An exerciser begins with his arm in the relaxed vertical position OA, at which the elastic band is unstretched. He then rotates his arm to the horizontal position OB. The elastic modulus of the band is k = 60 N/m – that is, 60 N of force is required to stretch the band each additional meter of elongation. Determine the moment about O of the force which the band exerts on the hand B.

An exerciser begins with his arm in the relaxed vertical position OA, at which the elastic band is unstretched. He then rotates his arm to the horizontal position OB. The elastic modulus of the band is k = 60 N/m – that is, 60 N of force is required to stretch the band each additional meter of elongation. Determine the moment about O of the force which the band exerts on the hand B.

An exerciser begins with his arm in the relaxed vertical … Read More...
MECET 423: Mechanics of Materials Chap. 7 HW Chap. 7 Homework Set 1. Consider the beam shown in the image below. Let F1 = 2 kN and F2 = 3 kN. Assume that points A, B and C represent pin connections and a wire rope connects points B and C. Consider the dimensions L1, L2, L3 and L4 to be 2 m, 4 m, 6 m, and 10 m, respectively. The beam is made from HSS 152 X 51 X 6.4 (Appendix A-9) and the longer side of the rectangle is vertical. What is the maximum normal stress (units: MPa) experienced by the beam? 2. Consider the beam and loading shown below. The beam has a total length of 12 ft. and a uniformly distributed load, w, of 125 lb./ft. The cross section of the beam is comprised of a standard steel channel (C6 X 13) which has a ½ in. plate of steel attached to its bottom. Determine the maximum normal stress in tension and compression that is experienced by this beam due to the described loading. MECET 423: Mechanics of Materials Chap. 7 HW 3. Consider the cantilever beam shown in the image below. The beam is experiencing a linearly varying distributed load with w1 = 50 lb./ft. and w2 = 10 lb./ft. The beam is to be made from ASTM A36 structural steel and is to be 8 ft. in length. Select the smallest standard schedule 40 steel pipe size (Appendix A-12) which will ensure a factor of safety of at least 3. 4. The beam shown below has been fabricated by combining two wooden boards into a T-section. The dimensions for these sizes can be found in Appendix A-4. The beam is 9 ft. in length overall and dimension L1 = 3 ft. Assume the beam is made from a wood which has an allowable bending stress of 1500 psi (in both tension and compression). What is the largest value of the force which can be applied? MECET 423: Mechanics of Materials Chap. 7 HW 5. The image below shows a hydraulic cylinder which is being utilized in a simple press-fit operation. As can be seen the cylinder is being suspended over the work piece using a cantilever beam. Note from the right view that there is a beam on either side of the cylinder. You may assume that each will be equally loaded by the cylinder. The beams are to be cut from AISI 1040 HR steel plate which has a thickness of 0.750 in. The proposed design includes the following dimensions (units: inch): H = 2.00, h = 1.00, r = 0.08, L1 = 8, and L2 = 18. Evaluate the design by calculating the resulting factor of safety with respect to the yield strength of the material at the location of the step if the total force generated by the cylinder is 1,000 lb. Also state whether or not yielding is predicted to occur. You may assume that bending in the thickness direction of the beams is negligible. 6. Consider the cantilever beam shown below. The beam has a length of 4 ft. and is made from a material whose design stress, σd, is equal to 10,000 psi. It is to carry a load of 200 lb. applied at its free end. The beam is to be designed as a beam of constant strength where the maximum normal stress experienced at each cross section is equal to the design normal stress. To achieve this the height will be held constant at 1.5 in. while the base will vary as a function of the position along the length of the beam. Determine the equation which describes the required length of the base as a function of the position along the length of the beam. For consistency, let the origin be located at point A and the positive x axis be directed toward the right. MECET 423: Mechanics of Materials Chap. 7 HW 7. Consider the overhanging beam shown in the image below. Assume that L = 5 ft. and L1 = 3 ft. The beam’s cross section is shown below. The centerline marks the horizontal centroidal axis. The moment of inertia about this axis is approx. 0.208 in4. Due to the geometry of the cross section and the material, the beam has different maximum allowable normal stresses in tension and compression. The design normal stress in tension is 24,000 psi while the design normal stress in compression is 18,000 psi. Using this data determine the maximum force, F, which can be applied to the beam.

