Typesetting math: 100% ANSWER: Correct Part D Determine the algebraic signs of the ball’s x velocity and y velocity at the ball’s maximum height. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s velocity at maximum height Typesetting math: 100% On the diagram below, sketch the vector representing the velocity of the basketball the instant it reaches its maximum height. ANSWER: ANSWER: Answer Requested PSS 4.1 Projectile Motion Problems Learning Goal: Typesetting math: 100% To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 9.00 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 17.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: and . Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are , . is the same for the horizontal and vertical components of the motion. Find from one component, and then use that value for the other component. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly. Visualize Part A Which diagram represents an accurate sketch of the rock’s trajectory? Hint 1. The launch angle In a projectile’s motion, the angle of the initial velocity above the horizontal is called the launch angle. ANSWER: m/s  d m g m/s2 ax = 0 ay = −g xf = xi +vixt, yf = yi +viyt− g(t 1 2 )2 vfx = vix = constant, and vfy = viy − gt t t v i Typesetting math: 100% Typesetting math: 100% Correct Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile’s acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn’t change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? ANSWER: ay At ground level below the point where the rock is launched At the point where the rock strikes the ground At the peak of the trajectory At the point where the rock is released At ground level below the peak of the trajectory Typesetting math: 100% Correct It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below: Solve Part C Find the height from which the rock was launched. Express your answer in meters to three significant figures. yi yi Typesetting math: 100% Hint 1. How to approach the problem The time needed to move horizontally to the final position = 17.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for . Hint 2. Find the time spent in the air How long ( ) is the rock in the air? Express your answer in seconds to three significant figures. Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time the rock spent in the air? ANSWER: Hint 2. Find the x component of the initial velocity What is the x component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: t xf = d m yi t yi t t xf = xi + vixt yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt vix = 7.79 m/s Typesetting math: 100% Hint 3. Find the y component of the initial velocity What is the y component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: Answer Requested Assess Part D A second rock is thrown straight upward with a speed 4.500 . If this rock takes 2.181 to fall to the ground, from what height was it released? Express your answer in meters to three significant figures. Hint 1. Identify the known variables What are the values of , , , and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: t = 2.18 s viy = 4.50 m/s yi = 13.5 m m/s s H yf viy t a Typesetting math: 100% Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, . ANSWER: ANSWER: Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 4.500 , and fell the same vertical distance of 13.5 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly. ± Arrow Hits Apple An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at the same height above the ground as it was shot. Use for the magnitude of the acceleration due to gravity. Part A , , , = 0,4.500,2.181,-yf viy t a 9.810 m, m/s, s, m/s2 H yi = H yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt = − 2g( − ) v2f y v2i y yf yi H = 13.5 m viy = m/s m  = 45 D = 220 m g = 9.8 m/s2 Typesetting math: 100% Find , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables and for the initial components of velocity. Then use kinematics to relate them and solve for . What is the vertical component of the initial velocity? Express your answer symbolically in terms of and . Hint 1. Find Find the horizontal component of the initial velocity. Express your answer symbolically in terms of and given symbolic quantities. ANSWER: Hint 2. Find What is the vertical component of the initial velocity? Express your answer symbolically in terms of . ANSWER: ANSWER: ta vy0 vx0 ta vy0 ta D vx0 vx0 ta vx0 = D ta vy0 vy0 vx0 vy0 = vx0 vy0 = D ta Typesetting math: 100% Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find in terms of and . Give your answer in terms of and . Hint 1. Find When applied to the y-component of velocity, in this problem the formula for with constant acceleration is What is , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of only. ANSWER: ANSWER: Hint 3. Put the algebra together to find symbolically. If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial component to find an expression for Express your answer symbolically in terms of given variables. ANSWER: ta vy0 g vy0 g vy(ta) v(t) −g vy(t) = vy0 − g t vy(ta ) vy0 vy(ta) = −vy0 ta = 2vy0 g ta D ta g ta ta2 t2 = a 2D g Typesetting math: 100% ANSWER: Answer Requested Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: Answer Requested ANSWER: ta = 6.7 s tf = 1.1 s td = 5.6 s Typesetting math: 100% Answer Requested Video Tutor: Ball Fired Upward from Accelerating Cart 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 Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? Hint 1. Determine how long the ball is in the air How will doubling the initial upward speed of the ball change the time the ball spends in the air? A kinematic equation may be helpful here. The time in the air will ANSWER: be cut in half. stay the same. double. quadruple. Typesetting math: 100% Hint 2. Determine the appropriate kinematic expression Which of the following kinematic equations correctly describes the horizontal distance between the ball and the cart at the moment the ball lands? The cart’s initial horizontal velocity is , its horizontal acceleration is , and is the time elapsed between launch and impact. ANSWER: ANSWER: Correct The ball will spend twice as much time in the air ( , where is the ball’s initial upward velocity), so it will land four times farther behind the cart: (where is the cart’s horizontal acceleration). Video Tutor: Ball Fired Upward from Moving Cart 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. d v0x ax t d = v0x t d = 1 2 axv0x t2 d = v0x t+ 1 2 axt2 d = 1 2 axt2 the same distance twice as far half as far four times as far by a factor not listed above t = 2v0y/g v0y d = 1 2 axt2 ax Typesetting math: 100% Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance. Hint 1. How to approach the problem While the crate is on the plane, it shares the plane’s velocity. What is the crate’s velocity immediately after it is released? Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate’s horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that’s not very realistic in this situation.) ANSWER: Correct At the moment it is released, the crate shares the plane’s horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell. Score Summary: Your score on this assignment is 0%. Before the plane is directly over the target After the plane has flown over the target When the plane is directly over the target Typesetting math: 100% You received 0 out of a possible total of 0 points. Typesetting math: 100%

