Two forces are applied to a structure. F1 has a magnitude of 950 N and its direction makes angles of 45° with respect to the x-axis, 120° with respect to the y-axis, and 30° with respect to the z-axis. F2 is given as (10i -50j + 50k) N. Find the resultant vector 1 2 R  F  F in terms of its rectangular components, find the magnitude of R , and determine the unit vector in the direction of R ( R  ).

Two forces are applied to a structure. F1 has a magnitude of 950 N and its direction makes angles of 45° with respect to the x-axis, 120° with respect to the y-axis, and 30° with respect to the z-axis. F2 is given as (10i -50j + 50k) N. Find the resultant vector 1 2 R  F  F in terms of its rectangular components, find the magnitude of R , and determine the unit vector in the direction of R ( R  ).

Two forces are applied to a structure. F1 has a … 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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Chapter 9 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t)

Chapter 9 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t)

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Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 Assignment 4 – Noise and Correlation 1. If a signal is measured as 2.5 V and the noise is 28 mV (28 × 10−3 V), what is the SNR in dB? 2. A single sinusoidal signal is found with some noise. If the RMS value of the noise is 0.5 V and the SNR is 10 dB, what is the RMS amplitude of the sinusoid? 3. The file signal_noise.mat contains a variable x that consists of a 1.0-V peak sinusoidal signal buried in noise. What is the SNR for this signal and noise? Assume that the noise RMS is much greater than the signal RMS. Note: “signal_noise.mat” and other files used in these assignments can be downloaded from the content area of Brightspace, within the “Data Files for Exercises” folder. These files can be opened in Matlab by copying into the active folder and double-clicking on the file or using the Matlab load command using the format: load(‘signal_noise.mat’). To discover the variables within the files use the Matlab who command. 4. An 8-bit ADC converter that has an input range of ±5 V is used to convert a signal that ranges between ±2 V. What is the SNR of the input if the input noise equals the quantization noise of the converter? Hint: Refer to Equation below to find the quantization noise: 5. The file filter1.mat contains the spectrum of a fourth-order lowpass filter as variable x in dB. The file also contains the corresponding frequencies of x in variable freq. Plot the spectrum of this filter both as dB versus log frequency and as linear amplitude versus linear frequency. The frequency axis should range between 10 and 400 Hz in both plots. Hint: Use Equation below to convert: Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 6. Generate one cycle of the square wave similar to the one shown below in a 500-point MATLAB array. Determine the RMS value of this waveform. [Hint: When you take the square of the data array, be sure to use a period before the up arrow so that MATLAB does the squaring point-by-point (i.e., x.^2).]. 7. A resistor produces 10 μV noise (i.e., 10 × 10−6 V noise) when the room temperature is 310 K and the bandwidth is 1 kHz (i.e., 1000 Hz). What current noise would be produced by this resistor? 8. A 3-ma current flows through both a diode (i.e., a semiconductor) and a 20,000-Ω (i.e., 20-kΩ) resistor. What is the net current noise, in? Assume a bandwidth of 1 kHz (i.e., 1 × 103 Hz). Which of the two components is responsible for producing the most noise? 9. Determine if the two signals, x and y, in file correl1.mat are correlated by checking the angle between them. 10. Modify the approach used in Practice Problem 3 to find the angle between short signals: Do not attempt to plot these vectors as it would require a 6-dimensional plot!

Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 Assignment 4 – Noise and Correlation 1. If a signal is measured as 2.5 V and the noise is 28 mV (28 × 10−3 V), what is the SNR in dB? 2. A single sinusoidal signal is found with some noise. If the RMS value of the noise is 0.5 V and the SNR is 10 dB, what is the RMS amplitude of the sinusoid? 3. The file signal_noise.mat contains a variable x that consists of a 1.0-V peak sinusoidal signal buried in noise. What is the SNR for this signal and noise? Assume that the noise RMS is much greater than the signal RMS. Note: “signal_noise.mat” and other files used in these assignments can be downloaded from the content area of Brightspace, within the “Data Files for Exercises” folder. These files can be opened in Matlab by copying into the active folder and double-clicking on the file or using the Matlab load command using the format: load(‘signal_noise.mat’). To discover the variables within the files use the Matlab who command. 4. An 8-bit ADC converter that has an input range of ±5 V is used to convert a signal that ranges between ±2 V. What is the SNR of the input if the input noise equals the quantization noise of the converter? Hint: Refer to Equation below to find the quantization noise: 5. The file filter1.mat contains the spectrum of a fourth-order lowpass filter as variable x in dB. The file also contains the corresponding frequencies of x in variable freq. Plot the spectrum of this filter both as dB versus log frequency and as linear amplitude versus linear frequency. The frequency axis should range between 10 and 400 Hz in both plots. Hint: Use Equation below to convert: Biomedical Signal and Image Processing (4800_420_001) Assigned on September 12th, 2017 6. Generate one cycle of the square wave similar to the one shown below in a 500-point MATLAB array. Determine the RMS value of this waveform. [Hint: When you take the square of the data array, be sure to use a period before the up arrow so that MATLAB does the squaring point-by-point (i.e., x.^2).]. 7. A resistor produces 10 μV noise (i.e., 10 × 10−6 V noise) when the room temperature is 310 K and the bandwidth is 1 kHz (i.e., 1000 Hz). What current noise would be produced by this resistor? 8. A 3-ma current flows through both a diode (i.e., a semiconductor) and a 20,000-Ω (i.e., 20-kΩ) resistor. What is the net current noise, in? Assume a bandwidth of 1 kHz (i.e., 1 × 103 Hz). Which of the two components is responsible for producing the most noise? 9. Determine if the two signals, x and y, in file correl1.mat are correlated by checking the angle between them. 10. Modify the approach used in Practice Problem 3 to find the angle between short signals: Do not attempt to plot these vectors as it would require a 6-dimensional plot!

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PHET ElectroMagnetism Key to this Document Instructions are in black. Experimental questions that you need to solve through experimentation with an online animation are in green highlighted. Important instructions are in red highlighted. Items that need a response from you are in yellow highlighted. Please put your answers to this activity in RED. Part I- Comparing Permanent Magnets and Electromagnets: 1. Select the simulation “Magnets and Electromagnets.” It is at this link: http://phet.colorado.edu/new/simulations/sims.php?sim=Magnets_and_Electromagnets 2. Move the compass slowly along a semicircular path above the bar magnet until you’ve put it on the opposite side of the bar magnet. Describe what happens to the compass needle. 3. Move the compass along a semicircular path below the bar magnet until you’ve put it on the opposite side of the bar magnet. Describe what happens to the compass needle. 4. What do you suppose the compass needles drawn all over the screen tell you? 5. Use page 10 in your book to look up what it looks like when scientists use a drawing to represent a magnetic field. Describe the field around a bar magnet here. 6. Put the compass to the left or right of the magnet. Click “flip polarity” and notice what happens to the compass. Using the compass needle as your observation tool, describe the effect that flipping the poles of the magnet has on the magnetic field. 7. Click on the electromagnet tab along the top of the simulation window. Place the compass on the left side of the coil so that the compass center lies along the axis of the coil. <--like this 8. Move the compass along a semicircular path above the coil until you’ve put it on the opposite side of the coil. Then do the same below the coil. Notice what happens to the compass needle. Compare this answer to the answer you got to Number 2 and 3. 9. Compare the shape of the magnetic field of a bar magnet to the magnetic field of an electromagnet. 10. Use the voltage slider to change the direction of the current and investigate the shape of the magnetic field the coil using the compass after you’ve let the compass stabilize. Summarize, the effect that the direction of current has on the shape of the magnetic field around an electrified coil of wires. 11. What happens to the current in the coil when you set the voltage of the battery to zero? 12. What happens to the magnetic field around the coil when you set the voltage of the battery to zero? Part II – Investigating relationships- No Answers are written on this document after this point. All three data tables, graphs and conclusion statements go on the Google Spreadsheet that you can download from Ms. Pogge’s website. Experimental Question #1: How does distance affect the strength of the magnetic field around an electromagnet? 1. Using the Electromagnet simulation, click on “Show Field Meter.” 2. Set the battery voltage to 10V where the positive is on the right of the battery (slide the switch all the way to the right). 3. Magnetic field strength (symbol B on the top line of the meter) is measured in gauss (G). You’ll only need to record the value on the top line of the Field Meter. 4. Position zero will be right on top of the coil. Negative number positions will be to the left and positive number positions to the right of the coil. 5. Move the field meter one compass needle to the right and record the value of B at position 1. 6. This data table below will be used to help you fill in the first spreadsheet you downloaded from Ms. Pogge’s website. You will end up with 3 data tables, 3 graphs and 3 conclusion statements in your document, one for each mini-experiment you are doing. a. NOTE: Be sure to take all of your values along the horizontal axis of the coil. You’ll know you’re on the axis because the B-y measurement of the magnetic field is zero along the axis. Compass position (no units) Magnetic Field Strength ( )<--Fill in units! -5 (5 needles to the left of coil) Don’t fill in the table here...do it on the Google Spreadsheet you downloaded -4 -3 -2 -1 0 (middle of coil) 1 2 3 4 5 (5 needles to right of coil) 7. In your Google Spreadsheet: Graph the compass position on the horizontal (x) axis and magnetic field magnitude on the vertical (y) axis. 8. Make sure to label the axes and title the graph. Share this spreadsheet with your teacher. 9. Analyze your graph to discover how the two variables are related, and report the relationship between magnetic field strength and position using 1-3 complete sentences. Experimental Question #2: How does the number of coils affect the strength of the magnetic field around an electromagnet? Design an experiment to test how field strength varies with the number of coils. Enter your data, graph your results and write your conclusion statement on the Google Spreadsheet. Experimental Question #3: How does the amount of current affect the strength of the magnetic field around an electromagnet? Design an experiment to test how field strength varies with the Current. (Recall that voltage is directly proportional to current….Ohm’s Law.) Enter your data, graph your results and write your conclusion statement on the Google Spreadsheet.