MECET 423: Mechanics of Materials Chap. 7 HW Chap. 7 Homework Set 1. Consider the beam shown in the image below. Let F1 = 2 kN and F2 = 3 kN. Assume that points A, B and C represent pin connections and a wire rope connects points B and C. Consider the dimensions L1, L2, L3 and L4 to be 2 m, 4 m, 6 m, and 10 m, respectively. The beam is made from HSS 152 X 51 X 6.4 (Appendix A-9) and the longer side of the rectangle is vertical. What is the maximum normal stress (units: MPa) experienced by the beam? 2. Consider the beam and loading shown below. The beam has a total length of 12 ft. and a uniformly distributed load, w, of 125 lb./ft. The cross section of the beam is comprised of a standard steel channel (C6 X 13) which has a ½ in. plate of steel attached to its bottom. Determine the maximum normal stress in tension and compression that is experienced by this beam due to the described loading. MECET 423: Mechanics of Materials Chap. 7 HW 3. Consider the cantilever beam shown in the image below. The beam is experiencing a linearly varying distributed load with w1 = 50 lb./ft. and w2 = 10 lb./ft. The beam is to be made from ASTM A36 structural steel and is to be 8 ft. in length. Select the smallest standard schedule 40 steel pipe size (Appendix A-12) which will ensure a factor of safety of at least 3. 4. The beam shown below has been fabricated by combining two wooden boards into a T-section. The dimensions for these sizes can be found in Appendix A-4. The beam is 9 ft. in length overall and dimension L1 = 3 ft. Assume the beam is made from a wood which has an allowable bending stress of 1500 psi (in both tension and compression). What is the largest value of the force which can be applied? MECET 423: Mechanics of Materials Chap. 7 HW 5. The image below shows a hydraulic cylinder which is being utilized in a simple press-fit operation. As can be seen the cylinder is being suspended over the work piece using a cantilever beam. Note from the right view that there is a beam on either side of the cylinder. You may assume that each will be equally loaded by the cylinder. The beams are to be cut from AISI 1040 HR steel plate which has a thickness of 0.750 in. The proposed design includes the following dimensions (units: inch): H = 2.00, h = 1.00, r = 0.08, L1 = 8, and L2 = 18. Evaluate the design by calculating the resulting factor of safety with respect to the yield strength of the material at the location of the step if the total force generated by the cylinder is 1,000 lb. Also state whether or not yielding is predicted to occur. You may assume that bending in the thickness direction of the beams is negligible. 6. Consider the cantilever beam shown below. The beam has a length of 4 ft. and is made from a material whose design stress, σd, is equal to 10,000 psi. It is to carry a load of 200 lb. applied at its free end. The beam is to be designed as a beam of constant strength where the maximum normal stress experienced at each cross section is equal to the design normal stress. To achieve this the height will be held constant at 1.5 in. while the base will vary as a function of the position along the length of the beam. Determine the equation which describes the required length of the base as a function of the position along the length of the beam. For consistency, let the origin be located at point A and the positive x axis be directed toward the right. MECET 423: Mechanics of Materials Chap. 7 HW 7. Consider the overhanging beam shown in the image below. Assume that L = 5 ft. and L1 = 3 ft. The beam’s cross section is shown below. The centerline marks the horizontal centroidal axis. The moment of inertia about this axis is approx. 0.208 in4. Due to the geometry of the cross section and the material, the beam has different maximum allowable normal stresses in tension and compression. The design normal stress in tension is 24,000 psi while the design normal stress in compression is 18,000 psi. Using this data determine the maximum force, F, which can be applied to the beam.

info@checkyourstudy.com Whatsapp +919911743277
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

please email info@checkyourstudy.com
Based on molecular mass and dipole moment of the five compounds in the table below, which should have the highest boiling point? A) 3 2 3 CH CH CH B) 3 3 CH OCH C) 3 CH Cl D) 3 CH CHO E) 3 CH CN 30)

Based on molecular mass and dipole moment of the five compounds in the table below, which should have the highest boiling point? A) 3 2 3 CH CH CH B) 3 3 CH OCH C) 3 CH Cl D) 3 CH CHO E) 3 CH CN 30)

Question 3 An electric dipole moment, consists of two equal but opposite charges, + q and -q, separated by a distance, d. The magnitude of the dipole moment is p = qd and the electric dipole moment vector points from the negative towards the positive charge of the configuration. Consider a water molecule. The measured value of the electrical dipole moment of the water molecule is about . An ionized hydrogen atom happens to be 3.5 nanometers away from the center of the water molecule on the side away from its two hydrogen atoms and along the line defined by the molecule’s dipole moment. (Take the center of the water molecule to be the center of its oxygen atom. Also consider the water molecule to be essentially a small, nearly a point object). How large is the force on the ionized hydrogen atom and what is its direction? A. , attractive B. , repulsive C. , attractive D. , repulsive + E. , attractive