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) An aluminum wire (1 inch in diameter, 1000 m long) is used to pull a cable car up a mountain. If the weight of the unloaded cable car is 1 ton, and if there are 6 people in the car, and each weighs an average of 150 pounds: a) calculate the stress (in pounds per sq. inch), b) calculate the strain of the wire when it begins to lift the cable car, c) calculate the change in length of the cable. d) determine the change in the diameter of the wire (Poisson’s ratio is 0.31 for Al) – be sure to indicate the proper sign (positive for increase, negative for decrease) State your assumptions. EAl = 10 x 106 psi (pounds per square inch). [hint: be careful with the units]

) An aluminum wire (1 inch in diameter, 1000 m long) is used to pull a cable car up a mountain. If the weight of the unloaded cable car is 1 ton, and if there are 6 people in the car, and each weighs an average of 150 pounds: a) calculate the stress (in pounds per sq. inch), b) calculate the strain of the wire when it begins to lift the cable car, c) calculate the change in length of the cable. d) determine the change in the diameter of the wire (Poisson’s ratio is 0.31 for Al) – be sure to indicate the proper sign (positive for increase, negative for decrease) State your assumptions. EAl = 10 x 106 psi (pounds per square inch). [hint: be careful with the units]

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Chapter 1 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Curved Motion Diagram The motion diagram shown in the figure represents a pendulum released from rest at an angle of 45 from the vertical. The dots in the motion diagram represent the positions of the pendulum bob at eleven moments separated by equal time intervals. The green arrows represent the average velocity between adjacent dots. Also given is a “compass rose” in which directions are labeled with the letters of the alphabet.  Part A What is the direction of the acceleration of the object at moment 5? Enter the letter of the arrow with this direction from the compass rose in the figure. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again Part B What is the direction of the acceleration of the object at moments 0 and 10? Enter the letters corresponding to the arrows with these directions from the compass rose in the figure, separated by commas. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again PSS 1.1 Motion Diagrams Learning Goal: To practice Problem-Solving Strategy 1.1 for motion diagram problems. A car is traveling with constant velocity along a highway. The driver notices he is late for work, so he stomps down on the gas pedal and the car begins to speed up. The car has just achieved double its directions at time step 0, time step 10 = initial velocity when the driver spots a police officer behind him and applies the brakes. The car then slows down, coming to rest at a stoplight ahead. Draw a complete motion diagram for this situation. PROBLEM-SOLVING STRATEGY 1.1 Motion diagrams MODEL: Represent the moving object as a particle. Make simplifying assumptions when interpreting the problem statement. VISUALIZE: A complete motion diagram consists of: The position of the object in each frame of the film, shown as a dot. Use five or six dots to make the motion clear but without overcrowding the picture. More complex motions may need more dots. The average velocity vectors, found by connecting each dot in the motion diagram to the next with a vector arrow. There is one velocity vector linking each set of two position dots. Label the row of velocity vectors . The average acceleration vectors, found using Tactics Box 1.3. There is one acceleration vector linking each set of two velocity vectors. Each acceleration vector is drawn at the dot between the two velocity vectors it links. Use to indicate a point at which the acceleration is zero. Label the row of acceleration vectors . Model It is appropriate to use the particle model for the car. You should also make some simplifying assumptions. v 0 a Part A The car’s motion can be divided into three different stages: its motion before the driver realizes he’s late, its motion after the driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? During each of the three different stages of its motion, the car is moving with constant A. acceleration. B. During each of the three different stages of its motion, the car is moving with constant velocity. C. The highway is straight (i.e., there are no curves). D. The highway is level (i.e., there are no hills or valleys). Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD. ANSWER: Correct In addition to the assumptions listed above, in the rest of this problem assume that the car is moving in a straight line to the right. Visualize Part B In the three diagrams shown to the left, the position of the car at five subsequent instants of time is represented by black dots, and the car’s average velocity is represented by green arrows. Which of these diagrams best describes the position and the velocity of the car before the driver notices he is late? ANSWER: Correct Part C Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver hits the gas, but before he notices the police officer? ANSWER: Correct A B C A B C Part D Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver notices the police officer? ANSWER: Correct Part E Which of the diagrams shown below most accurately depicts the average acceleration vectors of the car during the events described in the problem introduction? ANSWER: A B C Correct You can now draw a complete motion diagram for the situation described in this problem. Your diagram should look like this: Measurements in SI Units Familiarity with SI units will aid your study of physics and all other sciences. Part A What is the approximate height of the average adult in centimeters? Hint 1. Converting between feet and centimeters The distance from your elbow to your fingertips is typically about 50 . A B C cm ANSWER: Correct If you’re not familiar with metric units of length, you can use your body to develop intuition for them. The average height of an adult is 5 6.4 . The distance from elbow to fingertips on the average adult is about 50 . Ten (1 ) is about the width of this adult’s little finger and 10 is about the width of the average hand. Part B Approximately what is the mass of the average adult in kilograms? Hint 1. Converting between pounds and kilograms Something that weighs 1 has a mass of about . ANSWER: Correct Something that weighs 1 has a mass of about . This is a useful conversion to keep in mind! ± A Trip to Europe 100 200 300 cm cm cm feet inches cm mm cm cm pound 1 kg 2 80 500 1200 kg kg kg pound (1/2) kg Learning Goal: To understand how to use dimensional analysis to solve problems. Dimensional analysis is a useful tool for solving problems that involve unit conversions. Since unit conversion is not limited to physics problems