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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Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 57 5/9/2014 8:02 PM Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 57 5/9/2014 8:02 PM Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? = = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 57 5/9/2014 8:02 PM Hint 1. What is the general quadratic equation? The general quadratic equation is , where , , and are constants. Depending on the value of the discriminant, , the equation may have two real valued 1. solutions if , 2. one real valued solution if , or 3. two complex valued solutions if . In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. No; there is no chance he is going to get aboard. Yes; he will get a second chance Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 57 5/9/2014 8:02 PM ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: C and E E and F A and F C and D B and D C and D A and F E and F A and B E and D Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 57 5/9/2014 8:02 PM Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 2. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. ANSWER: A and F A and E D and B C and D E and F Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 57 5/9/2014 8:02 PM Correct Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. = 5 positive negative Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 57 5/9/2014 8:02 PM ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth’s gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, , is constant. 1. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of its velocity, , changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time corresponds to the moment just after the ball is launched from position and . Its launch velocity, also called the initial velocity, is . Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time with velocity . Its position at this moment is denoted by or since it is at its maximum height. The other point, at time with velocity , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is , also known as ( since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence . Part A , = -2,-5 , Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 57 5/9/2014 8:02 PM How do the speeds , , and (at times ,

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 57 5/9/2014 8:02 PM Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 57 5/9/2014 8:02 PM Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? = = Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 57 5/9/2014 8:02 PM Hint 1. What is the general quadratic equation? The general quadratic equation is , where , , and are constants. Depending on the value of the discriminant, , the equation may have two real valued 1. solutions if , 2. one real valued solution if , or 3. two complex valued solutions if . In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. No; there is no chance he is going to get aboard. Yes; he will get a second chance Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 57 5/9/2014 8:02 PM ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: C and E E and F A and F C and D B and D C and D A and F E and F A and B E and D Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 57 5/9/2014 8:02 PM Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 2. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. ANSWER: A and F A and E D and B C and D E and F Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 57 5/9/2014 8:02 PM Correct Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. = 5 positive negative Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 57 5/9/2014 8:02 PM ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth’s gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, , is constant. 1. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of its velocity, , changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time corresponds to the moment just after the ball is launched from position and . Its launch velocity, also called the initial velocity, is . Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time with velocity . Its position at this moment is denoted by or since it is at its maximum height. The other point, at time with velocity , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is , also known as ( since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence . Part A , = -2,-5 , Extra Credit http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 57 5/9/2014 8:02 PM How do the speeds , , and (at times ,