PHET ElectroMagnetism Key to this Document Instructions are in black. Experimental questions that you need to solve through experimentation with an online animation are in green highlighted. Important instructions are in red highlighted. Items that need a response from you are in yellow highlighted. Please put your answers to this activity in RED. Part I- Comparing Permanent Magnets and Electromagnets: 1. Select the simulation “Magnets and Electromagnets.” It is at this link: http://phet.colorado.edu/new/simulations/sims.php?sim=Magnets_and_Electromagnets 2. Move the compass slowly along a semicircular path above the bar magnet until you’ve put it on the opposite side of the bar magnet. Describe what happens to the compass needle. 3. Move the compass along a semicircular path below the bar magnet until you’ve put it on the opposite side of the bar magnet. Describe what happens to the compass needle. 4. What do you suppose the compass needles drawn all over the screen tell you? 5. Use page 10 in your book to look up what it looks like when scientists use a drawing to represent a magnetic field. Describe the field around a bar magnet here. 6. Put the compass to the left or right of the magnet. Click “flip polarity” and notice what happens to the compass. Using the compass needle as your observation tool, describe the effect that flipping the poles of the magnet has on the magnetic field. 7. Click on the electromagnet tab along the top of the simulation window. Place the compass on the left side of the coil so that the compass center lies along the axis of the coil. <--like this 8. Move the compass along a semicircular path above the coil until you’ve put it on the opposite side of the coil. Then do the same below the coil. Notice what happens to the compass needle. Compare this answer to the answer you got to Number 2 and 3. 9. Compare the shape of the magnetic field of a bar magnet to the magnetic field of an electromagnet. 10. Use the voltage slider to change the direction of the current and investigate the shape of the magnetic field the coil using the compass after you’ve let the compass stabilize. Summarize, the effect that the direction of current has on the shape of the magnetic field around an electrified coil of wires. 11. What happens to the current in the coil when you set the voltage of the battery to zero? 12. What happens to the magnetic field around the coil when you set the voltage of the battery to zero? Part II – Investigating relationships- No Answers are written on this document after this point. All three data tables, graphs and conclusion statements go on the Google Spreadsheet that you can download from Ms. Pogge’s website. Experimental Question #1: How does distance affect the strength of the magnetic field around an electromagnet? 1. Using the Electromagnet simulation, click on “Show Field Meter.” 2. Set the battery voltage to 10V where the positive is on the right of the battery (slide the switch all the way to the right). 3. Magnetic field strength (symbol B on the top line of the meter) is measured in gauss (G). You’ll only need to record the value on the top line of the Field Meter. 4. Position zero will be right on top of the coil. Negative number positions will be to the left and positive number positions to the right of the coil. 5. Move the field meter one compass needle to the right and record the value of B at position 1. 6. This data table below will be used to help you fill in the first spreadsheet you downloaded from Ms. Pogge’s website. You will end up with 3 data tables, 3 graphs and 3 conclusion statements in your document, one for each mini-experiment you are doing. a. NOTE: Be sure to take all of your values along the horizontal axis of the coil. You’ll know you’re on the axis because the B-y measurement of the magnetic field is zero along the axis. Compass position (no units) Magnetic Field Strength ( )<--Fill in units! -5 (5 needles to the left of coil) Don’t fill in the table here...do it on the Google Spreadsheet you downloaded -4 -3 -2 -1 0 (middle of coil) 1 2 3 4 5 (5 needles to right of coil) 7. In your Google Spreadsheet: Graph the compass position on the horizontal (x) axis and magnetic field magnitude on the vertical (y) axis. 8. Make sure to label the axes and title the graph. Share this spreadsheet with your teacher. 9. Analyze your graph to discover how the two variables are related, and report the relationship between magnetic field strength and position using 1-3 complete sentences. Experimental Question #2: How does the number of coils affect the strength of the magnetic field around an electromagnet? Design an experiment to test how field strength varies with the number of coils. Enter your data, graph your results and write your conclusion statement on the Google Spreadsheet. Experimental Question #3: How does the amount of current affect the strength of the magnetic field around an electromagnet? Design an experiment to test how field strength varies with the Current. (Recall that voltage is directly proportional to current….Ohm’s Law.) Enter your data, graph your results and write your conclusion statement on the Google Spreadsheet.

Assignment 3 Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 2.68 As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 36.0 . Part A How fast is the watermelon going when it passes Superman? Express your answer with the appropriate units. ANSWER: Correct Problem 2.63 A motorist is driving at when she sees that a traffic light ahead has just turned red. She knows that this light stays red for , and she wants to reach the light just as it turns green again. It takes her to step on the brakes and begin slowing. Part A What is her speed as she reaches the light at the instant it turns green? Express your answer with the appropriate units. ANSWER: m/s 72.0 ms 20 m/s 200 m 15 s 1.0 s 5.71 ms Correct Conceptual Question 4.1 Part A At this instant, is the particle in the figurespeeding up, slowing down, or traveling at constant speed? ANSWER: Correct Part B Is this particle curving to the right, curving to the left, or traveling straight? Speeding up Slowing down Traveling at constant speed ANSWER: Correct Conceptual Question 4.2 Part A At this instant, is the particle in the following figure speeding up, slowing down, or traveling at constant speed? ANSWER: Curving to the right Curving to the left Traveling straight Correct Part B Is this particle curving upward, curving downward, or traveling straight? ANSWER: Correct Problem 4.8 A particle’s trajectory is described by and , where is in s. Part A What is the particle’s speed at ? ANSWER: The particle is speeding up. The particle is slowing down. The particle is traveling at constant speed. The particle is curving upward. The particle is curving downward. The particle is traveling straight. x = ( 1 −2 ) m 2 t3 t2 y = ( 1 −2t) m 2 t2 t t = 0 s v = 2 m/s Correct Part B What is the particle’s speed at ? Express your answer using two significant figures. ANSWER: Correct Part C What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: Correct Part D What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: t = 5.0s v = 18 m/s t = 0 s  = -90  counterclockwise from the +x axis. t = 5.0s  = 9.7  counterclockwise from the +x axis. Correct Problem 4.9 A rocket-powered hockey puck moves on a horizontal frictionless table. The figure shows the graph of and the figure shows the graph of , the x- and y-components of the puck’s velocity, respectively. The puck starts at the origin. Part A In which direction is the puck moving at = 3 ? Give your answer as an angle from the x-axis. Express your answer using two significant figures. ANSWER: Correct Part B vx vy t s = 51   above the x-axis How far from the origin is the puck at 5 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.13 A rifle is aimed horizontally at a target 51.0 away. The bullet hits the target 1.50 below the aim point. You may want to review ( pages 91 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A What was the bullet’s flight time? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the bullet’s trajectory, including where it leaves the gun and where it hits the target. You can assume that the gun was held parallel to the ground. Label the distances given in the problem. Choose an x-y coordinate system, making sure to label the origin. It is conventional to have x in the horizontal direction and y in the vertical direction. What is the y coordinate when the bullet leaves the gun? What is the y coordinate when it hits the target? What is the initial velocity in the y direction? What is the acceleration in the y direction? What is the equation that describes the motion in the vertical y direction as a function of time? Can you use the equation for to determine the time of flight? Why was it not necessary to include the motion in the x direction? s s = 180 cm m cm y(t) y(t) ANSWER: Correct Part B What was the bullet’s speed as it left the barrel? Express your answer with the appropriate units. Hint 1. How to approach the problem In the coordinate system introduced in Part A, what are the x coordinates when the bullet leaves the gun and when it hits the target? Is there any acceleration in the x direction? What is the equation that describes the motion in the horizontal x direction as a function of time? Can you use the equation for to determine the initial velocity? ANSWER: Correct Introduction to Projectile Motion Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really no different than the one-dimensional motions that you have already studied. One of the most often used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more managable, it is common to break up such a quantity into its x component and its y component . 5.53×10−2 s x(t) x(t) 922 ms v vx vy Consider a particle with initial velocity that has magnitude 12.0 and is directed 60.0 above the negative x axis. Part A What is the x component of ? Express your answer in meters per second. ANSWER: Correct Part B What is the y component of ? Express your answer in meters per second. ANSWER: Correct Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C Look at this applet. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x-component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle’s shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: v m/s degrees vx v vx = -6.00 m/s vy v vy = 10.4 m/s Correct As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation . Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation . Now, consider this applet. Two balls are simultaneously dropped from a height of 5.0 . Part D How long does it take for the balls to reach the ground? Use 10 for the magnitude of the acceleration due to gravity. Express your answer in seconds to two significant figures. Hint 1. How to approach the problem The balls are released from rest at a height of 5.0 at time . Using these numbers and basic kinematics, you can determine the amount of time it takes for the balls to reach the ground. ANSWER: Correct This situation, which you have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion where the horizontal component of the initial velocity is zero. Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity. g = 10 m/s2 y = y0 + v0 t + (1/2)at2 x = x0 + v0 t m tg m/s2 m t = 0 s tg = 1.0 s Part E Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Remember that the two balls are released, starting a horizontal distance of 3.0 apart. Express your answer in meters per second to two significant figures. Hint 1. How to approach the problem Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that, in a time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is . ANSWER: Correct You can adjust the horizontal speeds in this applet. Notice that regardless of what horizontal speeds you give to the balls, they continue to move vertically in the same way (i.e., they are at the same y coordinate at the same time). Problem 4.12 A ball thrown horizontally at 27 travels a horizontal distance of 49 before hitting the ground. Part A From what height was the ball thrown? Express your answer using two significant figures with the appropriate units. ANSWER: vx m vx tg = 1.0 s m x0 = 0.0 m vx = 3.0 m/s m/s m h = 16 m Correct Enhanced EOC: Problem 4.20 The figure shows the angular-velocity-versus-time graph for a particle moving in a circle. You may want to review ( page ) . For help with math skills, you may want to review: The Definite Integral Part A How many revolutions does the object make during the first 3.5 ? Express your answer using two significant figures. You did not open hints for this part. ANSWER: s n = Incorrect; Try Again Problem 4.26 To withstand “g-forces” of up to 10 g’s, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a “human centrifuge.” 10 g’s is an acceleration of . Part A If the length of the centrifuge arm is 10.0 , at what speed is the rider moving when she experiences 10 g’s? Express your answer with the appropriate units. ANSWER: Correct Problem 4.28 Your roommate is working on his bicycle and has the bike upside down. He spins the 60.0 -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. Part A What is the pebble’s speed? Express your answer with the appropriate units. ANSWER: Correct 98 m/s2 m 31.3 ms cm 5.65 ms Part B What is the pebble’s acceleration? Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.43 On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 13 at an angle 50 above the horizontal. You may want to review ( pages 90 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A How much farther did the ball travel on the moon than it would have on earth? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the golf ball, showing its starting and ending points. Choose a coordinate system, and label the origin. It is conventional to let x be the horizontal direction and y the vertical direction. What is the initial velocity in the x and y directions? What is the acceleration in the x and y directions on the moon and on the earth? What are the equations for and as a function of time, and , respectively? What is the y coordinate when the golf ball hits the ground? Can you use this information to determine the time of flight on the moon and on the earth? 107 m s2 m/s  x y x(t) y(t) Once you have the time of flight, how can you use the equation to determine the total distance traveled? Compare the distance traveled on the moon to the distance traveled on the earth . ANSWER: Correct Part B For how much more time was the ball in flight? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the equation describing as a function of time? What is the initial x component of the ball’s velocity? How are the initial x component of the ball’s velocity and the distance traveled related to the time of flight? What is the difference between the time of flight on the moon and on earth? ANSWER: Correct Problem 4.42 In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 at a 40.0 angle from the horizontal. The shot leaves her hand at a height of 1.8 above the ground. x(t) L = 85 m x(t) x t = 10 s m/s  m Part A How far does the shot travel? Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part B Repeat the calculation of part (a) for angles of 42.5 , 45.0 , and 47.5 . Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part C Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part D x = 16.36 m    x(42.5 ) = 16.39 m x(45.0 ) = 16.31 m Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part E At what angle of release does she throw the farthest? ANSWER: Correct Problem 4.44 A ball is thrown toward a cliff of height with a speed of 32 and an angle of 60 above horizontal. It lands on the edge of the cliff 3.2 later. Part A How high is the cliff? Express your answer to two significant figures and include the appropriate units. ANSWER: x(47.5 ) = 16.13 m 40.0 42.5 45.0 47.5 h m/s  s h = 39 m Answer Requested Part B What was the maximum height of the ball? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the ball’s impact speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 4.58 A typical laboratory centrifuge rotates at 3600 . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 from the axis of rotation? Express your answer with the appropriate units. hmax = 39 m v = 16 ms rpm cm ANSWER: Correct Part B For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 and stopped in a 1.7-ms-long encounter with a hard floor? Express your answer with the appropriate units. ANSWER: Correct Problem 4.62 Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The radius of the earth is , and the altitude of a geosynchronous orbit is ( 22000 miles). Part A What is the speed of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct a = 1.42×104 m s2 m a = 2610 m s2 6.37 × 106m 3.58 × 107m  v = 3070 ms Part B What is the magnitude of the acceleration of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 89.5%. You received 103.82 out of a possible total of 116 points. a = 0.223 m s2

Assignment 3 Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 2.68 As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 36.0 . Part A How fast is the watermelon going when it passes Superman? Express your answer with the appropriate units. ANSWER: Correct Problem 2.63 A motorist is driving at when she sees that a traffic light ahead has just turned red. She knows that this light stays red for , and she wants to reach the light just as it turns green again. It takes her to step on the brakes and begin slowing. Part A What is her speed as she reaches the light at the instant it turns green? Express your answer with the appropriate units. ANSWER: m/s 72.0 ms 20 m/s 200 m 15 s 1.0 s 5.71 ms Correct Conceptual Question 4.1 Part A At this instant, is the particle in the figurespeeding up, slowing down, or traveling at constant speed? ANSWER: Correct Part B Is this particle curving to the right, curving to the left, or traveling straight? Speeding up Slowing down Traveling at constant speed ANSWER: Correct Conceptual Question 4.2 Part A At this instant, is the particle in the following figure speeding up, slowing down, or traveling at constant speed? ANSWER: Curving to the right Curving to the left Traveling straight Correct Part B Is this particle curving upward, curving downward, or traveling straight? ANSWER: Correct Problem 4.8 A particle’s trajectory is described by and , where is in s. Part A What is the particle’s speed at ? ANSWER: The particle is speeding up. The particle is slowing down. The particle is traveling at constant speed. The particle is curving upward. The particle is curving downward. The particle is traveling straight. x = ( 1 −2 ) m 2 t3 t2 y = ( 1 −2t) m 2 t2 t t = 0 s v = 2 m/s Correct Part B What is the particle’s speed at ? Express your answer using two significant figures. ANSWER: Correct Part C What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: Correct Part D What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: t = 5.0s v = 18 m/s t = 0 s  = -90  counterclockwise from the +x axis. t = 5.0s  = 9.7  counterclockwise from the +x axis. Correct Problem 4.9 A rocket-powered hockey puck moves on a horizontal frictionless table. The figure shows the graph of and the figure shows the graph of , the x- and y-components of the puck’s velocity, respectively. The puck starts at the origin. Part A In which direction is the puck moving at = 3 ? Give your answer as an angle from the x-axis. Express your answer using two significant figures. ANSWER: Correct Part B vx vy t s = 51   above the x-axis How far from the origin is the puck at 5 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.13 A rifle is aimed horizontally at a target 51.0 away. The bullet hits the target 1.50 below the aim point. You may want to review ( pages 91 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A What was the bullet’s flight time? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the bullet’s trajectory, including where it leaves the gun and where it hits the target. You can assume that the gun was held parallel to the ground. Label the distances given in the problem. Choose an x-y coordinate system, making sure to label the origin. It is conventional to have x in the horizontal direction and y in the vertical direction. What is the y coordinate when the bullet leaves the gun? What is the y coordinate when it hits the target? What is the initial velocity in the y direction? What is the acceleration in the y direction? What is the equation that describes the motion in the vertical y direction as a function of time? Can you use the equation for to determine the time of flight? Why was it not necessary to include the motion in the x direction? s s = 180 cm m cm y(t) y(t) ANSWER: Correct Part B What was the bullet’s speed as it left the barrel? Express your answer with the appropriate units. Hint 1. How to approach the problem In the coordinate system introduced in Part A, what are the x coordinates when the bullet leaves the gun and when it hits the target? Is there any acceleration in the x direction? What is the equation that describes the motion in the horizontal x direction as a function of time? Can you use the equation for to determine the initial velocity? ANSWER: Correct Introduction to Projectile Motion Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really no different than the one-dimensional motions that you have already studied. One of the most often used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more managable, it is common to break up such a quantity into its x component and its y component . 