# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

10 Pages

### OLDHW24

Course: PHY 58235, Spring 2009
School: University of Texas
Rating:

Word Count: 2482

#### Document Preview

24 oldhomewk PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am we obtain Question 1, chap -1, sect -1. part 1 of 2 10 points Four particles with masses 3 kg, 3 kg, 3 kg, and 8 kg are connected by rigid rods of negligible mass as shown. 8 kg 3 kg y Iz = [m1 + m2 + m3 + m4 ] r 2 = [(3 kg) + (3 kg) + (3 kg) + (8 kg)] (3.60555 m)2 = 221 kg m2 . 1 Question 2, chap -1, sect -1. part 2 of 2 10 points 6m O x Find the...

Register Now

#### Unformatted Document Excerpt

Coursehero >> Texas >> University of Texas >> PHY 58235

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
24 oldhomewk PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am we obtain Question 1, chap -1, sect -1. part 1 of 2 10 points Four particles with masses 3 kg, 3 kg, 3 kg, and 8 kg are connected by rigid rods of negligible mass as shown. 8 kg 3 kg y Iz = [m1 + m2 + m3 + m4 ] r 2 = [(3 kg) + (3 kg) + (3 kg) + (8 kg)] (3.60555 m)2 = 221 kg m2 . 1 Question 2, chap -1, sect -1. part 2 of 2 10 points 6m O x Find the rotational energy of the system. Correct answer: 7072 J (tolerance 1 %). Explanation: The rotational energy of the system is 3 kg 4m 3 kg K= 1 I 2 2 1 = (221 kg m2 ) (8 rad/s)2 2 = 7072 J . The origin is at the center of the rectangle, which is 4 m wide and 6 m long. If the system rotates in the xy plane about the z axis (origin, O) with an angular speed of 8 rad/s, calculate the moment of inertia of the system about the z axis. Correct answer: 221 kg m2 (tolerance 1 %). Explanation: Let : m1 = 3 kg , m2 = 3 kg , m3 = 3 kg , m4 = 8 kg , w = 4 m, = 6 m, From I= j 2 mj rj , Question 3, chap 12, sect 5. part 1 of 2 10 points The rigid object shown is rotated about an axis perpendicular to the paper and through center point O. The total kinetic energy of the object as it rotates is 6.2 J. 60 kg 4m O 8m 60 kg What is the moment of inertia of the object? Neglect the mass of the connecting rods and treat the masses as point masses. Correct answer: 5760 kg m2 (tolerance 1 %). Explanation: 4m 30 kg 8m top left bottom left bottom right top right and 30 kg where in this case all distances are equal to r= w 2 2 + 2 2 2 (6 m) (4 m) + = 2 2 = 3.60555 m , 2 oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am 2 Let : M = 30 kg , L = 4 m , and K = 6.2 J . Basic Concepts: I= 2 mi ri an angle of 59 with the horizontal. The rod is released from rest at an angle of 59 with the horizontal, as shown in the gure below The acceleration of gravity is 9.8 m/s2 . Hint: The moment of inertia of the rod 1 about its center-of-mass is Icm = m 2 . 12 Solution: The moment of inertia of the rigid body is given by I= 2 mi ri 59 = 2 2 M (L)2 + 2 M (2 L)2 = 12 M L2 = 12 (30 kg) (4 m)2 = 5760 kg m2 . Question 4, chap 12, sect 5. part 2 of 2 10 points What is the angular velocity of the object? Correct answer: 0.046398 rad/s (tolerance 1 %). Explanation: Since the rotational kinetic energy is given 1 by KR = I 2 , we may then solve for . 2 = = 2 KR I 2 (6.2 J) (5760 kg m2 ) Solution: O 2m 1.4 kg 2m What is the angular speed of the rod at the instant the rod is in a horizontal position? Correct answer: 2.78461 rad/s (tolerance 1 %). Explanation: Let : = 2 m , d = = 2 m, = 59 , and m = 1.4 kg . Basic Concepts: KR = 1 I 2 2 =rF =I Ki + Ui = Kf + Uf . = 0.046398 rad/s . Question 5, chap 12, sect 5. part 1 of 1 10 points A uniform rod of mass 1.4 kg is 2 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 2 m from the center of mass of the rod. Initially the rod makes I = Icm + m d2 1 = m 2 + m 2 12 13 m 2 = 12 13 (1.4 kg) (2 m)2 = 12 = 6.06667 kg m2 . (1) oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am Since the rod is uniform, its center of mass is located a distance from the pivot. The vertical height of the center of mass above horizontal is sin . Using conservation of energy and substituting I from Eq. 1, we have Kf = Ui 1 I 2 = m g sin 2 1. I = 4 3 5 2. I = 3 3 3. I = 2 5 4. I = 4 m L2 correct m L2 m L2 m L2 3 13 m 2 2 = m g sin 24 24 g sin 2 = 13 24 g sin = 13 = 24 (9.8 m/s2 ) sin(59 ) 13 (2 m) 5. None of these. (2) 6. I = 13 m L2 12 7. I = L2 Explanation: Basic Concepts: The moment of inertia, the center of mass and the energy conservation. The rotational kinetic energy is Krot = 1 I 2 . 2 Solution: The moment of inertia of the rod, Irod , with respect to the pivot point is Irod = 1 m L2 , 3 = 2.78461 rad/s . Question 6, chap 12, sect 5. part 1 of 2 10 points Consider a rod of length L and mass m which is pivoted at one end. The moment of inertia of the rod about an end is Irod = 1 m L2 . An object with mass m is attached 3 to the free end of the rod. The acceleration of gravity g = 9.8 m/s2 . Note: Contrary to the diagram shown below, consider the mass at the end of the rod to be a point particle. and the moment of inertia Im of the mass m with respect to the pivot point is Imass = m L2 . Then, the moment of inertia of the system I is I = Irod + Im 1 = m L2 + m L2 3 4 = m L2 . 3 Question 7, chap 12, sect 5. part 2 of 2 10 points m C L 25 m Note: The length C in the gure represents the location of the center-of-mass of the rod plus mass system (but is not drawn to scale). Determine the position of the center of mass from the pivot point; i.e., nd C . 1. C = 5 L 8 Determine the moment of inertia, I, of the system with respect to the pivot point. 2. None of these. oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am 7 L 8 1 4. C = L 2 3. C = 5. C = L 3 L correct 4 Explanation: 6. C = where m 4 dm , and dm = y dx , where mass is the areal density of the plate. area Solution: Let (x1 , y1 ) = (0 m, 0 m) (x2 , y2 ) = (2 m, 0 m) (x3 , y3 ) = (0 m, 4 m) . The equation for the hypotenuse is y y2 y3 y2 = . x x2 x3 x 2 The slope of the hypotenuse is s= y3 y2 x3 x2 4m0m = 2 . = 0m2m 1 The center of mass of the rod is L. Then, 2 the center of mass of the rod plus mass system is 1 L+L m 2 C= m+m 1 1 = L+ L 4 2 3 = L. 4 Question 8, chap 12, sect 5. part 1 of 2 10 points (1) Rewriting the equation, we have y = s (x x2 ) + y2 = (2) (x 2 m) + 0 m = s (x x2 ) , (2) A uniform at plate of metal is situated in the reference frame shown in the gure below. 5 4 3 2 1 0 0 12 3 4 5 6 7 8 9 10 x (m) (m) where y2 = 0 . The x-coordinate of the center of mass is x2 xcm = x2 x1 x1 x y dx x2 y dx x s (x x2 ) dx y = x1 x2 s (x x2 ) dx x1 x2 Calculate the x coordinate of the center of mass of the metal plate. Correct answer: 0.666667 m (tolerance 1 %). Explanation: Basic Concept: The center of mass xcoordinate is x dm x m , x (x x2 ) dx = 0 x2 (x x2 ) dx 0 13 1 x (x2 ) x2 3 2 = 12 x (x2 ) x 2 13 1 x2 (x2 ) x2 2 2 =3 12 x (x2 ) x2 22 x2 0 oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am = x3 2 3 x2 2 1 = x2 3 1 = (2 m) 3 = 0.666667 m . = 0.25 kg/m2 , (3) 5 Vertical Center-of-Mass: Using Eq. 3, the y-coordinate of the center of mass of the metal plate is 1 y = y3 3 1 = (4 m) 3 = 1.33333 m . Question 9, chap 12, sect 5. part 2 of 2 10 points Given: The mass of the plate is m = 1 kg . Calculate the moment of inertia of the triangle with the y-axis as the axis of rotation. Correct answer: 0.666667 kg m2 (tolerance 1 %). Explanation: m Areal Density: The areal density = A of the plate is m = x2 y dx x1 y3 from Eq. 1. The area of a x2 1 1 triangle is A = (base) (height) = x2 y3 , 2 2 which agrees with Eq. 4. Conventional Solution: The moment of x=x