Question 3 An electric dipole moment, consists of two equal but opposite charges, + q and -q, separated by a distance, d. The magnitude of the dipole moment is p = qd and the electric dipole moment vector points from the negative towards the positive charge of the configuration. Consider a water molecule. The measured value of the electrical dipole moment of the water molecule is about . An ionized hydrogen atom happens to be 3.5 nanometers away from the center of the water molecule on the side away from its two hydrogen atoms and along the line defined by the molecule’s dipole moment. (Take the center of the water molecule to be the center of its oxygen atom. Also consider the water molecule to be essentially a small, nearly a point object). How large is the force on the ionized hydrogen atom and what is its direction? A. , attractive B. , repulsive C. , attractive D. , repulsive + E. , attractive

Question 3 An electric dipole moment, consists of two … Read More...
Replace the loading system by an equivalent resultant force and couple moment acting at point A. Assume F1 = 950 N and F2 =300 N.

Assignment 8 Due: 11:59pm on Friday, April 4, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 10.3 Part A If a particle’s speed increases by a factor of 5, by what factor does its kinetic energy change? ANSWER: Correct Conceptual Question 10.11 A spring is compressed 1.5 . Part A How far must you compress a spring with twice the spring constant to store the same amount of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct = 25 K2 K1 cm x = 1.1 cm Problem 10.2 The lowest point in Death Valley is below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 . Part A What is the change in potential energy of an energetic 80 hiker who makes it from the floor of Death Valley to the top of Mt.Whitney? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.3 Part A At what speed does a 1800 compact car have the same kinetic energy as a 1.80×104 truck going 21.0 ? Express your answer with the appropriate units. ANSWER: Correct Problem 10.5 A boy reaches out of a window and tosses a ball straight up with a speed of 13 . The ball is 21 above the ground as he releases it. 85m m kg U = 3.5×106 J kg kg km/hr vc = 66.4 km hr m/s m Part A Use energy to find the ball’s maximum height above the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Use energy to find the ball’s speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Use energy to find the speed of impact on the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Hmax = 30 m v = 13 ms v = 24 ms Problem 10.8 A 59.0 skateboarder wants to just make it to the upper edge of a “quarter pipe,” a track that is one-quarter of a circle with a radius of 2.30 . Part A What speed does he need at the bottom? Express your answer with the appropriate units. ANSWER: Correct Problem 10.12 A 1500 car traveling at 12 suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll. Part A kg m 6.71 ms kg m/s What will be the car’s speed as it coasts into the gas station on the other side of the valley? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object’s kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object’s kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where “i” denotes the “initial” moment and “f” denotes the “final” moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: v = 6.8 ms K = (1/2)mv2 U = mgh Ki + Ui = Kf + Uf v Correct Part B At the top point of the flight, what can be said about the projectile’s kinetic and potential energy? ANSWER: Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system “Earth-ball.” Although we will often talk about “the gravitational potential energy of an elevated object,” it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch __________. ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases. Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum. Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level–but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: Correct You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: is negative is positive is zero depends on the choice of the “zero level” of potential energy U = 0 hmax v g hmax = v2 2g h 0.5v v g h = 3 v2 8g Correct Part F What is the speed of the object at the height of ? Express your answer in terms of and . Use three significant figures in the numeric coefficient. Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. ANSWER: Correct Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball’s flight. Express your answer in terms of , , and . Hint 1. Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of “final” moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. u (1/2)hmax v g h = 0 v (1/2)mv2 h (1/2)hmax (1/4)mv2 h = (1/2)hmax v u = 0.707v v hmax v g Kf Express your answer in terms of , , and . Hint 1. Find the speed at the maximum height The speed of the ball at the maximum height is __________. ANSWER: ANSWER: ANSWER: Correct Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. v m 0 v v cos v sin v tan Kf = 0.5m(vcos( ))2 hmax = (vsin( ))2 2g v hmax v g ANSWER: Correct Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: Correct The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Problem 10.14 A 12- -long spring is attached to the ceiling. When a 2.2 mass is hung from it, the spring stretches to a length of 17 . Part A What is the spring constant ? Express your answer to two significant figures and include the appropriate units. hmax = v2 2g v hmax v g hmax = v2 2g Ki + Ui = Kf + Uf cm kg cm k ANSWER: Correct Part B How long is the spring when a 3.0 mass is suspended from it? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.17 A 6.2 mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in . You may want to review ( pages 255 – 257) . For help with math skills, you may want to review: Solving Algebraic Equations = 430 k Nm kg y = 19 cm kg Part A What does the spring scale read just before the mass touches the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass before it touches the scale. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? ANSWER: Correct Part B The scale reads 22 when the lower spring has been compressed by 2.7 . What is the value of the spring constant for the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? Use these to determine the force on the mass by the spring, taking note of the directions from your picture. How is the spring constant related to the force by the spring and the compression of the spring? Check your units. ANSWER: F = 61 N N cm k = 1400 k Nm Correct Part C At what compression length will the scale read zero? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces on the mass. When the scale reads zero, what is the force on the mass due to the scale? What is the gravitational force on the mass? What is the force on the mass by the spring? How is the compression length related to the force by the spring and the spring constant? Check your units. ANSWER: Correct Problem 10.18 Part A How far must you stretch a spring with = 800 to store 180 of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: y = 4.2 cm k N/m J Correct Problem 10.22 A 15 runaway grocery cart runs into a spring with spring constant 230 and compresses it by 57 . Part A What was the speed of the cart just before it hit the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Spring Gun A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measured from the equilibrium position of the spring. s = 0.67 m kg N/m cm v = 2.2 ms H Part A The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to . Hint 1. Potential energy of the spring The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial. Hint 2. Potential energy of the ball At the highest point in the ball’s trajectory, all of the spring’s potential energy has been converted into gravitational potential energy of the ball. ANSWER: Correct A Bullet Is Fired into a Wooden Block A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed . H height = H 4 mb vi mw vf Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of , , and . Hint 1. Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy vi mw mb ptotal vf ptotal = (mw + mb)vf ptotal vi ptotal = mbvi ANSWER: Correct Problem 10.31 Ball 1, with a mass of 150 and traveling at 15.0 , collides head on with ball 2, which has a mass of 340 and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Express your answer with the appropriate units. ANSWER: Correct Part B Express your answer with the appropriate units. ANSWER: Correct Part C vf = mb vi mb+mw g m/s g (vfx) = -5.82 1 ms (vfx) = 9.18 2 ms What are the final velocities of each ball if the collision is perfectly inelastic? Express your answer with the appropriate units. ANSWER: Correct Part D Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.43 A package of mass is released from rest at a warehouse loading dock and slides down the = 2.2 – high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. You may want to review ( pages 265 – 269) . For help with math skills, you may want to review: Solving Algebraic Equations (vfx) = 4.59 1 ms (vfx) = 4.59 2 ms m h m 2m Part A Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are two parts to this problem: the block sliding down the frictionless incline and the collision. What conservation laws are valid in each part? In terms of , what are the kinetic and potential energies of the block at the top of the incline? What is the potential energy of the same block at the bottom just before the collision? What are the kinetic energy and velocity of block just before the collision? What is conserved during the collision? What is the total momentum of the two blocks before the collision? What is the momentum of the two blocks stuck together after the collision? What is the velocity of the two blocks after the collision? ANSWER: Correct Part B Suppose the collision between the packages is perfectly elastic. To what height does the package of mass rebound? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are three parts to this problem: the block sliding down the incline, the collision, and mass going back up the incline. What conservation laws are valid in each part? m m v = 2.2 ms m m What is an elastic collision? For an elastic collision, how are the initial and final velocities related when one of the masses is initially at rest? Using the velocity of just before the collision from Part A, what is the velocity of just after the collision in this case? What are the kinetic and potential energies of mass just after the collision? What is the kinetic energy of mass at its maximum rebound height? Using conservation of energy, what is the potential energy of mass at its maximum height? What is the maximum height? ANSWER: Correct Problem 10.35 A cannon tilted up at a 35.0 angle fires a cannon ball at 79.0 from atop a 21.0 -high fortress wall. Part A What is the ball’s impact speed on the ground below? Express your answer with the appropriate units. ANSWER: Correct Problem 10.45 A 1000 safe is 2.5 above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 . m m m m m h = 24 cm $m/s m vf = 81.6 ms kg m cm Part A What is the spring constant of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.49 A 100 block on a frictionless table is firmly attached to one end of a spring with = 21 . The other end of the spring is anchored to the wall. A 30 ball is thrown horizontally toward the block with a speed of 6.0 . Part A If the collision is perfectly elastic, what is the ball’s speed immediately after the collision? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the maximum compression of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.5×105 k Nm g k N/m g m/s v = 3.2 ms Correct Part C Repeat part A for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Repeat part B for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.4%. You received 120.28 out of a possible total of 121 points. x = 0.19 m v = 1.4 ms x = 0.11 m Assignment 8 Due: 11:59pm on Friday, April 4, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 10.3 Part A If a particle’s speed increases by a factor of 5, by what factor does its kinetic energy change? ANSWER: Correct Conceptual Question 10.11 A spring is compressed 1.5 . Part A How far must you compress a spring with twice the spring constant to store the same amount of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct = 25 K2 K1 cm x = 1.1 cm Problem 10.2 The lowest point in Death Valley is below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 . Part A What is the change in potential energy of an energetic 80 hiker who makes it from the floor of Death Valley to the top of Mt.Whitney? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.3 Part A At what speed does a 1800 compact car have the same kinetic energy as a 1.80×104 truck going 21.0 ? Express your answer with the appropriate units. ANSWER: Correct Problem 10.5 A boy reaches out of a window and tosses a ball straight up with a speed of 13 . The ball is 21 above the ground as he releases it. 85m m kg U = 3.5×106 J kg kg km/hr vc = 66.4 km hr m/s m Part A Use energy to find the ball’s maximum height above the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Use energy to find the ball’s speed as it passes the window on its way down. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Use energy to find the speed of impact on the ground. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Hmax = 30 m v = 13 ms v = 24 ms Problem 10.8 A 59.0 skateboarder wants to just make it to the upper edge of a “quarter pipe,” a track that is one-quarter of a circle with a radius of 2.30 . Part A What speed does he need at the bottom? Express your answer with the appropriate units. ANSWER: Correct Problem 10.12 A 1500 car traveling at 12 suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll. Part A kg m 6.71 ms kg m/s What will be the car’s speed as it coasts into the gas station on the other side of the valley? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object’s kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object’s kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where “i” denotes the “initial” moment and “f” denotes the “final” moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: v = 6.8 ms K = (1/2)mv2 U = mgh Ki + Ui = Kf + Uf v Correct Part B At the top point of the flight, what can be said about the projectile’s kinetic and potential energy? ANSWER: Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system “Earth-ball.” Although we will often talk about “the gravitational potential energy of an elevated object,” it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch __________. ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases. Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum. Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level–but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: Correct You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: is negative is positive is zero depends on the choice of the “zero level” of potential energy U = 0 hmax v g hmax = v2 2g h 0.5v v g h = 3 v2 8g Correct Part F What is the speed of the object at the height of ? Express your answer in terms of and . Use three significant figures in the numeric coefficient. Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. ANSWER: Correct Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball’s flight. Express your answer in terms of , , and . Hint 1. Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of “final” moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. u (1/2)hmax v g h = 0 v (1/2)mv2 h (1/2)hmax (1/4)mv2 h = (1/2)hmax v u = 0.707v v hmax v g Kf Express your answer in terms of , , and . Hint 1. Find the speed at the maximum height The speed of the ball at the maximum height is __________. ANSWER: ANSWER: ANSWER: Correct Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. v m 0 v v cos v sin v tan Kf = 0.5m(vcos( ))2 hmax = (vsin( ))2 2g v hmax v g ANSWER: Correct Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: Correct The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Problem 10.14 A 12- -long spring is attached to the ceiling. When a 2.2 mass is hung from it, the spring stretches to a length of 17 . Part A What is the spring constant ? Express your answer to two significant figures and include the appropriate units. hmax = v2 2g v hmax v g hmax = v2 2g Ki + Ui = Kf + Uf cm kg cm k ANSWER: Correct Part B How long is the spring when a 3.0 mass is suspended from it? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.17 A 6.2 mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in . You may want to review ( pages 255 – 257) . For help with math skills, you may want to review: Solving Algebraic Equations = 430 k Nm kg y = 19 cm kg Part A What does the spring scale read just before the mass touches the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass before it touches the scale. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? ANSWER: Correct Part B The scale reads 22 when the lower spring has been compressed by 2.7 . What is the value of the spring constant for the lower spring? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces acting on the mass. What is the net force on the mass? What is the force on the mass due to gravity? What is the force on the mass due to the scale? Use these to determine the force on the mass by the spring, taking note of the directions from your picture. How is the spring constant related to the force by the spring and the compression of the spring? Check your units. ANSWER: F = 61 N N cm k = 1400 k Nm Correct Part C At what compression length will the scale read zero? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture showing the forces on the mass. When the scale reads zero, what is the force on the mass due to the scale? What is the gravitational force on the mass? What is the force on the mass by the spring? How is the compression length related to the force by the spring and the spring constant? Check your units. ANSWER: Correct Problem 10.18 Part A How far must you stretch a spring with = 800 to store 180 of energy? Express your answer to two significant figures and include the appropriate units. ANSWER: y = 4.2 cm k N/m J Correct Problem 10.22 A 15 runaway grocery cart runs into a spring with spring constant 230 and compresses it by 57 . Part A What was the speed of the cart just before it hit the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Spring Gun A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measured from the equilibrium position of the spring. s = 0.67 m kg N/m cm v = 2.2 ms H Part A The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to . Hint 1. Potential energy of the spring The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial. Hint 2. Potential energy of the ball At the highest point in the ball’s trajectory, all of the spring’s potential energy has been converted into gravitational potential energy of the ball. ANSWER: Correct A Bullet Is Fired into a Wooden Block A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed . H height = H 4 mb vi mw vf Part A Which of the following best describes this collision? Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: Correct Part B Which of the following quantities, if any, are conserved during this collision? Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part C What is the speed of the block/bullet system after the collision? Express your answer in terms of , , and . Hint 1. Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy vi mw mb ptotal vf ptotal = (mw + mb)vf ptotal vi ptotal = mbvi ANSWER: Correct Problem 10.31 Ball 1, with a mass of 150 and traveling at 15.0 , collides head on with ball 2, which has a mass of 340 and is initially at rest. Part A What are the final velocities of each ball if the collision is perfectly elastic? Express your answer with the appropriate units. ANSWER: Correct Part B Express your answer with the appropriate units. ANSWER: Correct Part C vf = mb vi mb+mw g m/s g (vfx) = -5.82 1 ms (vfx) = 9.18 2 ms What are the final velocities of each ball if the collision is perfectly inelastic? Express your answer with the appropriate units. ANSWER: Correct Part D Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 10.43 A package of mass is released from rest at a warehouse loading dock and slides down the = 2.2 – high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass , from the bottom of the chute. You may want to review ( pages 265 – 269) . For help with math skills, you may want to review: Solving Algebraic Equations (vfx) = 4.59 1 ms (vfx) = 4.59 2 ms m h m 2m Part A Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are two parts to this problem: the block sliding down the frictionless incline and the collision. What conservation laws are valid in each part? In terms of , what are the kinetic and potential energies of the block at the top of the incline? What is the potential energy of the same block at the bottom just before the collision? What are the kinetic energy and velocity of block just before the collision? What is conserved during the collision? What is the total momentum of the two blocks before the collision? What is the momentum of the two blocks stuck together after the collision? What is the velocity of the two blocks after the collision? ANSWER: Correct Part B Suppose the collision between the packages is perfectly elastic. To what height does the package of mass rebound? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem There are three parts to this problem: the block sliding down the incline, the collision, and mass going back up the incline. What conservation laws are valid in each part? m m v = 2.2 ms m m What is an elastic collision? For an elastic collision, how are the initial and final velocities related when one of the masses is initially at rest? Using the velocity of just before the collision from Part A, what is the velocity of just after the collision in this case? What are the kinetic and potential energies of mass just after the collision? What is the kinetic energy of mass at its maximum rebound height? Using conservation of energy, what is the potential energy of mass at its maximum height? What is the maximum height? ANSWER: Correct Problem 10.35 A cannon tilted up at a 35.0 angle fires a cannon ball at 79.0 from atop a 21.0 -high fortress wall. Part A What is the ball’s impact speed on the ground below? Express your answer with the appropriate units. ANSWER: Correct Problem 10.45 A 1000 safe is 2.5 above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 . m m m m m h = 24 cm$ m/s m vf = 81.6 ms kg m cm Part A What is the spring constant of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 10.49 A 100 block on a frictionless table is firmly attached to one end of a spring with = 21 . The other end of the spring is anchored to the wall. A 30 ball is thrown horizontally toward the block with a speed of 6.0 . Part A If the collision is perfectly elastic, what is the ball’s speed immediately after the collision? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the maximum compression of the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.5×105 k Nm g k N/m g m/s v = 3.2 ms Correct Part C Repeat part A for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Repeat part B for the case of a perfectly inelastic collision. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.4%. You received 120.28 out of a possible total of 121 points. x = 0.19 m v = 1.4 ms x = 0.11 m