but is part of our everyday life, correct use of conversion factors is essential to working through problems of practical importance. For example, dimensional analysis could be used in problems involving currency exchange. Say you want to calculate how many euros you get if you exchange 3600 ( ), given the exchange rate , that is, 1 to 1.20 . Begin by writing down the starting value, 3600 . This can also be written as a fraction: . Next, convert dollars to euros. This conversion involves multiplying by a simple conversion factor derived from the exchange rate: . Note that the “dollar” unit, , should appear on the bottom of this conversion factor, since appears on the top of the starting value. Finally, since dollars are divided by dollars, the units can be canceled and the final result is . Currency exchange is only one example of many practical situations where dimensional analysis may help you to work through problems. Remember that dimensional analysis involves multiplying a given value by a conversion factor, resulting in a value in the new units. The conversion factor can be the ratio of any two quantities, as long as the ratio is equal to one. You and your friends are organizing a trip to Europe. Your plan is to rent a car and drive through the major European capitals. By consulting a map you estimate that you will cover a total distance of 5000 . Consider the euro-dollar exchange rate given in the introduction and use dimensional analysis to work through these simple problems. Part A You select a rental package that includes a car with an average consumption of 6.00 of fuel per 100 . Considering that in Europe the average fuel cost is 1.063 , how much (in US dollars) will you spend in fuel on your trip? Express your answer numerically in US dollars to three significant figures. You did not open hints for this part. ANSWER: US dollars USD 1 EUR = 1.20 USD euro US dollars USD 3600 USD 1 1.00 EUR 1.20 USD USD USD ( )( ) = 3000 EUR 3600 USD 1 1.00 EUR 1.20 USD km liters km euros/liter Part B How many gallons of fuel would the rental car consume per mile? Express your answer numerically in gallons per mile to three significant figures. You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Cost of fuel = USD gallons/mile

Chapter 1 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Curved Motion Diagram The motion diagram shown in the figure represents a pendulum released from rest at an angle of 45 from the vertical. The dots in the motion diagram represent the positions of the pendulum bob at eleven moments separated by equal time intervals. The green arrows represent the average velocity between adjacent dots. Also given is a “compass rose” in which directions are labeled with the letters of the alphabet.  Part A What is the direction of the acceleration of the object at moment 5? Enter the letter of the arrow with this direction from the compass rose in the figure. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again Part B What is the direction of the acceleration of the object at moments 0 and 10? Enter the letters corresponding to the arrows with these directions from the compass rose in the figure, separated by commas. Type Z if the acceleration vector has zero length. You did not open hints for this part. ANSWER: Incorrect; Try Again PSS 1.1 Motion Diagrams Learning Goal: To practice Problem-Solving Strategy 1.1 for motion diagram problems. A car is traveling with constant velocity along a highway. The driver notices he is late for work, so he stomps down on the gas pedal and the car begins to speed up. The car has just achieved double its directions at time step 0, time step 10 = initial velocity when the driver spots a police officer behind him and applies the brakes. The car then slows down, coming to rest at a stoplight ahead. Draw a complete motion diagram for this situation. PROBLEM-SOLVING STRATEGY 1.1 Motion diagrams MODEL: Represent the moving object as a particle. Make simplifying assumptions when interpreting the problem statement. VISUALIZE: A complete motion diagram consists of: The position of the object in each frame of the film, shown as a dot. Use five or six dots to make the motion clear but without overcrowding the picture. More complex motions may need more dots. The average velocity vectors, found by connecting each dot in the motion diagram to the next with a vector arrow. There is one velocity vector linking each set of two position dots. Label the row of velocity vectors . The average acceleration vectors, found using Tactics Box 1.3. There is one acceleration vector linking each set of two velocity vectors. Each acceleration vector is drawn at the dot between the two velocity vectors it links. Use to indicate a point at which the acceleration is zero. Label the row of acceleration vectors . Model It is appropriate to use the particle model for the car. You should also make some simplifying assumptions. v 0 a Part A The car’s motion can be divided into three different stages: its motion before the driver realizes he’s late, its motion after the driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? During each of the three different stages of its motion, the car is moving with constant A. acceleration. B. During each of the three different stages of its motion, the car is moving with constant velocity. C. The highway is straight (i.e., there are no curves). D. The highway is level (i.e., there are no hills or valleys). Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD. ANSWER: Correct In addition to the assumptions listed above, in the rest of this problem assume that the car is moving in a straight line to the right. Visualize Part B In the three diagrams shown to the left, the position of the car at five subsequent instants of time is represented by black dots, and the car’s average velocity is represented by green arrows. Which of these diagrams best describes the position and the velocity of the car before the driver notices he is late? ANSWER: Correct Part C Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver hits the gas, but before he notices the police officer? ANSWER: Correct A B C A B C Part D Which of the diagrams shown to the left best describes the position and the velocity of the car after the driver notices the police officer? ANSWER: Correct Part E Which of the diagrams shown below most accurately depicts the average acceleration vectors of the car during the events described in the problem introduction? ANSWER: A B C Correct You can now draw a complete motion diagram for the situation described in this problem. Your diagram should look like this: Measurements in SI Units Familiarity with SI units will aid your study of physics and all other sciences. Part A What is the approximate height of the average adult in centimeters? Hint 1. Converting between feet and centimeters The distance from your elbow to your fingertips is typically about 50 . A B C cm ANSWER: Correct If you’re not familiar with metric units of length, you can use your body to develop intuition for them. The average height of an adult is 5 6.4 . The distance from elbow to fingertips on the average adult is about 50 . Ten (1 ) is about the width of this adult’s