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Doppler Shift 73 Because of the Doppler Effect, light emitted by an object can appear to change wavelength due to its motion toward or away from an observer. When the observer and the source of light are moving toward each other, the light is shifted to shorter wavelengths (blueshifted). When the observer and the source of light are moving away from each other, the light is shifted to longer wavelengths (redshifted). Part I: Motion of Source Star is not . rnovrng r ABCD 1) Consider the situations shown (A—D). a) In which situation will the observer receive light that is shifted to shorter wavelengths? b) Will this light be blueshifted or redshifted for this case? c) What direction is the star moving relative to the observer for this case? 2) Consider the situations shown (A—D). a) In which situation will the observer receive light that is shifted to longer wavelengths? b) Will this light be blueshifted or redshifted for this case? c) What direction is the star moving relative to the observer for this case? . 74 Doppler Shift 3) In which of the srtuations shown (A—D) will theobserver receive light that Is not Doppler Shifted at all? Explain your reasoning. – 4) Imagine our solar system Is moving In the Milky Way toward a group of three stars. Star A is a blue star that is slightly closer to us than the other two. Star B is a red star that is farthest away from us. Star C is a yellow star that is halfway between Stars A end B. a) Which of these three stars, if any, will give off light that appears to be blueshifted? Explain your reasoning. . / b) Which of these three stars, if any, will give off light that appears to be redshifted? Explain your reasoning. c) Which of these three stars, if any, will give off light that appears to have no shift? Explain your reasoning. — 5) You overhear two students discussing the topic of Doppler Shift. Student 1: Since Betelgeuse is a red star, it must be going away from us, and since Rigel is a blue star it must be coming toward us. Student 2: 1 disagree, the color of the star does not tell you if it is moving. You have to look at the shift in wavelength of the lines in the star’s absorption spectrum to determine whether it’s moving toward or away from you. Do you agree or disagree with either or both of the students? Explain your reasoning. 5 Part II: Shift in Absorption Spectra When we study an astronomical object like a star or galaxy, we examine the spectrum of light it gives off. Since the lines of a spectrum occur at specific wavelengths we can determine that an object is moving when we see that the lines have been shifted to either longer or shorter wavelengths. For the absorption line spectra shown on the next page, short-wavelength light (the blue end of the spectrum) is shown on the left-hand side and long-wavelength light (the red end of the spectrum) is shown on the right-hand side. Doppler Shift 75 For the three absorption line spectra shown below (A, B, and C), one of the spectra corresponds to a star that is not moving relative to you, one of the spectra is from a star that is moving toward you, and one of the spectra is from a star that is moving away from you. A B Blue J___ ..‘ C 6) Which of the three spectra above corresponds with the star moving toward you? Explain your reasoning. If two sources of llght are moving relative to an observer, the light from the star that is moving faster will appear to undergo a greater Doppler Consider the four spectra at the right. The spectrum labeled F is an absorption line spectrum from a star that is at rest. Again, note that short-wavelength (blue) light is shown on the left-hand side of each spectrum and long-wavelength (red) light is shown on the right-hand side of each spectrum. 7) Which of the three spectra corresponds with the star moving away from you? Explain your reasoning. Part 111: Size of Shift and Speed Blue Red . – 76 Doppler Shift 8) Which of the four spectra would be from the star that is moving the fastest? Would this star be moving toward or away from the observer? 9) Of the stars that are moving, which spectra would be from the star that is moving the slowest? Describe the motion of this star, – (fJ 1O)An Important line In the absorption spectrum of stars occurs at a wavelength of 656 nm for stars at rest. Irna me that you observe five stars (H—L) from Earth and discover that this Important absorption line Is measured at the wavelength shown in the table below for each of the five stars, Star Wavelength of Absorption Line H 649nm I 660 nm J 656nrn K 658nrn L 647nm a) Which of the stars are gMng off light that appears blueshifted? Explain your reasoning. b) Which of the stars are gMng off light that appears redshifted? Explain your reasoning. d) Which star is moving the fastest? Is it moving toward or away from the observer? Explain your reasoning. , . . c) Which star is giving off light that appears shifted by the greatest amount? Is this light shifted to longer or shorter wavelengths? Explain your reasoning. a) Which planets will receive a radio signal that Is redshifted? Explain your reasoning. b) Which planets wfll receive a radio signal that is shifted to shorter wavelengths? Explain your reasoning. a a . ii) The figure at right shows a spaceprobe and five planets. The motion of the spaceprobe is indicated by the arrow. The spaceprobe is continuously broadcasting a radio signal in all directions. 4 C E not to scale c) Will all the planets receive radio signals from the spaceprobe that are Doppler shifted? Explain your reasoning. d) How will the size of the Doppler Shift in the radio signals detected at Planets A and B compare? Explain your reasoning. Cats r , ‘, e) How Will the slz of 1h Dupler Shift in the radio signals deteed °lane E and B compare? Explain your reasoning. ‘