please email info@checkyourstudy.com
From the moment the Prime Minister mobilized the troops what was her stated objective? A. To broker a peace deal between the Argentines and the inhabitants of the Falklands. B. To restore the Islands to their previously free state under the British Administration. C. To negotiate a peace talk in which two separate states will be created. D. To cede the Falkland Islands to the Argentines after evacuating all the British citizens. E. To wipe out the Argentine resistance to British rule.

From the moment the Prime Minister mobilized the troops what was her stated objective? A. To broker a peace deal between the Argentines and the inhabitants of the Falklands. B. To restore the Islands to their previously free state under the British Administration. C. To negotiate a peace talk in which two separate states will be created. D. To cede the Falkland Islands to the Argentines after evacuating all the British citizens. E. To wipe out the Argentine resistance to British rule.

From the moment the Prime Minister mobilized the troops what … Read More...
determine the resultant moment produced by the forces about point O. Assume F1 = 510 N, F2 = 610 N . Express your answer to there significant figures and include the appropriate units.

determine the resultant moment produced by the forces about point O. Assume F1 = 510 N, F2 = 610 N . Express your answer to there significant figures and include the appropriate units.

determine the resultant moment produced by the forces about point … Read More...
CHM114: Exam #2 CHM 114, S2015 Exam #2, Version C 16 March 2015 Instructor: O. Graudejus Points: 100 Print Name Sign Name Student I.D. # 1. You are responsible for the information on this page. Please read it carefully. 2. Code your name and 10 digit affiliate identification number on the separate scantron answer sheet. Use only a #2 pencil 3. If you enter your ASU ID incorrectly on the scantron, a 3 point penalty will be assessed. 4. Do all calculations on the exam pages. Do not make any unnecessary marks on the answer sheet. 5. This exam consists of 25 multiple choice questions worth 4 points each and a periodic table. Make sure you have them all. 6. Choose the best answer to each of the questions and answer it on the computer-graded answer sheet. Read all responses before making a selection. 7. Read the directions carefully for each problem. 8. Avoid even casual glances at other students’ exams. 9. Stop writing and hand in your scantron answer sheet and your test promptly when instructed. LATE EXAMS MAY HAVE POINTS DEDUCTED. 10. You will have 50 minutes to complete the exam. 11. If you leave early, please do so quietly. 12. Work the easiest problems first. 13. A periodic table is attached as the last page to this exam. 14. Answers will be posted online this afternoon. Potentially useful information: K = ºC + 273.15 RH=2.18·10-18 J R=8.314 J·K-1·mol-1 1Å=10-10 m c=3·108 m/s Ephoton=h·n=h·c/l h=6.626·10-34 Js Avogadro’s Number = 6.022 × 1023 particles/mole DH°rxn =  n DHf° (products) –  n DHf° (reactants) ) 1 1 ( 2 2 f i H n n DE = R − \ -2- CHM114: Exam #2 1) Which one of the following is an incorrect orbital notation? A) 2s B) 2p C) 3f D) 3d E) 4s 2) The energy of a photon that has a frequency of 8.21 1015s 1 − × is __________ J. A) 8.08 10 50 − × B) 1.99 10 25 − × C) 5.44 10 18 − × D) 1.24×1049 E) 1.26 10 19 − × 3) The ground state electron configuration of Ga is __________. A) 1s22s23s23p64s23d104p1 B) 1s22s22p63s23p64s24d104p1 C) 1s22s22p63s23p64s23d104p1 D) 1s22s22p63s23p64s23d104d1 E) [Ar]4s23d11 4) Of the bonds N–N, N=N, and NN, the N-N bond is __________. A) strongest/shortest B) weakest/longest C) strongest/longest D) weakest/shortest E) intermediate in both strength and length 5) Of