little finger and 10 is about the width of the average hand. Part B Approximately what is the mass of the average adult in kilograms? Hint 1. Converting between pounds and kilograms Something that weighs 1 has a mass of about . ANSWER: Correct Something that weighs 1 has a mass of about . This is a useful conversion to keep in mind! ± A Trip to Europe 100 200 300 cm cm cm feet inches cm mm cm cm pound 1 kg 2 80 500 1200 kg kg kg pound (1/2) kg Learning Goal: To understand how to use dimensional analysis to solve problems. Dimensional analysis is a useful tool for solving problems that involve unit conversions. Since unit conversion is not limited to physics problems but is part of our everyday life, correct use of conversion factors is essential to working through problems of practical importance. For example, dimensional analysis could be used in problems involving currency exchange. Say you want to calculate how many euros you get if you exchange 3600 ( ), given the exchange rate , that is, 1 to 1.20 . Begin by writing down the starting value, 3600 . This can also be written as a fraction: . Next, convert dollars to euros. This conversion involves multiplying by a simple conversion factor derived from the exchange rate: . Note that the “dollar” unit, , should appear on the bottom of this conversion factor, since appears on the top of the starting value. Finally, since dollars are divided by dollars, the units can be canceled and the final result is . Currency exchange is only one example of many practical situations where dimensional analysis may help you to work through problems. Remember that dimensional analysis involves multiplying a given value by a conversion factor, resulting in a value in the new units. The conversion factor can be the ratio of any two quantities, as long as the ratio is equal to one. You and your friends are organizing a trip to Europe. Your plan is to rent a car and drive through the major European capitals. By consulting a map you estimate that you will cover a total distance of 5000 . Consider the euro-dollar exchange rate given in the introduction and use dimensional analysis to work through these simple problems. Part A You select a rental package that includes a car with an average consumption of 6.00 of fuel per 100 . Considering that in Europe the average fuel cost is 1.063 , how much (in US dollars) will you spend in fuel on your trip? Express your answer numerically in US dollars to three significant figures. You did not open hints for this part. ANSWER: US dollars USD 1 EUR = 1.20 USD euro US dollars USD 3600 USD 1 1.00 EUR 1.20 USD USD USD ( )( ) = 3000 EUR 3600 USD 1 1.00 EUR 1.20 USD km liters km euros/liter Part B How many gallons of fuel would the rental car consume per mile? Express your answer numerically in gallons per mile to three significant figures. You did not open hints for this part. ANSWER: Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Cost of fuel = USD gallons/mile

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– 1 – Fall 2015 EECS 338 Assignment 2 Due: Oct. 1st, 2015 G. Ozsoyoglu Concurrent Programming with Semaphores; 140 points (100 pts) 1. Priority-based Searchers/Inserters/Deleters Problem without starvation. Three types of processes, namely, searchers, inserters, and deleters share access to a singly linked list L, and perform search, insert, or delete operations, respectively. The list L does not have duplicate values. a) Searchers merely search the list L, and report success (i.e., item searched is in L) or no-success (i.e., item searched is not in L) to a log file. Hence they can execute concurrently with each other. b) Inserters add new items to the end of the list L, and report success (i.e., item is not in L, and successfully inserted into L) or no-success (i.e., item is already in L, and no insertion takes place) to a log file. Insertions must be mutually exclusive to preclude two inserters from inserting new items at about the same time. However, one insert can proceed in parallel with any number of searches. c) Deleters remove items from anywhere in the list, and report success (i.e., the item is found in L and deleted) or no-success (i.e., item is not in L, and could not be deleted) to a log file. At most one deleter can access the list L at a time, and the deletion must be mutually exclusive with searches and insertions. d) Initial start. Searcher, inserter, and deleter processes are initially launched as follows. A user process that needs a search/insertion/deletion operation to the list L first forks a process, and then, in the forked process, performs an execv into a searcher/ inserter/deleter process. e) Log maintenance. Upon start, each searcher/inserter/deleter writes to a log file, recording the time of insertion, process id, process type (i.e., searcher, inserter, or deleter), and the item that is being searched/inserted/deleted. f) Termination. Upon successful or unsuccessful completion, each searcher/inserter/deleter writes to the same log file, recording the time and the result of its execution. g) Priority-based service between three types. Searchers, inserters, and deleters perform their search, insert, delete operations, respectively, on a priority basis (not on a first-come-first-serve (FCFS) basis) between separate process types (i.e., searchers, inserters, deleters) as follows. Searchers search with the highest priority; inserters insert with the second highest priority (except that one inserter can proceed in parallel with any number of searchers), and deleters delete with the lowest priority. h) FCFS service within a single type. Processes of the same type are serviced FCFS. As an example, among multiple inserters, the order of insertions into L is FCFS. Similarly, among multiple deleters, the order of deletions into L is FCFS. Note that, among searchers, while the start of search among searchers is FCFS, due to concurrent searcher execution, the completions of multiple searchers may not be FCFS. i) Starvation avoidance. In addition to the above priority-based search/insert/delete operations, the following starvation-avoidance rule is enforced. o After 10 consecutive searchers search the list L, if there is at least one waiting inserter or deleter then newly arriving searchers are blocked until (a) all waiting inserters are first serviced FCFS, and, then (b) all waiting deleters are serviced FCFS. Then, both the standard priority-based service between process types and the FCFS service within a process type resume. You are to specify a semaphore-based algorithm to synchronize searcher, inserter and deleter processes. Note:  Explain your algorithm.  Make sure to state any assumptions you make in your solution.  Specify the initial states of your variables and semaphores.  Specify whether your semaphores are binary or nonbinary.  Do not bother specifying algorithms for sequential tasks: simply specify a well-defined function/procedure (i.e., one with well-defined input/output/functional specification). – 2 – (40 pts) 2. Four-of-a-Kind Problem is defined as follows.  There is a deck of 24 cards, split into 6 different kinds, 4 cards of each kind.  There are 4 players (i.e., processes) ??,0≤?