5.53×10−2 s x(t) x(t) 922 ms v vx vy Consider a particle with initial velocity that has magnitude 12.0 and is directed 60.0 above the negative x axis. Part A What is the x component of ? Express your answer in meters per second. ANSWER: Correct Part B What is the y component of ? Express your answer in meters per second. ANSWER: Correct Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C Look at this applet. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x-component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle’s shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: v m/s degrees vx v vx = -6.00 m/s vy v vy = 10.4 m/s Correct As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation . Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation . Now, consider this applet. Two balls are simultaneously dropped from a height of 5.0 . Part D How long does it take for the balls to reach the ground? Use 10 for the magnitude of the acceleration due to gravity. Express your answer in seconds to two significant figures. Hint 1. How to approach the problem The balls are released from rest at a height of 5.0 at time . Using these numbers and basic kinematics, you can determine the amount of time it takes for the balls to reach the ground. ANSWER: Correct This situation, which you have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion where the horizontal component of the initial velocity is zero. Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity. g = 10 m/s2 y = y0 + v0 t + (1/2)at2 x = x0 + v0 t m tg m/s2 m t = 0 s tg = 1.0 s Part E Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Remember that the two balls are released, starting a horizontal distance of 3.0 apart. Express your answer in meters per second to two significant figures. Hint 1. How to approach the problem Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that, in a time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is . ANSWER: Correct You can adjust the horizontal speeds in this applet. Notice that regardless of what horizontal speeds you give to the balls, they continue to move vertically in the same way (i.e., they are at the same y coordinate at the same time). Problem 4.12 A ball thrown horizontally at 27 travels a horizontal distance of 49 before hitting the ground. Part A From what height was the ball thrown? Express your answer using two significant figures with the appropriate units. ANSWER: vx m vx tg = 1.0 s m x0 = 0.0 m vx = 3.0 m/s m/s m h = 16 m Correct Enhanced EOC: Problem 4.20 The figure shows the angular-velocity-versus-time graph for a particle moving in a circle. You may want to review ( page ) . For help with math skills, you may want to review: The Definite Integral Part A How many revolutions does the object make during the first 3.5 ? Express your answer using two significant figures. You did not open hints for this part. ANSWER: s n = Incorrect; Try Again Problem 4.26 To withstand “g-forces” of up to 10 g’s, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a “human centrifuge.” 10 g’s is an acceleration of . Part A If the length of the centrifuge arm is 10.0 , at what speed is the rider moving when she experiences 10 g’s? Express your answer with the appropriate units. ANSWER: Correct Problem 4.28 Your roommate is working on his bicycle and has the bike upside down. He spins the 60.0 -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. Part A What is the pebble’s speed? Express your answer with the appropriate units. ANSWER: Correct 98 m/s2 m 31.3 ms cm 5.65 ms Part B What is the pebble’s acceleration? Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.43 On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 13 at an angle 50 above the horizontal. You may want to review ( pages 90 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A How much farther did the ball travel on the moon than it would have on earth? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the golf ball, showing its starting and ending points. Choose a coordinate system, and label the origin. It is conventional to let x be the horizontal direction and y the vertical direction. What is the initial velocity in the x and y directions? What is the acceleration in the x and y directions on the moon and on the earth? What are the equations for and as a function of time, and , respectively? What is the y coordinate when the golf ball hits the ground? Can you use this information to determine the time of flight on the moon and on the earth? 107 m s2 m/s  x y x(t) y(t) Once you have the time of flight, how can you use the equation to determine the total distance traveled? Compare the distance traveled on the moon to the distance traveled on the earth . ANSWER: Correct Part B For how much more time was the ball in flight? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the equation describing as a function of time? What is the initial x component of the ball’s velocity? How are the initial x component of the ball’s velocity and the distance traveled related to the time of flight? What is the difference between the time of flight on the moon and on earth? ANSWER: Correct Problem 4.42 In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 at a 40.0 angle from the horizontal. The shot leaves her hand at a height of 1.8 above the ground. x(t) L = 85 m x(t) x t = 10 s m/s  m Part A How far does the shot travel? Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part B Repeat the calculation of part (a) for angles of 42.5 , 45.0 , and 47.5 . Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part C Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part D x = 16.36 m    x(42.5 ) = 16.39 m x(45.0 ) = 16.31 m Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part E At what angle of release does she throw the farthest? ANSWER: Correct Problem 4.44 A ball is thrown toward a cliff of height with a speed of 32 and an angle of 60 above horizontal. It lands on the edge of the cliff 3.2 later. Part A How high is the cliff? Express your answer to two significant figures and include the appropriate units. ANSWER: x(47.5 ) = 16.13 m 40.0 42.5 45.0 47.5 h m/s  s h = 39 m Answer Requested Part B What was the maximum height of the ball? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the ball’s impact speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 4.58 A typical laboratory centrifuge rotates at 3600 . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 from the axis of rotation? Express your answer with the appropriate units. hmax = 39 m v = 16 ms rpm cm ANSWER: Correct Part B For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 and stopped in a 1.7-ms-long encounter with a hard floor? Express your answer with the appropriate units. ANSWER: Correct Problem 4.62 Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The radius of the earth is , and the altitude of a geosynchronous orbit is ( 22000 miles). Part A What is the speed of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct a = 1.42×104 m s2 m a = 2610 m s2 6.37 × 106m 3.58 × 107m  v = 3070 ms Part B What is the magnitude of the acceleration of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 89.5%. You received 103.82 out of a possible total of 116 points. a = 0.223 m s2

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Assignment 10 Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 12.3 Part A The figure shows three rotating disks, all of equal mass. Rank in order, from largest to smallest, their rotational kinetic energies to . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 21 4/11/2014 1:13 PM Incorrect; Try Again Conceptual Question 12.6 You have two steel solid spheres. Sphere 2 has twice the radius of sphere 1. Part A By what factor does the moment of inertia of sphere 2 exceed the moment of inertia of sphere 1? ANSWER: Correct Problem 12.2 A high-speed drill reaches 2500 in 0.59 . Part A What is the drill’s angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B Through how many revolutions does it turn during this first 0.59 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Constant Angular Acceleration in the Kitchen = 32 = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 21 4/11/2014 1:13 PM Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. Part A What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in degrees per second per second. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). ± A Spinning Electric Fan An electric fan is turned off, and its angular velocity decreases uniformly from 540 to 250 in a time interval of length 4.40 . Part A Find the angular acceleration in revolutions per second per second. You did not open hints for this part. ANSWER: Part B Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A. = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 21 4/11/2014 1:13 PM You did not open hints for this part. ANSWER: Part C How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A? You did not open hints for this part. ANSWER: Problem 12.8 A 100 ball and a 230 ball are connected by a 34- -long, massless, rigid rod. The balls rotate about their center of mass at 130 . Part A What is the speed of the 100 ball? Express your answer to two significant figures and include the appropriate units. ANSWER: Problem 12.10 A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through its center with 0.200 of kinetic energy. Part A What is the speed of a point on the rim? = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 21 4/11/2014 1:13 PM Express your answer with the appropriate units. ANSWER: Problem 12.12 A drum major twirls a 95- -long, 470 baton about its center of mass at 150 . Part A What is the baton’s rotational kinetic energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord’s weight can be ignored. Part A Which of the following statements correctly describes the system shown in the figure? Check all that apply. = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 21 4/11/2014 1:13 PM You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Problem 