Doppler Shift 73 Because of the Doppler Effect, light emitted by an object can appear to change wavelength due to its motion toward or away from an observer. When the observer and the source of light are moving toward each other, the light is shifted to shorter wavelengths (blueshifted). When the observer and the source of light are moving away from each other, the light is shifted to longer wavelengths (redshifted). Part I: Motion of Source Star is not . rnovrng r ABCD 1) Consider the situations shown (A—D). a) In which situation will the observer receive light that is shifted to shorter wavelengths? b) Will this light be blueshifted or redshifted for this case? c) What direction is the star moving relative to the observer for this case? 2) Consider the situations shown (A—D). a) In which situation will the observer receive light that is shifted to longer wavelengths? b) Will this light be blueshifted or redshifted for this case? c) What direction is the star moving relative to the observer for this case? . 74 Doppler Shift 3) In which of the srtuations shown (A—D) will theobserver receive light that Is not Doppler Shifted at all? Explain your reasoning. – 4) Imagine our solar system Is moving In the Milky Way toward a group of three stars. Star A is a blue star that is slightly closer to us than the other two. Star B is a red star that is farthest away from us. Star C is a yellow star that is halfway between Stars A end B. a) Which of these three stars, if any, will give off light that appears to be blueshifted? Explain your reasoning. . / b) Which of these three stars, if any, will give off light that appears to be redshifted? Explain your reasoning. c) Which of these three stars, if any, will give off light that appears to have no shift? Explain your reasoning. — 5) You overhear two students discussing the topic of Doppler Shift. Student 1: Since Betelgeuse is a red star, it must be going away from us, and since Rigel is a blue star it must be coming toward us. Student 2: 1 disagree, the color of the star does not tell you if it is moving. You have to look at the shift in wavelength of the lines in the star’s absorption spectrum to determine whether it’s moving toward or away from you. Do you agree or disagree with either or both of the students? Explain your reasoning. 5 Part II: Shift in Absorption Spectra When we study an astronomical object like a star or galaxy, we examine the spectrum of light it gives off. Since the lines of a spectrum occur at specific wavelengths we can determine that an object is moving when we see that the lines have been shifted to either longer or shorter wavelengths. For the absorption line spectra shown on the next page, short-wavelength light (the blue end of the spectrum) is shown on the left-hand side and long-wavelength light (the red end of the spectrum) is shown on the right-hand side. Doppler Shift 75 For the three absorption line spectra shown below (A, B, and C), one of the spectra corresponds to a star that is not moving relative to you, one of the spectra is from a star that is moving toward you, and one of the spectra is from a star that is moving away from you. A B Blue J___ ..‘ C 6) Which of the three spectra above corresponds with the star moving toward you? Explain your reasoning. If two sources of llght are moving relative to an observer, the light from the star that is moving faster will appear to undergo a greater Doppler Consider the four spectra at the right. The spectrum labeled F is an absorption line spectrum from a star that is at rest. Again, note that short-wavelength (blue) light is shown on the left-hand side of each spectrum and long-wavelength (red) light is shown on the right-hand side of each spectrum. 7) Which of the three spectra corresponds with the star moving away from you? Explain your reasoning. Part 111: Size of Shift and Speed Blue Red . – 76 Doppler Shift 8) Which of the four spectra would be from the star that is moving the fastest? Would this star be moving toward or away from the observer? 9) Of the stars that are moving, which spectra would be from the star that is moving the slowest? Describe the motion of this star, – (fJ 1O)An Important line In the absorption spectrum of stars occurs at a wavelength of 656 nm for stars at rest. Irna me that you observe five stars (H—L) from Earth and discover that this Important absorption line Is measured at the wavelength shown in the table below for each of the five stars, Star Wavelength of Absorption Line H 649nm I 660 nm J 656nrn K 658nrn L 647nm a) Which of the stars are gMng off light that appears blueshifted? Explain your reasoning. b) Which of the stars are gMng off light that appears redshifted? Explain your reasoning. d) Which star is moving the fastest? Is it moving toward or away from the observer? Explain your reasoning. , . . c) Which star is giving off light that appears shifted by the greatest amount? Is this light shifted to longer or shorter wavelengths? Explain your reasoning. a) Which planets will receive a radio signal that Is redshifted? Explain your reasoning. b) Which planets wfll receive a radio signal that is shifted to shorter wavelengths? Explain your reasoning. a a . ii) The figure at right shows a spaceprobe and five planets. The motion of the spaceprobe is indicated by the arrow. The spaceprobe is continuously broadcasting a radio signal in all directions. 4 C E not to scale c) Will all the planets receive radio signals from the spaceprobe that are Doppler shifted? Explain your reasoning. d) How will the size of the Doppler Shift in the radio signals detected at Planets A and B compare? Explain your reasoning. Cats r , ‘, e) How Will the slz of 1h Dupler Shift in the radio signals deteed °lane E and B compare? Explain your reasoning. ‘