≤3; each player can hold 4 cards.  Between each pair of adjacent (i.e., seated next to each other) players, there is a pile of cards.  The game begins by o someone dealing four cards to each player, and putting two cards on the pile between each pair of adjacent players, and o ?0 starting the game. If ?0 has four-of-a-kind, ?0 wins. Whoever gets four-of-a-kind first wins.  Players take turns to play clockwise. That is, ?0 plays, ?1 plays, ?2 plays, ?3 plays, ?0 plays, etc.  Each player behaves as follows. o So long as no one has won, keep playing. o If it is my turn and no one has won:  Check for Four-of-a-Kind. If yes, claim victory. Otherwise discard a card into the pile on the right; pick up a card from the pile on the left; and, check again: If Four-of-a-Kind, claim victory; otherwise revise turn so that the next player plays and wait for your turn.  There are no ties; when a player has claimed victory, all other players stop (when their turns to play come up). You are to specify a semaphore-based algorithm to the Four-of-a-Kind problem. Note:  Explain your algorithm.  Make sure to state any assumptions you make in your solution.  Specify the initial states of your variables and semaphores.  Specify whether your semaphores are binary or nonbinary.  Do not bother specifying algorithms for sequential tasks: simply specify a well-defined function/procedure (i.e., one with well-defined input/output/functional specification). P1 P0 P2 P3 pile 1 pile 2 pile 3 pile 0

– 1 – Fall 2015 EECS 338 Assignment 2 Due: Oct. 1st, 2015 G. Ozsoyoglu Concurrent Programming with Semaphores; 140 points (100 pts) 1. Priority-based Searchers/Inserters/Deleters Problem without starvation. Three types of processes, namely, searchers, inserters, and deleters share access to a singly linked list L, and perform search, insert, or delete operations, respectively. The list L does not have duplicate values. a) Searchers merely search the list L, and report success (i.e., item searched is in L) or no-success (i.e., item searched is not in L) to a log file. Hence they can execute concurrently with each other. b) Inserters add new items to the end of the list L, and report success (i.e., item is not in L, and successfully inserted into L) or no-success (i.e., item is already in L, and no insertion takes place) to a log file. Insertions must be mutually exclusive to preclude two inserters from inserting new items at about the same time. However, one insert can proceed in parallel with any number of searches. c) Deleters remove items from anywhere in the list, and report success (i.e., the item is found in L and deleted) or no-success (i.e., item is not in L, and could not be deleted) to a log file. At most one deleter can access the list L at a time, and the deletion must be mutually exclusive with searches and insertions. d) Initial start. Searcher, inserter, and deleter processes are initially launched as follows. A user process that needs a search/insertion/deletion operation to the list L first forks a process, and then, in the forked process, performs an execv into a searcher/ inserter/deleter process. e) Log maintenance. Upon start, each searcher/inserter/deleter writes to a log file, recording the time of insertion, process id, process type (i.e., searcher, inserter, or deleter), and the item that is being searched/inserted/deleted. f) Termination. Upon successful or unsuccessful completion, each searcher/inserter/deleter writes to the same log file, recording the time and the result of its execution. g) Priority-based service between three types. Searchers, inserters, and deleters perform their search, insert, delete operations, respectively, on a priority basis (not on a first-come-first-serve (FCFS) basis) between separate process types (i.e., searchers, inserters, deleters) as follows. Searchers search with the highest priority; inserters insert with the second highest priority (except that one inserter can proceed in parallel with any number of searchers), and deleters delete with the lowest priority. h) FCFS service within a single type. Processes of the same type are serviced FCFS. As an example, among multiple inserters, the order of insertions into L is FCFS. Similarly, among multiple deleters, the order of deletions into L is FCFS. Note that, among searchers, while the start of search among searchers is FCFS, due to concurrent searcher execution, the completions of multiple searchers may not be FCFS. i) Starvation avoidance. In addition to the above priority-based search/insert/delete operations, the following starvation-avoidance rule is enforced. o After 10 consecutive searchers search the list L, if there is at least one waiting inserter or deleter then newly arriving searchers are blocked until (a) all waiting inserters are first serviced FCFS, and, then (b) all waiting deleters are serviced FCFS. Then, both the standard priority-based service between process types and the FCFS service within a process type resume. You are to specify a semaphore-based algorithm to synchronize searcher, inserter and deleter processes. Note:  Explain your algorithm.  Make sure to state any assumptions you make in your solution.  Specify the initial states of your variables and semaphores.  Specify whether your semaphores are binary or nonbinary.  Do not bother specifying algorithms for sequential tasks: simply specify a well-defined function/procedure (i.e., one with well-defined input/output/functional specification). – 2 – (40 pts) 2. Four-of-a-Kind Problem is defined as follows.  There is a deck of 24 cards, split into 6 different kinds, 4 cards of each kind.  There are 4 players (i.e., processes) ??,0≤?≤3; each player can hold 4 cards.  Between each pair of adjacent (i.e., seated next to each other) players, there is a pile of cards.  The game begins by o someone dealing four cards to each player, and putting two cards on the pile between each pair of adjacent players, and o ?0 starting the game. If ?0 has four-of-a-kind, ?0 wins. Whoever gets four-of-a-kind first wins.  Players take turns to play clockwise. That is, ?0 plays, ?1 plays, ?2 plays, ?3 plays, ?0 plays, etc.  Each player behaves as follows. o So long as no one has won, keep playing. o If it is my turn and no one has won:  Check for Four-of-a-Kind. If yes, claim victory. Otherwise discard a card into the pile on the right; pick up a card from the pile on the left; and, check again: If Four-of-a-Kind, claim victory; otherwise revise turn so that the next player plays and wait for your turn.  There are no ties; when a player has claimed victory, all other players stop (when their turns to play come up). You are to specify a semaphore-based algorithm to the Four-of-a-Kind problem. Note:  Explain your algorithm.  Make sure to state any assumptions you make in your solution.  Specify the initial states of your variables and semaphores.  Specify whether your semaphores are binary or nonbinary.  Do not bother specifying algorithms for sequential tasks: simply specify a well-defined function/procedure (i.e., one with well-defined input/output/functional specification). P1 P0 P2 P3 pile 1 pile 2 pile 3 pile 0

– 1 – Fall 2015 EECS 338 Assignment 2 Due: … Read More...