12.18 Part A In the figure , what is the magnitude of net torque about the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B What is the direction of net torque about the axle? ANSWER: The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero. = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 21 4/11/2014 1:13 PM Problem 12.22 An athlete at the gym holds a 3.5 steel ball in his hand. His arm is 78 long and has a mass of 3.6 . Assume the center of mass of the arm is at the geometrical center of the arm. Part A What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B What is the magnitude of the torque about his shoulder if he holds his arm straight, but below horizontal? Express your answer to two significant figures and include the appropriate units. ANSWER: Parallel Axis Theorem The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center of mass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that passes through point p, and is the mass of the object. Part A Suppose a uniform slender rod has length and mass . The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by . Find , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express in terms of and . Use fractions rather than decimal numbers in your answer. Clockwise Counterclockwise = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 21 4/11/2014 1:13 PM You did not open hints for this part. ANSWER: Part B Now consider a cube of mass with edges of length . The moment of inertia of the cube about an axis through its center of mass and perpendicular to one of its faces is given by . Find , the moment of inertia about an axis p through one of the edges of the cube Express in terms of and . Use fractions rather than decimal numbers in your answer. You did not open hints for this part. ANSWER: Problem 12.26 Starting from rest, a 12- -diameter compact disk takes 2.9 to reach its operating angular velocity of 2000 . Assume that the angular acceleration is constant. The disk’s moment of inertia is . Part A How much torque is applied to the disk? Express your answer to two significant figures and include the appropriate units. = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 21 4/11/2014 1:13 PM ANSWER: Part B How many revolutions does it make before reaching full speed? Express your answer using two significant figures. ANSWER: Problem 12.23 An object’s moment of inertia is 2.20 . Its angular velocity is increasing at the rate of 3.70 . Part A What is the total torque on the object? ANSWER: Problem 12.31 A 5.1 cat and a 2.5 bowl of tuna fish are at opposite ends of the 4.0- -long seesaw. = = rev Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 9 of 21 4/11/2014 1:13 PM Part A How far to the left of the pivot must a 3.8 cat stand to keep the seesaw balanced? Express your answer to two significant figures and include the appropriate units. ANSWER: Static Equilibrium of the Arm You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass , length and its center of mass is a distance from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance from the scapula. The deltoid muscle makes an angle of with the horizontal, as shown. Use throughout the problem. Part A Find the tension in the deltoid muscle. Express the tension in newtons, to the nearest integer. You did not open hints for this part. ANSWER: = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 10 of 21 4/11/2014 1:13 PM Part B Using the conditions for static equilibrium, find the magnitude of the vertical component of the force exerted by the scapula on the humerus (where the humerus attaches to the rest of the body). Express your answer in newtons, to the nearest integer. You did not open hints for this part. ANSWER: Part C Now find the magnitude of the horizontal component of the force exerted by the scapula on the humerus. Express your answer in newtons, to the nearest integer. ANSWER: ± Moments around a Rod A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored for this problem. There are three forces that are applied to the rod at different points and angles: , , and . Note that the dimensions of the bent rod are in centimeters in the figure, although the answers are requested in SI units (kilograms, meters, seconds). = N = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 11 of 21 4/11/2014 1:13 PM Part A If and , what does the magnitude of have to be for there to be rotational equilibrium? Answer numerically in newtons to two significant figures. You did not open hints for this part. ANSWER: Part B If the L-shaped rod has a moment of inertia , , , and again , how long a time would it take for the object to move through ( /4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures. You did not open hints for this part. ANSWER: Part C Now consider the situation in which and , but now a force with nonzero magnitude is acting on the rod. What does have to be to obtain equilibrium? Give a numerical answer, without trigonometric functions, in newtons, to two significant figures. You did not open hints for this part. ANSWER: = N = s = N Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 12 of 21 4/11/2014 1:13 PM Problem 12.32 A car tire is 55.0 in diameter. The car is traveling at a speed of 24.0 . Part A What is the tire’s rotation frequency, in rpm? Express your answer to three significant figures and include the appropriate units. ANSWER: Part B What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropriate units. ANSWER: Part C What is the speed of a point at the bottom edge of the tire? Express your answer as an integer and include the appropriate units. ANSWER: Problem 12.33 A 460 , 8.00-cm-diameter solid cylinder rolls across the floor at 1.30 . Part A What is the can’s kinetic energy? Express your answer with the appropriate units. Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 13 of 21 4/11/2014 1:13 PM ANSWER: Problem 12.45 Part A What is the magnitude of the angular momentum of the 780 rotating bar in the figure ? ANSWER: Part B What is the direction of the angular momentum of the bar ? ANSWER: Problem 12.46 into the page out of the page Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 14 of 21 4/11/2014 1:13 PM Part A What is the magnitude of the angular momentum of the 2.20 , 4.60-cm-diameter rotating disk in the figure ? ANSWER: Part B What is its direction? ANSWER: Problem 12.60 A 3.0- -long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.46. x direction -x direction y direction -y direction z direction -z direction Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 15 of 21 4/11/2014 1:13 PM Part A What is the minimum angle the ladder can make with the floor without slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Problem 12.61 The 3.0- -long, 90 rigid beam in the following figure is supported at each end. An 70 student stands 2.0 from support 1. Part A How much upward force does the support 1 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 16 of 21 4/11/2014 1:13 PM Part B How much upward force does the support 2 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Enhanced EOC: Problem 12.63 A 44 , 5.5- -long beam is supported, but not attached to, the two posts in the figure . A 22 boy starts walking along the beam. You may want to review ( pages 330 – 334) . For help with math skills, you may want to review: The Vector Cross Product Part A How close can he get to the right end of the beam without it falling over? Express your answer to two significant figures and include the appropriate units. You did not open hints for this part. ANSWER: = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 17 of 21 4/11/2014 1:13 PM Problem 12.68 Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel’s energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.6 diameter and a mass of 270 . Its maximum angular velocity is 1500 . Part A A motor spins up the flywheel with a constant torque of 54 . How long does it take the flywheel to reach top speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B How much energy is stored in the flywheel? Express your answer to two significant figures and include the appropriate units. ANSWER: Part C The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.2 . What is the average power delivered to the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: = = = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 18 of 21 4/11/2014 1:13 PM Part D How much torque does the flywheel exert on the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: Problem 12.71 The 3.30 , 40.0-cm-diameter disk in the figure is spinning at 350 . Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.10 ? Express your answer with the appropriate units. ANSWER: Problem 12.74 A 5.0 , 60- -diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 19 of 21 4/11/2014 1:13 PM Part A What is the magnitude of the cylinder’s initial angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B What is the magnitude of the cylinder’s angular velocity when it is directly below the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Problem 12.82 A 45 figure skater is spinning on the toes of her skates at 0.90 . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 , 20 average diameter, 160 tall) plus two rod-like arms (2.5 each, 67 long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 , 20- -diameter, 200- -tall cylinder. Part A What is her new rotation frequency, in revolutions per second? Express your answer to two significant figures and include the appropriate units. ANSWER: = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 20 of 21 4/11/2014 1:13 PM Score Summary: Your score on this assignment is 4.0%. You received 7.84 out of a possible total of 198 points. = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?disp