  ANSWERS Part 1 1 C is the answer because … Read More...
Initial Data Collection After implementing your intervention/innovation, you may have noted that data collection isn’t exactly a linear process. Sometimes you need to go back and get more information, and sometimes you find yourself asking additional questions (that’s ok). In Chapter 6, Fichtman Dana and Yendol-Hoppey provide four steps to data analysis: 1. providing a description of the data; 2. making sense of what you have (and don’t have); 3. interpreting your data by creating statements about how the data informs an answer to the original question; 4. implications of the data. For this assignment, please develop responses to the first two steps using the following points as your guide: ● Please describe the data you’ve collected. ○ What did you see as you inquired? What was happening? ○ What are your initial insights into the data? ● Next, please explain how you have organized your data (“chronologically, by key events, or some combination of organizing units?”). ○ Have you provided the reader with evidence that you’ve looked at your inquiry from a number of angles and have collected trustworthy data? ○ Have you provided evidence of data triangulation? ○ What further questions do you have after your initial data collection? ○ How will you collect more information to satisfy your next questions? Assignment: Initial Data Collection (Due Week 2 Sunday, 11:59 p.m.) After implementing your intervention/innovation, you may have noted that data collection isn’t exactly a linear process. Sometimes you need to go back and get more information, and sometimes you find yourself asking additional questions (that’s ok). In Chapter 6, Fichtman Dana and Yendol-Hoppey provide four steps to data analysis: 1. providing a description of the data; 2. making sense of what you have (and don’t have); 3. interpreting your data by creating statements about how the data informs an answer to the original question; 4. implications of the data. For this assignment, please develop responses to the first two steps using the following points as your guide: ● Please describe the data you’ve collected. ○ What did you see as you inquired? What was happening? ○ What are your initial insights into the data? ● Next, please explain how you have organized your data (“chronologically, by key events, or some combination of organizing units?”). ○ Have you provided the reader with evidence that you’ve looked at your inquiry from a number of angles and have collected trustworthy data? ○ Have you provided evidence of data triangulation? ○ What further questions do you have after your initial data collection? ○ How will you collect more information to satisfy your next questions? Module 2 – Data Collection, Part 2 Module 2 continues to examine the data you are collecting with respect to issues of validity, reliability, trustworthiness, and sufficiency. Please continue to collect data relevant to your inquiry and begin to think about how you will code this data into meaningful organizing principles. Be sure to continuously write memos about your process as a sort of idea journal that you can continually draw from when writing your assignments. Required Readings: Dana, N. F. & Yendol-Hoppey, D. – Revisit Chapter 6 Assignments: For assignment details refer to the “Assignments for the Course” section in this syllabus or the submission link within Blackboard. Assignment: Initial Data Collection (Due Week 2 Sunday, 11:59 p.m.)

Initial Data Collection After implementing your intervention/innovation, you may have noted that data collection isn’t exactly a linear process. Sometimes you need to go back and get more information, and sometimes you find yourself asking additional questions (that’s ok). In Chapter 6, Fichtman Dana and Yendol-Hoppey provide four steps to data analysis: 1. providing a description of the data; 2. making sense of what you have (and don’t have); 3. interpreting your data by creating statements about how the data informs an answer to the original question; 4. implications of the data. For this assignment, please develop responses to the first two steps using the following points as your guide: ● Please describe the data you’ve collected. ○ What did you see as you inquired? What was happening? ○ What are your initial insights into the data? ● Next, please explain how you have organized your data (“chronologically, by key events, or some combination of organizing units?”). ○ Have you provided the reader with evidence that you’ve looked at your inquiry from a number of angles and have collected trustworthy data? ○ Have you provided evidence of data triangulation? ○ What further questions do you have after your initial data collection? ○ How will you collect more information to satisfy your next questions? Assignment: Initial Data Collection (Due Week 2 Sunday, 11:59 p.m.) After implementing your intervention/innovation, you may have noted that data collection isn’t exactly a linear process. Sometimes you need to go back and get more information, and sometimes you find yourself asking additional questions (that’s ok). In Chapter 6, Fichtman Dana and Yendol-Hoppey provide four steps to data analysis: 1. providing a description of the data; 2. making sense of what you have (and don’t have); 3. interpreting your data by creating statements about how the data informs an answer to the original question; 4. implications of the data. For this assignment, please develop responses to the first two steps using the following points as your guide: ● Please describe the data you’ve collected. ○ What did you see as you inquired? What was happening? ○ What are your initial insights into the data? ● Next, please explain how you have organized your data (“chronologically, by key events, or some combination of organizing units?”). ○ Have you provided the reader with evidence that you’ve looked at your inquiry from a number of angles and have collected trustworthy data? ○ Have you provided evidence of data triangulation? ○ What further questions do you have after your initial data collection? ○ How will you collect more information to satisfy your next questions? Module 2 – Data Collection, Part 2 Module 2 continues to examine the data you are collecting with respect to issues of validity, reliability, trustworthiness, and sufficiency. Please continue to collect data relevant to your inquiry and begin to think about how you will code this data into meaningful organizing principles. Be sure to continuously write memos about your process as a sort of idea journal that you can continually draw from when writing your assignments. Required Readings: Dana, N. F. & Yendol-Hoppey, D. – Revisit Chapter 6 Assignments: For assignment details refer to the “Assignments for the Course” section in this syllabus or the submission link within Blackboard. Assignment: Initial Data Collection (Due Week 2 Sunday, 11:59 p.m.)