the atoms below, __________ is the most electronegative. A) Br B) O C) Cl D) N E) F 6) Of the following, __________ cannot accommodate more than an octet of electrons. A) P B) O C) S D) Cl E) I -3- CHM 114: Exam #2 7) Which electron configuration represents a violation of Hund’s Rule? A) B) C) D) E) 8) A tin atom has 50 electrons. Electrons in the _____ subshell experience the highest effective nuclear charge. A) 1s B) 3p C) 3d D) 5s E) 5p 9) In ionic compounds, the lattice energy_____ as the magnitude of the ion charges _____ and the radii _____. A) increases, decrease, increase B) increases, increase, increase C) decreases, increase, increase D) increases, increase, decrease E) increases, decrease, decrease 10) Which of the following ionic compounds has the highest lattice energy? A) LiF B) MgO C) CsF D) CsI E) LiI -4- CHM 114: Exam #2 11) For which one of the following reactions is the value of H°rxn equal to Hf° for the product? A) 2 C (s, graphite) + 2 H2 (g)  C2H4 (g) B) N2 (g) + O2 (g)  2 NO (g) C) 2 H2 (g) + O2 (g)  2 H2O (l) D) 2 H2 (g) + O2 (g)  2 H2O (g) E) all of the above 12) Given the data in the table below, H rxn D ° for the reaction 3 2 3 PCl (g) + 3HCl(g)®3Cl (g) + PH (g) is __________ kJ. A) -570.37 B) -385.77 C) 570.37 D) 385.77 E) The f DH° of 2 Cl (g) is needed for the calculation. 13) Given the following reactions (1) 2 2 2NO® N +O H = -180 kJ (2) 2 2 2NO+O ®2NO H = -112 kJ the enthalpy of the reaction of nitrogen with oxygen to produce nitrogen dioxide 2 2 2 N + 2O ®2NO is __________ kJ. A) 68 B) -68 C) -292 D) 292 E) -146 14) Of the following transitions in the Bohr hydrogen atom, the __________ transition results in the absorption of the lowest-energy photon. A) n = 1  n = 6 B) n = 6  n = 1 C) n = 6  n = 5 D) n = 3  n = 6 E) n = 1  n = 4 -5- CHM 114: Exam #2 15) Which equation correctly represents the electron affinity of calcium? A) Ca (g)  Ca+ (g) + e- B) Ca (g)  Ca- (g) + e- C) Ca (g) + e-  Ca- (g) D) Ca- (g)  Ca (g) + e- E) Ca+ (g) + e-  Ca (g) 16) Which of the following does not have eight valence electrons? A) Ca+ B) Rb+ C) Xe D) Br− E) All of the above have eight valence electrons. 17) The specific heat of liquid bromine is 0.226 J/g · K. The molar heat capacity (in J/mol-K) of liquid bromine is __________. A) 707 B) 36.1 C) 18.1 D) 9.05 E) 0.226 18) Given the electronegativities below, which covalent single bond is least polar? Element: H C N O F Electronegativity: 2.1 2.5 3.0 3.5 4.0 A) C-H B) C-F C) O-H D) O-C E) F-H 19) The bond length in an HCl molecule is 1.27 Å and the measured dipole moment is 1.08 D. What is the magnitude (in units of e) of the negative charge on Cl in HCl? (1 debye = 3.34 10 30 coulomb-meters − × ; e=1.6 10 19 coulombs − × ) A) 1.6 10 19 − × B) 0.057 C) 0.18 D) 1 E) 0.22 -6- CHM 114: Exam #2 20) The F-B-F bond angle in the BF3 molecule is approximately __________. A) 90° B) 109.5° C) 120° D) 180° E) 60° 21) Which isoelectronic series is correctly arranged in order of increasing radius? A) K+ < Ca2+ < Ar < Cl- B) Cl- < Ar < K+ < Ca2+ C) Ca2+ < Ar < K+ < Cl- D) Ca2+ < K+ < Ar < Cl- E) Ca2+ < K+ < Cl- < Ar 22) What is the electron configuration for the Fe2+ ion? A) [Ar]4s03d6 B) [Ar]4s23d4 C) [Ar]4s03d8 D) [Ar]4s23d8 E) [Ar]4s63d2 23) The formal charge on carbon in the Lewis structure of the NCS - ion is __________: A) -1 B) +1 C) +2 D) 0 E) +3 -7- CHM 114: Exam #2 24) Using the table of bond dissociation energies, the H for the following gas-phase reaction is __________ kJ. A) 291 B) 2017 C) -57 D) -356 E) -291 25) According to VSEPR theory, if there are six electron domains in the valence shell of an atom, they will be arranged in a(n) __________ geometry. A) octahedral B) linear C) tetrahedral D) trigonal planar E) trigonal bipyramidal -8- CHM 114: Exam #2