inertia Iy cm of the triangle about a y-axis through its center of mass x-coordinate can be accomplished in two steps. where s = x=x First: The moment of inertia Iy cm about its center of mass (xcm , ycm ) is determined. Second: The parallel axis theorem is used. The moment of inertia about the center of x=x mass Iy cm is x=x Iy cm x2 = x1 (x xcm )2 dm x2 = x1 (x xcm )2 y dx (x xcm )2 (x x2 ) dx x3 [2 xcm + x2 ] x2 + +[x2 2 xcm x2 ] x cm [x2 x2 ] dx cm x2 = s 0 x2 = s 0 = s = s 0 m x2 (x x2 ) dx m x 14 1 x [2 xcm + x2 ] x3 4 3 1 + [x2 + 2 xcm x2 ] x 2 cm x2 0 = = = = = 2 12 x (x2 ) x 2 0 m 12 y3 x (x2 ) x2 x2 22 m m = 1 y3 2 1 x x2 y 3 2 x2 2 2 2m x2 y 3 2 (1 kg) (2 m) (4 m) [x2 x2 ] x cm 14 1 = s x [2 xcm + x2 ] x3 2 42 3 1 + x2 + 2 xcm x2 x2 2 2 cm x2 x2 cm 2 = (4) y3 14 x x2 12 2 1 1 2 + xcm x3 x2 x2 2 cm 3 2 12 2 x2 xcm x2 + x2 =m cm 6 3 2m x2 y 3 s oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am =m 12 22 12 x x+ x 62 92 92 (5) Question 10, chap 12, sect 5. part 1 of 1 10 points 6 1 m x2 = 2 18 1 (1 kg) (2 m)2 = 18 = 0.222222 kg m2 . using Eq. 1 for s , Eq. 4 for , and Eq. 3 for xcm . Using the parallel axis theorem, we have x=0 x=x Iy = Iy cm + m x2 cm 1 = m x2 + m x2 2 cm 18 12 12 =m x+ x 18 2 9 2 1 = m x2 2 6 1 = (1 kg) (2 m)2 6 Assume: The masses are point particles; e.g., neglect the contribution due to moments of inertia about their center of mass. Three spherical masses are located in a plane at the positions shown in the gure below. A has mass 32.8 kg, B has mass 10.6 kg, and C has mass 35.9 kg. Three Masses in a Plane 10 9 8 y Distance (m) 7 6 5 4 3 2 1 0 A C B = 0.666667 kg m2 . Standard Solution: The moment of inerx=0 tia Iy of the triangle about the x-coordinate is x=0 Iy x2 x1 x2 dm x2 = x1 x2 y dx x2 (x x2 ) dx x x2 = s 0 2 14 1 3 = s x (x2 ) x 4 3 0 14 14 = s x x 42 32 y3 1 2m x4 = x2 y 3 x2 12 2 1 = m x2 2 6 1 = (1 kg) (2 m)2 6 3 4 5 6 7 8 9 10 x Distance (m) Figure: Drawn to scale. Calculate the moment of inertia (of the three masses) with respect to an axis perpendicular to the xy plane and passing through x = 3.5 m and y = 6.5 m . Correct answer: 425.125 kg m2 (tolerance 3 %). Explanation: Basic Concepts: The moment of inertia is I= i 2 mi ri , 2 mi (x2 + yi ) . i i 0 1 2 (1) (2) = 0.666667 kg m2 , using Eq. 1 for s and Eq. 4 for . Note: This problem has a dierent triangle for most students. = Solution: Using the knot (3.5 m, 6.5 m) as the coordinate origin (see gure below), we oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am have xa = (0.5 m) (3.5 m) = 3 m , ya = (7 m) (6.5 m) = 0.5 m , ra = 2 x2 + ya = 3.04138 m , a 2 2 = ma [x2 + ya ] + mb [x2 + ya ] a b 2 + mc [x2 + ya ] c = (32.8 kg) [(3 m)2 + (0.5 m)2 ] + (10.6 kg) [(2.5 m)2 + (0.5 m)2 ] + (35.9 kg) [(1 m)2 + (0.5 m)2 ] = 425.125 kg m2 . 7 xb = (6 m) (3.5 m) = 2.5 m , yb = (5.5 m) (6.5 m) = 1 m , rb = 2 x2 + yb = 2.69258 m , b xc = (4.5 m) (3.5 m) = 1 m , yc = (7 m) (6.5 m) = 0.5 m , rc = x2 c 2 + yc = 1.11803 m , Question 11, chap 13, sect 1. part 1 of 1 10 points Given: A circular shaped object with an inner radius of 14 cm and an outer radius of 28 cm. There are three forces (acting perpendicular to the axis of rotation) whose magnitudes are 13 N, 22 N, and 15 N acting on the object, as shown in the gure. The force of magnitude 22 N is 21 below horizontal. 13 N 28 cm 21 22 N 15 N 6 kg Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 4.76 N m (tolerance 1 %). (1) 14 cm Three Masses in a Plane 3 2 1A y Distance (m) 0 -1 -2 -3 -4 -5 -6 -3 -2 -1 0 1 2 3 4 5 x Distance (m) Figure: Drawn to scale. Using Eqs. 1, we have I= i 2 2 2 = ma ra + mb rb + mc rc = (32.8 kg) (3.04138 m)2 + (10.6 kg) (2.69258 m)2 + (35.9 kg) (1.11803 m)2 = 425.125 kg m2 . 