Chapter 6 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v  FG T F  = = net i F  i 0 ASSESS: Check if your result has the correct units, is reasonable, and answers the question. Model The trunk is moving at a constant velocity. This means that you can model it as a particle in dynamic equilibrium and apply the strategy above. Furthermore, you can ignore the masses of the ropes and the ring because it is reasonable to assume that their combined weight is much less than the weight of the trunk. Visualize Part A The most convenient coordinate system for this problem is one in which the y axis is vertical and the ropes both lie in the xy plane, as shown below. Identify the forces acting on the trunk, and then draw a free-body diagram of the trunk in the diagram below. The black dot represents the trunk as it is lifted by the students. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded. ANSWER: Part B This question will be shown after you complete previous question(s). Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). A Gymnast on a Rope A gymnast of mass 70.0 hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value for the acceleration of gravity. Part A Calculate the tension in the rope if the gymnast hangs motionless on the rope. Express your answer in newtons. You did not open hints for this part. ANSWER: Part B Calculate the tension in the rope if the gymnast climbs the rope at a constant rate. Express your answer in newtons. You did not open hints for this part. kg 9.81m/s2 T T = N T ANSWER: Part C Calculate the tension in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: Part D Calculate the tension in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: T = N T m/s2 T = N T m/s2 T = N Applying Newton’s 2nd Law Learning Goal: To learn a systematic approach to solving Newton’s 2nd law problems using a simple example. Once you have decided to solve a problem using Newton’s 2nd law, there are steps that will lead you to a solution. One such prescription is the following: Visualize the problem and identify special cases. Isolate each body and draw the forces acting on it. Choose a coordinate system for each body. Apply Newton’s 2nd law to each body. Write equations for the constraints and other given information. Solve the resulting equations symbolically. Check that your answer has the correct dimensions and satisfies special cases. If numbers are given in the problem, plug them in and check that the answer makes sense. Think about generalizations or simplfications of the problem. As an example, we will apply this procedure to find the acceleration of a block of mass that is pulled up a frictionless plane inclined at angle with respect to the horizontal by a perfect string that passes over a perfect pulley to a block of mass that is hanging vertically. Visualize the problem and identify special cases First examine the problem by drawing a picture and visualizing the motion. Apply Newton’s 2nd law, , to each body in your mind. Don’t worry about which quantities are given. Think about the forces on each body: How are these consistent with the direction of the acceleration for that body? Can you think of any special cases that you can solve quickly now and use to test your understanding later? m2  m1 F = ma One special case in this problem is if , in which case block 1 would simply fall freely under the acceleration of gravity: . Part A Consider another special case in which the inclined plane is vertical ( ). In this case, for what value of would the acceleration of the two blocks be equal to zero? Express your answer in terms of some or all of the variables and . ANSWER: Isolate each body and draw the forces acting on it A force diagram should include only real forces that act on the body and satisfy Newton’s 3rd law. One way to check if the forces are real is to detrmine whether they are part of a Newton’s 3rd law pair, that is, whether they result from a physical interaction that also causes an opposite force on some other body, which may not be part of the problem. Do not decompose the forces into components, and do not include resultant forces that are combinations of other real forces like centripetal force or fictitious forces like the “centrifugal” force. Assign each force a symbol, but don’t start to solve the problem at this point. Part B Which of the four drawings is a correct force diagram for this problem? = 0 m2 = −g a 1 j ^  = /2 m1 m2 g m1 = ANSWER: Choose a coordinate system for each body Newton’s 2nd law, , is a vector equation. To add or subtract vectors it is often easiest to decompose each vector into components. Whereas a particular set of vector components is only valid in a particular coordinate system, the vector equality holds in any coordinate system, giving you freedom to pick a coordinate system that most simplifies the equations that result from the component equations. It’s generally best to pick a coordinate system where the acceleration of the system lies directly on one of the coordinate axes. If there is no acceleration, then pick a coordinate system with as many unknowns as possible along the coordinate axes. Vectors that lie along the axes appear in only one of the equations for each component, rather than in two equations with trigonometric prefactors. Note that it is sometimes advantageous to use different coordinate systems for each body in the problem. In this problem, you should use Cartesian coordinates and your axes should be stationary with respect to the inclined plane. Part C Given the criteria just described, what orientation of the coordinate axes would be best to use in this problem? In the answer options, “tilted” means with the x axis oriented parallel to the plane (i.e., at angle to the horizontal), and “level” means with the x axis horizontal. ANSWER: Apply Newton’s 2nd law to each body a b c d F  = ma  tilted for both block 1 and block 2 tilted for block 1 and level for block 2 level for block 1 and tilted for block 2 level for both block 1 and block 2 Part D What is , the sum of the x components of the forces acting on block 2? Take forces acting up the incline to be positive. Express your answer in terms of some or all of the variables tension , , the magnitude of the acceleration of gravity , and . You did not open hints for this part. ANSWER: Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Lifting a Bucket A 6- bucket of water is being pulled straight up by a string at a constant speed. F2x T m2 g  m2a2x =F2x = kg Part A What is the tension in the rope? ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Friction Force on a Dancer on a Drawbridge A dancer is standing on one leg on a drawbridge that is about to open. The coefficients of static and kinetic friction between the drawbridge and the dancer’s foot are and , respectively. represents the normal force exerted on the dancer by the bridge, and represents the gravitational force exerted on the dancer, as shown in the drawing . For all the questions, we can assume that the bridge is a perfectly flat surface and lacks the curvature characteristic of most bridges. about 42 about 60 about 78 0 because the bucket has no acceleration. N N N N μs μk n F  g Part A Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component of the friction force, ? Express your answer in terms of some or all of the variables , , and/or . You did not open hints for this part. ANSWER: Part B The drawbridge then starts to rise and the dancer continues to stand on one leg. The drawbridge stops just at the point where the dancer is on the verge of slipping. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. F  f n μs μk Ff = Ff n μs μk  You did not open hints for this part. ANSWER: Part C Then, because the bridge is old and poorly designed, it falls a little bit and then jerks. This causes the person to start to slide down the bridge at a constant speed. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. ANSWER: Part D The bridge starts to come back down again. The dancer stops sliding. However, again because of the age and design of the bridge it never makes it all the way down; rather it stops half a meter short. This half a meter corresponds to an angle degree (see the diagram, which has the angle exaggerated). What is the force of friction now? Express your answer in terms of some or all of the variables , , and . Ff = Ff n μs μk  Ff =   1 Ff  n Fg You did not open hints for this part. ANSWER: Kinetic Friction Ranking Task Below are eight crates of different mass. The crates are attached to massless ropes, as indicated in the picture, where the ropes are marked by letters. Each crate is being pulled to the right at the same constant speed. The coefficient of kinetic friction between each crate and the surface on which it slides is the same for all eight crates. Ff = Part A Rank the ropes on the basis of the force each exerts on the crate immediately to its left. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Pushing a Block Learning Goal: To understand kinetic and static friction. A block of mass lies on a horizontal table. The coefficient of static friction between the block and the table is . The coefficient of kinetic friction is , with . Part A m μs μk μk < μs If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from the table), what is the magnitude of the force due to friction? You did not open hints for this part. ANSWER: Part B Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force must you be pushing the block just before the block begins to move? Express the magnitude of in terms of some or all the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Part C Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force? Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. Ffriction = F F μs μk m g F = μs μk m g ANSWER: Part D Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration of the block after it begins to move. Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Ffriction = a μs μk m g a =

Chapter 6 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v  FG T F  = = net i F  i 0 ASSESS: Check if your result has the correct units, is reasonable, and answers the question. Model The trunk is moving at a constant velocity. This means that you can model it as a particle in dynamic equilibrium and apply the strategy above. Furthermore, you can ignore the masses of the ropes and the ring because it is reasonable to assume that their combined weight is much less than the weight of the trunk. Visualize Part A The most convenient coordinate system for this problem is one in which the y axis is vertical and the ropes both lie in the xy plane, as shown below. Identify the forces acting on the trunk, and then draw a free-body diagram of the trunk in the diagram below. The black dot represents the trunk as it is lifted by the students. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded. ANSWER: Part B This question will be shown after you complete previous question(s). Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). A Gymnast on a Rope A gymnast of mass 70.0 hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value for the acceleration of gravity. Part A Calculate the tension in the rope if the gymnast hangs motionless on the rope. Express your answer in newtons. You did not open hints for this part. ANSWER: Part B Calculate the tension in the rope if the gymnast climbs the rope at a constant rate. Express your answer in newtons. You did not open hints for this part. kg 9.81m/s2 T T = N T ANSWER: Part C Calculate the tension in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: Part D Calculate the tension in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: T = N T m/s2 T = N T m/s2 T = N Applying Newton’s 2nd Law Learning Goal: To learn a systematic approach to solving Newton’s 2nd law problems using a simple example. Once you have decided to solve a problem using Newton’s 2nd law, there are steps that will lead you to a solution. One such prescription is the following: Visualize the problem and identify special cases. Isolate each body and draw the forces acting on it. Choose a coordinate system for each body. Apply Newton’s 2nd law to each body. Write equations for the constraints and other given information. Solve the resulting equations symbolically. Check that your answer has the correct dimensions and satisfies special cases. If numbers are given in the problem, plug them in and check that the answer makes sense. Think about generalizations or simplfications of the problem. As an example, we will apply this procedure to find the acceleration of a block of mass that is pulled up a frictionless plane inclined at angle with respect to the horizontal by a perfect string that passes over a perfect pulley to a block of mass that is hanging vertically. Visualize the problem and identify special cases First examine the problem by drawing a picture and visualizing the motion. Apply Newton’s 2nd law, , to each body in your mind. Don’t worry about which quantities are given. Think about the forces on each body: How are these consistent with the direction of the acceleration for that body? Can you think of any special cases that you can solve quickly now and use to test your understanding later? m2  m1 F = ma One special case in this problem is if , in which case block 1 would simply fall freely under the acceleration of gravity: . Part A Consider another special case in which the inclined plane is vertical ( ). In this case, for what value of would the acceleration of the two blocks be equal to zero? Express your answer in terms of some or all of the variables and . ANSWER: Isolate each body and draw the forces acting on it A force diagram should include only real forces that act on the body and satisfy Newton’s 3rd law. One way to check if the forces are real is to detrmine whether they are part of a Newton’s 3rd law pair, that is, whether they result from a physical interaction that also causes an opposite force on some other body, which may not be part of the problem. Do not decompose the forces into components, and do not include resultant forces that are combinations of other real forces like centripetal force or fictitious forces like the “centrifugal” force. Assign each force a symbol, but don’t start to solve the problem at this point. Part B Which of the four