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MAE 241 – Homework 2 Page 1 of 2 MAE 241 – Spring 2019 – Homework 2 Administered 1/18/2019 – Due 11PM, Sunday 1/27/2019 to Gradescope Problem 1 The average water head (vertical height of water column) maintained in Hoover dam reservoir is about 500 ft. Assume water density of 62.43 lb/ft3. a. Determine the maximum pressure at the bottom of reservoir. b. Find the power generation potential of the water at that pressure if the discharge rate is 500×103 ft3/s. Problem 2 The Vestas V164 is one of the largest wind turbines in the world, with diameter of 164 m. If the theoretical limit on the capacity of a wind turbine is 1/3rd of its power generation potential, determine the capacity of the turbine when it is placed in a location where the average wind speed is 10 m/s. Assume air density as 1.25 kg/m3. Problem 3 An automobile has a mass of 1200 kg. What is its kinetic energy, in kJ, relative to the road when traveling at a velocity of 50 km/h? If the vehicle accelerates to 100 km/h, what is the change in kinetic energy, in kJ? Problem 4 A 5 kg brick is dropped from a height of 12 m onto a spring with a spring constant 8 kN/m. If the spring has a unstretched length of 0.5m, find (a) the shortest length the spring will be compressed before recoil, and (b) the final length of spring once the whole system becomes static. Problem 5 A piping installation is used to transport 20 L/s of water from a reservoir (location 1) to a point of use (location 2) 20 meters above. The absolute pressure of water at the inlet of the installation is 110 kPa; the gauge pressure measured right before the point of use is 552 kPa. Determine the power input required, in kW. Assume that because the piping at locations (1) and (2) have the same diameter the average velocities of water are equal and the density of water is 1000 kg/m3. Problem 6 A system receives 10 MJ in the form of heat in a process and it produced 4 MJ of work. The system velocity changes from 10 m/s to 25 m/s. For a 50 kg mass of the system, determine the change in internal energy of the system. MAE 241 – Homework 2 Page 2 of 2 Problem 7 On a recent energy assessment performed to an industrial facility in Tempe by a team of ASU’s Industrial Assessment Center, the team evaluated a boiler whose rated input was 6 MBTUH (6 million BTU per hour). After measuring the composition of flue gases it was apparent that the boiler was not be operating at its best operating point; this suspicion was validated by determining that the combustion efficiency was equal to 0.65. As corrective measure the boiler received a tune up that increased the combustion efficiency to 0.8. The boiler operates 48 weeks per year continuously while the plant is in production. Taking the cost of energy to be $13 per MBTU, determine: a. The annual energy cost. b. The annual cost savings as a result of tuning up the boiler. c. List the assumptions used in your computations. Problem 8 Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as 𝐹𝑏 = 𝜌𝑎𝑖𝑟𝑔𝑉𝑏𝑎𝑙, will push the balloon upward. (a) If the balloon has a diameter of 15 m and carries eight people, 75 kg each, determine the acceleration of the balloon when it is first released. (b) The change in air density with altitude can be approximated up to 10km using a linear function 𝜌𝑎𝑖𝑟 = 1.173 − 8 × 10−5ℎ where ℎ is the altitude in m. At what theoretical altitude the balloon will stop climbing upwards? Assume the density of air is 1.173 kg/m3 at ground level, and neglect the weight of the ropes and the cage. Problem 9 A differential manometer is used to measure pressure difference between two fluid systems. Two parallel pipes carrying freshwater and seawater are connected to each other by a double U-tube differential manometer, as shown in Figure. (a) Determine the pressure difference between the two pipelines if ℎ = 10 cm. (b) If the pressure difference between the pipes is doubled, what will be the difference in heights (ℎ) of mercury? Take the density of seawater at that location to be 1035 kg/m3, and the specific gravity of the oil is 0.72. Assume all fluids are incompressible.