2 mi ri , C B 6 Explanation: Alternate Solution: Using Eqs. 2, we have I= i 2 mi [x2 + yi ] , i (2) Let : a = 14 cm , b = 28 cm , F1 = 13 N , F2 = 22 N , F3 = 15 N , and = 21 . oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am b F1 R = 20.6 cm , m1 = 1 kg , m2 = 1.64 kg , h = 1.2 m , v = R, 1 I = M R2 , and 2 1 1 Kdisk = I 2 = M v 2 . 2 4 Consider the free body diagrams T2 T1 8 F2 a The total torque is M F3 = a F2 b F1 b F3 = (14 cm) (22 N) (28 cm) (13 N) + (15 N) = 4.76 N m = 4.76 N m . Question 12, chap 13, sect 2. part 1 of 1 10 points An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on its rim, which is a distance 20.6 cm from the axle. The mass on the right is 1 kg and on the left is 1.64 kg. The acceleration of gravity is 9.8 m/s2 . 1.64 kg m2 g 1 kg m1 g a Basic Concepts: The net acceleration a = r is in the direction of the heavier mass m1 . The pulleys mass is concentrated on the rim, so I = m r2 For the pulley, net = counter clock = I a = mra. r (1) 20.6 cm 2.3 kg 1.64 kg 1.2 m 1 kg What is the magnitude of the linear acceleration a of the hanging masses? Correct answer: 1.26964 m/s2 (tolerance 1 %). Explanation: Let : M = 2.3 kg , T1 r T2 r = (m r 2 ) Dividing by r gives m a = T1 T2 For the mass m1 , Fnet = m1 a = m1 g T1 T1 = m1 g m1 a For the mass m2 , Fnet = m2 a = T2 m2 g T2 = m2 a + m2 g Substituting (2) and (3) into (1) gives ma = m1 g m1 a m2 a m2 g a (2) (3) oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am ma + m1 a + m2 a = m1 g m2 g (m1 m2 ) g a= . m + m1 + m2 Question 13, chap 13, sect 2. part 1 of 1 10 points A uniform 2 kg rod with length 42 m has a frictionless pivot at one end. The rod is released from rest at an angle of 28 beneath the horizontal. Note: The moment of inertia of a rod about 1 the center of mass is m L2 , where m is the 12 mass of the rod and L is the length of the rod. The moment of inertia of a rod about either 1 end is m L2 . 3 The acceleration of gravity is 9.8 m/s2 . 9 The rods moment of inertia about its end1 point is I = m L2 . Therefore, the angular 3 acceleration of the rod is I 3g = cos 2L 3 (9.8 m/s2 ) cos(28 ) = 2 (42 m) = 0.309032 rad/s2 . = Question 14, chap 13, sect 2. part 1 of 1 10 points A uniform rod of mass 1.3 kg is 6 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 6 m from the center of mass of the rod. The rod is released from rest at the horizontal position. The acceleration of gravity is 9.8 m/s2 . Hint: The moment of inertia of the rod 1 m 2 . about its center-of-mass is Icm = 12 6m 6m O 1.3 kg 21 m 42 2k m g 28 Note: The rod is initially at rest at 28 below the horizontal. What is the angular acceleration of the rod immediately after it is released? Correct answer: 0.309032 rad/s2 (tolerance 1 %). Explanation: Basic Concepts: The torque equation is = r F = r F = I . The torque exerted by gravity on the rod is 1 = m g L cos . 2 Let : m = 2 kg , L = 42 m , = 28 . 63 What is the angular acceleration of the rod at the instant the rod makes an angle of 63 with the horizontal? Correct answer: 0.684478 rad/s2 (tolerance 1 %). Explanation: and oldhomewk 24 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am 10 Let : = 6 m , d = = 6 m, = 63 , and m = 1.3 kg . Basic Concepts: 1 I 2 2 =rF =I KR = Solution: I = Icm + m d2 1 = m 2 + m 2 12 13 m 2 = 12 13 (1.3 kg) (6 m)2 = 12 = 50.7 kg m2 . Since the rod is uniform, its center of mass is located at . Recalling that the weight m g acts at the center of mass, the magnitude of the torque at = 63 is =rF = mg = m g sin(90 ) = m g cos Equating the above with = I and solving for gives m g cos = I . Therefore = m g cos I m g cos = 13 m 2 12 12 g cos = 13 12 (9.8 m/s2 ) cos(63 ) = 13 (6 m) = 0.684478 rad/s2 .
Textbooks related to the document above:
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