drawings is a correct force diagram for this problem? = 0 m2 = −g a 1 j ^  = /2 m1 m2 g m1 = ANSWER: Choose a coordinate system for each body Newton’s 2nd law, , is a vector equation. To add or subtract vectors it is often easiest to decompose each vector into components. Whereas a particular set of vector components is only valid in a particular coordinate system, the vector equality holds in any coordinate system, giving you freedom to pick a coordinate system that most simplifies the equations that result from the component equations. It’s generally best to pick a coordinate system where the acceleration of the system lies directly on one of the coordinate axes. If there is no acceleration, then pick a coordinate system with as many unknowns as possible along the coordinate axes. Vectors that lie along the axes appear in only one of the equations for each component, rather than in two equations with trigonometric prefactors. Note that it is sometimes advantageous to use different coordinate systems for each body in the problem. In this problem, you should use Cartesian coordinates and your axes should be stationary with respect to the inclined plane. Part C Given the criteria just described, what orientation of the coordinate axes would be best to use in this problem? In the answer options, “tilted” means with the x axis oriented parallel to the plane (i.e., at angle to the horizontal), and “level” means with the x axis horizontal. ANSWER: Apply Newton’s 2nd law to each body a b c d F  = ma  tilted for both block 1 and block 2 tilted for block 1 and level for block 2 level for block 1 and tilted for block 2 level for both block 1 and block 2 Part D What is , the sum of the x components of the forces acting on block 2? Take forces acting up the incline to be positive. Express your answer in terms of some or all of the variables tension , , the magnitude of the acceleration of gravity , and . You did not open hints for this part. ANSWER: Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Lifting a Bucket A 6- bucket of water is being pulled straight up by a string at a constant speed. F2x T m2 g  m2a2x =F2x = kg Part A What is the tension in the rope? ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Friction Force on a Dancer on a Drawbridge A dancer is standing on one leg on a drawbridge that is about to open. The coefficients of static and kinetic friction between the drawbridge and the dancer’s foot are and , respectively. represents the normal force exerted on the dancer by the bridge, and represents the gravitational force exerted on the dancer, as shown in the drawing . For all the questions, we can assume that the bridge is a perfectly flat surface and lacks the curvature characteristic of most bridges. about 42 about 60 about 78 0 because the bucket has no acceleration. N N N N μs μk n F  g Part A Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component of the friction force, ? Express your answer in terms of some or all of the variables , , and/or . You did not open hints for this part. ANSWER: Part B The drawbridge then starts to rise and the dancer continues to stand on one leg. The drawbridge stops just at the point where the dancer is on the verge of slipping. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. F  f n μs μk Ff = Ff n μs μk  You did not open hints for this part. ANSWER: Part C Then, because the bridge is old and poorly designed, it falls a little bit and then jerks. This causes the person to start to slide down the bridge at a constant speed. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. ANSWER: Part D The bridge starts to come back down again. The dancer stops sliding. However, again because of the age and design of the bridge it never makes it all the way down; rather it stops half a meter short. This half a meter corresponds to an angle degree (see the diagram, which has the angle exaggerated). What is the force of friction now? Express your answer in terms of some or all of the variables , , and . Ff = Ff n μs μk  Ff =   1 Ff  n Fg You did not open hints for this part. ANSWER: Kinetic Friction Ranking Task Below are eight crates of different mass. The crates are attached to massless ropes, as indicated in the picture, where the ropes are marked by letters. Each crate is being pulled to the right at the same constant speed. The coefficient of kinetic friction between each crate and the surface on which it slides is the same for all eight crates. Ff = Part A Rank the ropes on the basis of the force each exerts on the crate immediately to its left. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Pushing a Block Learning Goal: To understand kinetic and static friction. A block of mass lies on a horizontal table. The coefficient of static friction between the block and the table is . The coefficient of kinetic friction is , with . Part A m μs μk μk < μs If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from the table), what is the magnitude of the force due to friction? You did not open hints for this part. ANSWER: Part B Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force must you be pushing the block just before the block begins to move? Express the magnitude of in terms of some or all the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Part C Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force? Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. Ffriction = F F μs μk m g F = μs μk m g ANSWER: Part D Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration of the block after it begins to move. Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Ffriction = a μs μk m g a =

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Develop a any small project plan that include the following: 1. Name of your proposed project – a business related project 2. Description of the project 3. Goals and objectives 4. Scope statement that includes requirements/deliverable, constraints and assumptions 5. Work breakdown Structure (25 items) 6. Duration of each of the tasks 7. Draw a network diagram 8. Determine the project duration 9. Identify resources, hourly rate and assigned task 10. Identify total cost 11. Identify three to five risks on the project and develop a risk response plan for each one. 12. Develop a quality standard.

Develop a any small project plan that include the following: 1. Name of your proposed project – a business related project 2. Description of the project 3. Goals and objectives 4. Scope statement that includes requirements/deliverable, constraints and assumptions 5. Work breakdown Structure (25 items) 6. Duration of each of the tasks 7. Draw a network diagram 8. Determine the project duration 9. Identify resources, hourly rate and assigned task 10. Identify total cost 11. Identify three to five risks on the project and develop a risk response plan for each one. 12. Develop a quality standard.

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Chapter 6 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v  FG T F  = = net i F  i 0

Chapter 6 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v  FG T F  = = net i F  i 0

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