MAE 241 – Homework 2 Page 1 of 2 MAE 241 – Spring 2019 – Homework 2 Administered 1/18/2019 – Due 11PM, Sunday 1/27/2019 to Gradescope Problem 1 The average water head (vertical height of water column) maintained in Hoover dam reservoir is about 500 ft. Assume water density of 62.43 lb/ft3. a. Determine the maximum pressure at the bottom of reservoir. b. Find the power generation potential of the water at that pressure if the discharge rate is 500×103 ft3/s. Problem 2 The Vestas V164 is one of the largest wind turbines in the world, with diameter of 164 m. If the theoretical limit on the capacity of a wind turbine is 1/3rd of its power generation potential, determine the capacity of the turbine when it is placed in a location where the average wind speed is 10 m/s. Assume air density as 1.25 kg/m3. Problem 3 An automobile has a mass of 1200 kg. What is its kinetic energy, in kJ, relative to the road when traveling at a velocity of 50 km/h? If the vehicle accelerates to 100 km/h, what is the change in kinetic energy, in kJ? Problem 4 A 5 kg brick is dropped from a height of 12 m onto a spring with a spring constant 8 kN/m. If the spring has a unstretched length of 0.5m, find (a) the shortest length the spring will be compressed before recoil, and (b) the final length of spring once the whole system becomes static. Problem 5 A piping installation is used to transport 20 L/s of water from a reservoir (location 1) to a point of use (location 2) 20 meters above. The absolute pressure of water at the inlet of the installation is 110 kPa; the gauge pressure measured right before the point of use is 552 kPa. Determine the power input required, in kW. Assume that because the piping at locations (1) and (2) have the same diameter the average velocities of water are equal and the density of water is 1000 kg/m3. Problem 6 A system receives 10 MJ in the form of heat in a process and it produced 4 MJ of work. The system velocity changes from 10 m/s to 25 m/s. For a 50 kg mass of the system, determine the change in internal energy of the system. MAE 241 – Homework 2 Page 2 of 2 Problem 7 On a recent energy assessment performed to an industrial facility in Tempe by a team of ASU’s Industrial Assessment Center, the team evaluated a boiler whose rated input was 6 MBTUH (6 million BTU per hour). After measuring the composition of flue gases it was apparent that the boiler was not be operating at its best operating point; this suspicion was validated by determining that the combustion efficiency was equal to 0.65. As corrective measure the boiler received a tune up that increased the combustion efficiency to 0.8. The boiler operates 48 weeks per year continuously while the plant is in production. Taking the cost of energy to be $13 per MBTU, determine: a. The annual energy cost. b. The annual cost savings as a result of tuning up the boiler. c. List the assumptions used in your computations. Problem 8 Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as 𝐹𝑏 = 𝜌𝑎𝑖𝑟𝑔𝑉𝑏𝑎𝑙, will push the balloon upward. (a) If the balloon has a diameter of 15 m and carries eight people, 75 kg each, determine the acceleration of the balloon when it is first released. (b) The change in air density with altitude can be approximated up to 10km using a linear function 𝜌𝑎𝑖𝑟 = 1.173 − 8 × 10−5ℎ where ℎ is the altitude in m. At what theoretical altitude the balloon will stop climbing upwards? Assume the density of air is 1.173 kg/m3 at ground level, and neglect the weight of the ropes and the cage. Problem 9 A differential manometer is used to measure pressure difference between two fluid systems. Two parallel pipes carrying freshwater and seawater are connected to each other by a double U-tube differential manometer, as shown in Figure. (a) Determine the pressure difference between the two pipelines if ℎ = 10 cm. (b) If the pressure difference between the pipes is doubled, what will be the difference in heights (ℎ) of mercury? Take the density of seawater at that location to be 1035 kg/m3, and the specific gravity of the oil is 0.72. Assume all fluids are incompressible.

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In case the body have to stay in lower temperature for extended time period (more than 1 hour), how does the body regulate its response?

In case the body have to stay in lower temperature for extended time period (more than 1 hour), how does the body regulate its response?

Arterioles transporting blood to external capillaries beneath the surface of … Read More...
Clinical support services are diverse in any modern healthcare facility, but they all share some common elements from a management perspective. Select any one of the CSS departments of interest to yourself, and provide examples of performance measures that you would use in managing the service. Be sure to address all seven dimensions of performance for your department. Department: Emergency Medical System and Patient Transportation

Clinical support services are diverse in any modern healthcare facility, but they all share some common elements from a management perspective. Select any one of the CSS departments of interest to yourself, and provide examples of performance measures that you would use in managing the service. Be sure to address all seven dimensions of performance for your department. Department: Emergency Medical System and Patient Transportation

The performance measures must be developed and applied continuously for … Read More...