Assignment 10 Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 12.3 Part A The figure shows three rotating disks, all of equal mass. Rank in order, from largest to smallest, their rotational kinetic energies to . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 1 of 21 4/11/2014 1:13 PM Incorrect; Try Again Conceptual Question 12.6 You have two steel solid spheres. Sphere 2 has twice the radius of sphere 1. Part A By what factor does the moment of inertia of sphere 2 exceed the moment of inertia of sphere 1? ANSWER: Correct Problem 12.2 A high-speed drill reaches 2500 in 0.59 . Part A What is the drill’s angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B Through how many revolutions does it turn during this first 0.59 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Constant Angular Acceleration in the Kitchen = 32 = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 2 of 21 4/11/2014 1:13 PM Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. Part A What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in degrees per second per second. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). ± A Spinning Electric Fan An electric fan is turned off, and its angular velocity decreases uniformly from 540 to 250 in a time interval of length 4.40 . Part A Find the angular acceleration in revolutions per second per second. You did not open hints for this part. ANSWER: Part B Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A. = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 3 of 21 4/11/2014 1:13 PM You did not open hints for this part. ANSWER: Part C How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A? You did not open hints for this part. ANSWER: Problem 12.8 A 100 ball and a 230 ball are connected by a 34- -long, massless, rigid rod. The balls rotate about their center of mass at 130 . Part A What is the speed of the 100 ball? Express your answer to two significant figures and include the appropriate units. ANSWER: Problem 12.10 A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through its center with 0.200 of kinetic energy. Part A What is the speed of a point on the rim? = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 4 of 21 4/11/2014 1:13 PM Express your answer with the appropriate units. ANSWER: Problem 12.12 A drum major twirls a 95- -long, 470 baton about its center of mass at 150 . Part A What is the baton’s rotational kinetic energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord’s weight can be ignored. Part A Which of the following statements correctly describes the system shown in the figure? Check all that apply. = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 5 of 21 4/11/2014 1:13 PM You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Problem 12.18 Part A In the figure , what is the magnitude of net torque about the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B What is the direction of net torque about the axle? ANSWER: The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero. = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 6 of 21 4/11/2014 1:13 PM Problem 12.22 An athlete at the gym holds a 3.5 steel ball in his hand. His arm is 78 long and has a mass of 3.6 . Assume the center of mass of the arm is at the geometrical center of the arm. Part A What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B What is the magnitude of the torque about his shoulder if he holds his arm straight, but below horizontal? Express your answer to two significant figures and include the appropriate units. ANSWER: Parallel Axis Theorem The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center of mass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that passes through point p, and is the mass of the object. Part A Suppose a uniform slender rod has length and mass . The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by . Find , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express in terms of and . Use fractions rather than decimal numbers in your answer. Clockwise Counterclockwise = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 7 of 21 4/11/2014 1:13 PM You did not open hints for this part. ANSWER: Part B Now consider a cube of mass with edges of length . The moment of inertia of the cube about an axis through its center of mass and perpendicular to one of its faces is given by . Find , the moment of inertia about an axis p through one of the edges of the cube Express in terms of and . Use fractions rather than decimal numbers in your answer. You did not open hints for this part. ANSWER: Problem 12.26 Starting from rest, a 12- -diameter compact disk takes 2.9 to reach its operating angular velocity of 2000 . Assume that the angular acceleration is constant. The disk’s moment of inertia is . Part A How much torque is applied to the disk? Express your answer to two significant figures and include the appropriate units. = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 8 of 21 4/11/2014 1:13 PM ANSWER: Part B How many revolutions does it make before reaching full speed? Express your answer using two significant figures. ANSWER: Problem 12.23 An object’s moment of inertia is 2.20 . Its angular velocity is increasing at the rate of 3.70 . Part A What is the total torque on the object? ANSWER: Problem 12.31 A 5.1 cat and a 2.5 bowl of tuna fish are at opposite ends of the 4.0- -long seesaw. = = rev Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 9 of 21 4/11/2014 1:13 PM Part A How far to the left of the pivot must a 3.8 cat stand to keep the seesaw balanced? Express your answer to two significant figures and include the appropriate units. ANSWER: Static Equilibrium of the Arm You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass , length and its center of mass is a distance from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance from the scapula. The deltoid muscle makes an angle of with the horizontal, as shown. Use throughout the problem. Part A Find the tension in the deltoid muscle. Express the tension in newtons, to the nearest integer. You did not open hints for this part. ANSWER: = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 10 of 21 4/11/2014 1:13 PM Part B Using the conditions for static equilibrium, find the magnitude of the vertical component of the force exerted by the scapula on the humerus (where the humerus attaches to the rest of the body). Express your answer in newtons, to the nearest integer. You did not open hints for this part. ANSWER: Part C Now find the magnitude of the horizontal component of the force exerted by the scapula on the humerus. Express your answer in newtons, to the nearest integer. ANSWER: ± Moments around a Rod A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored for this problem. There are three forces that are applied to the rod at different points and angles: , , and . Note that the dimensions of the bent rod are in centimeters in the figure, although the answers are requested in SI units (kilograms, meters, seconds). = N = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 11 of 21 4/11/2014 1:13 PM Part A If and , what does the magnitude of have to be for there to be rotational equilibrium? Answer numerically in newtons to two significant figures. You did not open hints for this part. ANSWER: Part B If the L-shaped rod has a moment of inertia , , , and again , how long a time would it take for the object to move through ( /4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures. You did not open hints for this part. ANSWER: Part C Now consider the situation in which and , but now a force with nonzero magnitude is acting on the rod. What does have to be to obtain equilibrium? Give a numerical answer, without trigonometric functions, in newtons, to two significant figures. You did not open hints for this part. ANSWER: = N = s = N Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 12 of 21 4/11/2014 1:13 PM Problem 12.32 A car tire is 55.0 in diameter. The car is traveling at a speed of 24.0 . Part A What is the tire’s rotation frequency, in rpm? Express your answer to three significant figures and include the appropriate units. ANSWER: Part B What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropriate units. ANSWER: Part C What is the speed of a point at the bottom edge of the tire? Express your answer as an integer and include the appropriate units. ANSWER: Problem 12.33 A 460 , 8.00-cm-diameter solid cylinder rolls across the floor at 1.30 . Part A What is the can’s kinetic energy? Express your answer with the appropriate units. Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 13 of 21 4/11/2014 1:13 PM ANSWER: Problem 12.45 Part A What is the magnitude of the angular momentum of the 780 rotating bar in the figure ? ANSWER: Part B What is the direction of the angular momentum of the bar ? ANSWER: Problem 12.46 into the page out of the page Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 14 of 21 4/11/2014 1:13 PM Part A What is the magnitude of the angular momentum of the 2.20 , 4.60-cm-diameter rotating disk in the figure ? ANSWER: Part B What is its direction? ANSWER: Problem 12.60 A 3.0- -long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.46. x direction -x direction y direction -y direction z direction -z direction Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 15 of 21 4/11/2014 1:13 PM Part A What is the minimum angle the ladder can make with the floor without slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Problem 12.61 The 3.0- -long, 90 rigid beam in the following figure is supported at each end. An 70 student stands 2.0 from support 1. Part A How much upward force does the support 1 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 16 of 21 4/11/2014 1:13 PM Part B How much upward force does the support 2 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Enhanced EOC: Problem 12.63 A 44 , 5.5- -long beam is supported, but not attached to, the two posts in the figure . A 22 boy starts walking along the beam. You may want to review ( pages 330 – 334) . For help with math skills, you may want to review: The Vector Cross Product Part A How close can he get to the right end of the beam without it falling over? Express your answer to two significant figures and include the appropriate units. You did not open hints for this part. ANSWER: = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 17 of 21 4/11/2014 1:13 PM Problem 12.68 Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel’s energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.6 diameter and a mass of 270 . Its maximum angular velocity is 1500 . Part A A motor spins up the flywheel with a constant torque of 54 . How long does it take the flywheel to reach top speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B How much energy is stored in the flywheel? Express your answer to two significant figures and include the appropriate units. ANSWER: Part C The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.2 . What is the average power delivered to the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: = = = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 18 of 21 4/11/2014 1:13 PM Part D How much torque does the flywheel exert on the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: Problem 12.71 The 3.30 , 40.0-cm-diameter disk in the figure is spinning at 350 . Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.10 ? Express your answer with the appropriate units. ANSWER: Problem 12.74 A 5.0 , 60- -diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 19 of 21 4/11/2014 1:13 PM Part A What is the magnitude of the cylinder’s initial angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Part B What is the magnitude of the cylinder’s angular velocity when it is directly below the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Problem 12.82 A 45 figure skater is spinning on the toes of her skates at 0.90 . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 , 20 average diameter, 160 tall) plus two rod-like arms (2.5 each, 67 long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 , 20- -diameter, 200- -tall cylinder. Part A What is her new rotation frequency, in revolutions per second? Express your answer to two significant figures and include the appropriate units. ANSWER: = = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?displayM… 20 of 21 4/11/2014 1:13 PM Score Summary: Your score on this assignment is 4.0%. You received 7.84 out of a possible total of 198 points. = Assignment 10 http://session.masteringphysics.com/myct/assignmentPrintView?disp

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Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g               , (Eq.2) M slope m g       (Eq.3) M b f        Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g                Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060  m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm      To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s   0.4% 100 9.80 9.76 9.80 100 . . . %        Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2           Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol  (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30  Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T   L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s    Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the  is dropped for simplicity), Vo is V at t = 0 (try it), and  (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating  and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o    Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

Morgan Extra Pages Graphing with Excel to be carried out in a computer lab, 3rd floor Calloway Hall or elsewhere The Excel spreadsheet consists of vertical columns and horizontal rows; a column and row intersect at a cell. A cell can contain data for use in calculations of all sorts. The Name Box shows the currently selected cell (Fig. 1). In the Excel 2007 and 2010 versions the drop-down menus familiar in most software screens have been replaced by tabs with horizontally-arranged command buttons of various categories (Fig. 2) ___________________________________________________________________ Open Excel, click on the Microsoft circle, upper left, and Save As your surname. xlsx on the desktop. Before leaving the lab e-mail the file to yourself and/or save to a flash drive. Also e-mail it to your instructor. Figure 1. Parts of an Excel spreadsheet. Name Box Figure 2. Tabs. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 1: BASIC OPERATIONS Click Save often as you work. 1. Type the heading “Edge Length” in Cell A1 and double click the crack between the A and B column heading for automatic widening of column A. Similarly, write headings for columns B and C and enter numbers in Cells A2 and A3 as in Fig. 3. Highlight Cells A2 and A3 by dragging the cursor (chunky plus-shape) over the two of them and letting go. 2. Note that there are three types of cursor crosses: chunky for selecting, barbed for moving entries or blocks of entries from cell to cell, and tiny (appearing only at the little square in the lower-right corner of a cell). Obtain a tiny arrow for Cell A3 and perform a plus-drag down Column A until the cells are filled up to 40 (in Cell A8). Note that the two highlighted cells set both the starting value of the fill and the intervals. 3. Click on Cell B2 and enter a formula for face area of a cube as follows: type =, click on Cell A2, type ^2, and press Enter (note the formula bar in Fig. 4). 4. Enter the formula for cube volume in Cell C2 (same procedure, but “=, click on A2, ^3, Enter”). 5. Highlight Cells B2 and C2; plus-drag down to Row 8 (Fig. 5). Do the numbers look correct? Click on some cells in the newly filled area and notice how Excel steps the row designations as it moves down the column (it can do it for horizontal plusdrags along rows also). This is the major programming development that has led to the popularity of spreadsheets. Figure 3. Entries. Figure 4. A formula. Figure 5. Plus-dragging formulas. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 6. Now let’s graph the Face Area versus Edge Length: select Cells A1 through B8, choose the Insert tab, and click the Scatter drop-down menu and select “Scatter with only Markers” (Fig. 6). 7. Move the graph (Excel calls it a “chart”) that appears up alongside your number table and dress it up as follows: a. Note that some Chart Layouts have appeared above. Click Layout 1 and alter each title to read Face Area for the vertical axis, Edge Length for the horizontal and Face Area vs. Edge Length for the Graph Title. b. Activate the Excel Least squares routine, called “fitting a trendline” in the program: right click any of the data markers and click Add Trendline. Choose Power and also check “Display equation on chart” and “Display R-squared value on chart.” Fig. 7 shows what the graph will look like at this point. c. The titles are explicit, so the legend is unnecessary. Click on it and press the delete button to remove it. Figure 6. Creating a scatter graph. Figure 7. A graph with a fitted curve. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 8. Now let’s overlay the Volume vs. Edge Length curve onto the same graph (optional for 203L/205L): Make a copy of your graph by clicking on the outer white area, clicking ctrl-c (or right click, copy), and pasting the copy somewhere else (ctrl-v). If you wish, delete the trendline as in Fig. 8. a. Right click on the outer white space, choose Select Data and click the Add button. b. You can type in the cell ranges by hand in the dialog box that comes up, but it is easier to click the red, white, and blue button on the right of each space and highlight what you want to go in. Click the red, white, and blue of the bar that has appeared, and you will bounce back to the Add dialog box. Use the Edge Length column for the x’s and Volume for the y’s. c. Right-click on any volume data point and choose Format Data Series. Clicking Secondary Axis will place its scale on the right of the graph as in Fig. 8. d. Dress up your graph with two axis titles (Layout-Labels-Axis Titles), etc. Figure 8. Adding a second curve and y-axis to the graph Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2: INTERPRETING A LINEAR GRAPH Introduction: Many experiments are repeated a number of times with one of the parameters involved varied from run to run. Often the goal is to measure the rate of change of a dependent variable, rather than a particular value. If the dependent variable can be expressed as a linear function of the independent parameter, then the slope and yintercept of an appropriate graph will give the rate of change and a particular value, respectively. An example of such an experiment in PHYS.203L/205L is the first part of Lab 20, in which weights are added to the bottom of a suspended spring (Figure 9). This experiment shows that a spring exerts a force Fs proportional to the distance stretched y = (y-yo), a relationship known as Hooke’s Law: Fs = – k(y – yo) (Eq. 1) where k is called the Hooke’s Law constant. The minus sign shows that the spring opposes any push or pull on it. In Lab 20 Fs is equal to (- Mg) and y is given by the reading on a meter stick. Masses were added to the bottom of the spring in 50-g increments giving weights in newtons of 0.49, 0.98, etc. The weight pan was used as the pointer for reading y and had a mass of 50 g, so yo could not be directly measured. For convenient graphing Equation 1 can be rewritten: -(Mg) = – ky + kyo Or (Mg) = ky – kyo (Eq. 1′) Procedure 1. On your spreadsheet note the tabs at the bottom left and double-click Sheet1. Type in “Basics,” and then click the Sheet2 tab to bring up a fresh worksheet. Change the sheet name to “Linear Fit” and fill in data as in this table. Hooke’s Law Experiment y (m) -Fs = Mg (N) 0.337 0.49 0.388 0.98 0.446 1.47 0.498 1.96 0.550 2.45 2. Highlight the cells with the numbers, and graph (Mg) versus y as in Steps 6 and 7 of the Basics section. Your Trendline this time will be Linear of course. If you are having trouble remembering what’s versus what, “y” looks like “v”, so what comes before the “v” of “versus” goes on the y (vertical) axis. Yes, this graph is confusing: the horizontal (“x”) axis is distance y, and the “y” axis is something else. 3. Click on the Equation/R2 box on the graph and highlight just the slope, that is, only the number that comes before the “x.” Copy it (control-c is a fast way to Figure 9. A spring with a weight stretching it Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com do it) and paste it (control-v) into an empty cell. Do likewise for the intercept (including the minus sign). SAVE YOUR FILE! 5. The next steps use the standard procedure for obtaining information from linear data. Write the general equation for a straight line immediately below a hand-written copy of Equation 1′ then circle matching items: (Mg) = k y + (- k yo) (Eq. 1′) y = m x + b Note the parentheses around the intercept term of Equation 1′ to emphasize that the minus sign is part of it. Equating above and below, you can create two useful new equations: slope m = k (Eq. 2) y-intercept b = -kyo (Eq. 3) 6. Solve Equation 2 for k, that is, rewrite left to right. Then substitute the value for slope m from your graph, and you have an experimental value for the Hooke’s Law constant k. Next solve Equation 3 for yo, substitute the value for intercept b from your graph and the value of k that you just found, and calculate yo. 7. Examine your linear graph for clues to finding the units of the slope and the yintercept. Use these units to find the units of k and yo. 8. Present your values of k and yo with their units neatly at the bottom of your spreadsheet. 9. R2 in Excel, like r in our lab manual and Corr. in the LoggerPro software, is a measure of how well the calculated line matches the data points. 1.00 would indicate a perfect match. State how good a match you think was made in this case? 10. Do the Homework, Further Exercises on Interpreting Linear Graphs, on the following pages. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com Eq.1 M m f M a g               , (Eq.2) M slope m g       (Eq.3) M b f        Morgan Extra Pages Homework: Graph Interpretation Exercises EXAMPLE WITH COMPLETE SOLUTION In PHYS.203L and 205L we do Lab 9 Newton’s Second Law on Atwood’s Machine using a photogate sensor (Fig. 1). The Atwood’s apparatus can slow the rate of fall enough to be measured even with primitive timing devices. In our experiment LoggerPro software automatically collects and analyzes the data giving reliable measurements of g, the acceleration of gravity. The equation governing motion for Atwood’s Machine can be written: where a is the acceleration of the masses and string, g is the acceleration of gravity, M is the total mass at both ends of the string, m is the difference between the masses, and f is the frictional force at the hub of the pulley wheel. In this exercise you are given a graph of a vs. m obtained in this experiment with the values of M and the slope and intercept (Fig. 2). The goal is to extract values for acceleration of gravity g and frictional force f from this information. To analyze the graph we write y = mx + b, the general equation for a straight line, directly under Equation 1 and match up the various parameters: Equating above and below, you can create two new equations: and y m x b M m f M a g                Figure 1. The Atwood’s Machine setup (from the LoggerPro handout). Figure 2. Graph of acceleration versus mass difference; data from a Physics I experiment. Atwood’s Machine M = 0.400 kg a = 24.4 m – 0.018 R2 = 0.998 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1.40 0.000 0.010 0.020 0.030 0.040 0.050 0.060  m (kg) a (m/s2) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com 2 2 9.76 / 0.400 24.4 /( ) m s kg m kg s g Mm      To handle Equation 2 it pays to consider what the units of the slope are. A slope is “the rise over the run,“ so its units must be the units of the vertical axis divided by those of the horizontal axis. In this case: Now let’s solve Equation 2 for g and substitute the values of total mass M and of the slope m from the graph: Using 9.80 m/s2 as the Baltimore accepted value for g, we can calculate the percent error: A similar process with Equation 3 leads to a value for f, the frictional force at the hub of the pulley wheel. Note that the units of intercept b are simply whatever the vertical axis units are, m/s2 in this case. Solving Equation 3 for f: EXERCISE 1 The Picket Fence experiment makes use of LoggerPro software to calculate velocities at regular time intervals as the striped plate passes through the photogate (Fig. 3). The theoretical equation is v = vi + at (Eq. 4) where vi = 0 (the fence is dropped from rest) and a = g. a. Write Equation 4 with y = mx + b under it and circle matching factors as in the Example. b. What is the experimental value of the acceleration of gravity? What is its percent error from the accepted value for Baltimore, 9.80 m/s2? c. Does the value of the y-intercept make sense? d. How well did the straight Trendline match the data? 2 / 2 kg s m kg m s   0.4% 100 9.80 9.76 9.80 100 . . . %        Acc Exp Acc Error kg m s mN kg m s f Mb 7.2 10 / 7.2 0.400 ( 0.018 / ) 3 2 2           Figure 3. Graph of speed versus time as calculated by LoggerPro as a picket fence falls freely through a photogate. Picket Fence Drop y = 9.8224x + 0.0007 R2 = 0.9997 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 t (s) v (m/s) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 2 This is an electrical example from PHYS.204L/206L, potential difference, V, versus current, I (Fig. 4). The theoretical equation is V = IR (Eq. 5) and is known as “Ohm’s Law.” The unit symbols stand for volts, V, and Amperes, A. The factor R stands for resistance and is measured in units of ohms, symbol  (capital omega). The definition of the ohm is: V (Eq. 6) By coincidence the letter symbols for potential (a quantity ) and volts (its unit) are identical. Thus “voltage” has become the laboratory slang name for potential. a. Rearrange the Ohm’s Law equation to match y = mx + b.. b. What is the experimental resistance? c. Comment on the experimental intercept: is its value reasonable? EXERCISE 3 This graph (Fig. 5) also follows Ohm’s Law, but solved for current I. For this graph the experimenter held potential difference V constant at 15.0V and measured the current for resistances of 100, 50, 40, and 30  Solve Ohm’s Law for I and you will see that 1/R is the logical variable to use on the x axis. For units, someone once jokingly referred to a “reciprocal ohm” as a “mho,” and the name stuck. a. Rearrange Equation 5 solved for I to match y = mx + b. b. What is the experimental potential difference? c. Calculate the percent difference from the 15.0 V that the experimenter set on the power supply (the instrument used for such experiments). d. Comment on the experimental intercept: is its value reasonable? Figure 4. Graph of potential difference versus current; data from a Physics II experiment. The theoretical equation, V = IR, is known as “Ohm’s Law.” Ohm’s Law y = 0.628x – 0.0275 R2 = 0.9933 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 0.5 0.6 Current, I (A) Potential difference, V (V) Figure 5. Another application of Ohm’s Law: a graph of current versus the inverse of resistance, from a different electric circuit experiment. Current versus (1/Resistance) y = 14.727x – 0.2214 R2 = 0.9938 0 100 200 300 400 500 600 5 10 15 20 25 30 35 R-1 (millimhos) I (milliamperes) Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 4 The Atwood’s Machine experiment (see the solved example above) can be done in another way: keep mass difference m the same and vary the total mass M (Fig. 6). a. Rewrite Equation 1 and factor out (1/M). b. Equate the coefficient of (1/M) with the experimental slope and solve for acceleration of gravity g. c. Substitute the values for slope, mass difference, and frictional force and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? EXERCISE 5 In the previous two exercises the reciprocal of a variable was used to make the graph come out linear. In this one the trick will be to use the square root of a variable (Fig. 7). In PHYS.203L and 205L Lab 19 The Pendulum the theoretical equation is where the period T is the time per cycle, L is the length of the string, and g is the acceleration of gravity. a. Rewrite Equation 7 with the square root of L factored out and placed at the end. b. Equate the coefficient of √L with the experimental slope and solve for acceleration of gravity g. c. Substitute the value for slope and calculate the experimental of g. d. Derive the units of the slope and show that the units of g come out as they should. e. Is the value of the experimental intercept reasonable? 2 (Eq . 7) g T   L Figure 6. Graph of acceleration versus the reciprocal of total mass; data from a another Physics I experiment. Atwood’s Machine m = 0.020 kg f = 7.2 mN y = 0.1964x – 0.0735 R2 = 0.995 0.400 0.600 0.800 1.000 2.000 2.500 3.000 3.500 4.000 4.500 5.000 1/M (1/kg) a (m/s2) Effect of Pendulum Length on Period y = 2.0523x – 0.0331 R2 = 0.999 0.400 0.800 1.200 1.600 2.000 2.400 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 L1/2 (m1/2) T (s) Figure 7. Graph of period T versus the square root of pendulum length; data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 6 In Exercise 5 another approach would have been to square both sides of Equation 7 and plot T2 versus L. Lab 20 directs us to use that alternative. It involves another case of periodic or harmonic motion with a similar, but more complicated, equation for the period: where T is the period of the bobbing (Fig. 8), M is the suspended mass, ms is the mass of the spring, k is a measure of stiffness called the spring constant, and C is a dimensionless factor showing how much of the spring mass is effectively bobbing. a. Square both sides of Equation 8 and rearrange it to match y = mx + b. b. Write y = mx + b under your rearranged equation and circle matching factors as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating k and the second for finding C from the data of Fig. 9. d. A theoretical analysis has shown that for most springs C = 1/3. Find the percent error from that value. e. Derive the units of the slope and intercept; show that the units of k come out as N/m and that C is dimensionless. 2 (Eq . 8) k T M Cm s    Figure 8. In Lab 20 mass M is suspended from a spring which is set to bobbing up and down, a good approximation to simple harmonic motion (SHM), described by Equation 8. Lab 20: SHM of a Spring Mass of the spring, ms = 25.1 g y = 3.0185x + 0.0197 R2 = 0.9965 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 0 0.05 0.1 0.15 0.2 0.25 0.3 M (kg) T 2 2 Figure 9. Graph of the square of the period T2 versus suspended mass M data from a Physics I experiment. Click to buy NOW! PDF-XChange Viewer www.docu-track.com Click to buy NOW! PDF-XChange Viewer www.docu-track.com EXERCISE 7 This last exercise deals with an exponential equation, and the trick is to take the logarithm of both sides. In PHYS.204L/206L we do Lab 33 The RC Time Constant with theoretical equation: where V is the potential difference at time t across a circuit element called a capacitor (the  is dropped for simplicity), Vo is V at t = 0 (try it), and  (tau) is a characteristic of the circuit called the time constant. a. Take the natural log of both sides and apply the addition rule for logarithms of a product on the right-hand side. b. Noting that the graph (Fig. 10) plots lnV versus t, arrange your equation in y = mx + b order, write y = mx + b under it, and circle the parts as in the Example. c. Write two new equations analogous to Equations 2 and 3 in the Example. Use the first of the two for calculating  and the second for finding lnVo and then Vo. d. Note that the units of lnV are the natural log of volts, lnV. As usual derive the units of the slope and interecept and use them to obtain the units of your experimental V and t. V V e (Eq. 9) t o    Figure 10. Graph of a logarithm versus time; data from Lab 33, a Physics II experiment. Discharge of a Capacitor y = -9.17E-03x + 2.00E+00 R2 = 9.98E-01 0.00 0.50 1.00 1.50 2.00 2.50

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