University of Texas - PHY - 58235
oldhomewk 25 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am Question 1, chap 13, sect 1. part 1 of 1 10 points A rod can pivot at one end and is free to rotate without friction about a vertical axis, as shown. A force F is applied at the other end, at a
University of Texas - PHY - 58235
oldhomewk 26 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am Question 1, chap 13, sect 3. part 1 of 2 10 points A wooden block of mass M hangs from a rigid rod of length having negligible mass. The rod is pivoted at its upper end. A bullet of mass m tra
University of Texas - PHY - 58235
oldhomewk 27 PAPAGEORGE, MATT Due: Apr 1 2008, 4:00 am momentum I of the wheel is Question 1, chap 13, sect 4. part 1 of 3 10 points A 3 kg bicycle wheel rotating at a 2262 rev/min angular velocity has its shaft supported on one side, as shown in t
University of Texas - PHY - 58235
oldmidterm 04 PAPAGEORGE, MATT Due: Apr 29 2008, 4:00 am Question 1, chap 15, sect 1. part 1 of 1 10 points Hint: Write down equations for x(t) and v(t) and use sin2 + cos2 = 1 to calculate . A mass attached to a spring executes simple harmonic mot
University of Texas - PHY - 58235
Morby, Grant Quiz 4 Due: May 3 2006, 11:00 pm Inst: Drummond This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of
University of Texas - PHY - 58235
homework 28 PAPAGEORGE, MATT Due: Apr 8 2008, 4:00 am Question 1, chap 14, sect 2. part 1 of 3 10 points Consider a uniform ladder leaning against a smooth wall and resting on a smooth oor at point P . There is a rope stretched horizontally, with o
University of Texas - PHY - 58235
homework 29 PAPAGEORGE, MATT Due: Apr 10 2008, 4:00 am Question 1, chap 14, sect 1. part 1 of 4 10 points A uniform brick of length 38 cm is placed over the edge of a horizontal surface with the maximum overhang x possible without falling. 38 cm g
University of Texas - PHY - 58235
homework 30 PAPAGEORGE, MATT Due: Apr 12 2008, 4:00 amWith Particle Without Particle1Question 1, chap 15, sect 1. part 1 of 3 10 points A weight suspended from a spring is seen to bob up and down over a distance of 15 cm twice each second. Wha
University of Texas - PHY - 58235
homework 31 PAPAGEORGE, MATT Due: Apr 15 2008, 4:00 am Question 1, chap 15, sect 5. part 1 of 3 10 points A uniform disk of radius 0.1 m and 9.1 kg mass has a small hole a distance from the disks center that can serve as a pivot point. Since 0. 1
University of Texas - PHY - 58235
homework 32 PAPAGEORGE, MATT Due: Apr 17 2008, 4:00 am A sin( t) ; therefore Question 1, chap 15, sect 2. part 1 of 1 10 points Simple harmonic motion can be described using the equation y = A sin(k x t ) . Hint: sin() = sin . Consider the sim
University of Texas - PHY - 58235
homework 33 PAPAGEORGE, MATT Due: Apr 19 2008, 4:00 am Question 1, chap 16, sect 2. part 1 of 2 10 points A steel piano wire is 0.7 m long and has a mass of 60 g. It is stretched with a tension of 600 N. What is the speed of transverse waves on the
University of Texas - PHY - 58235
homework 34 PAPAGEORGE, MATT Due: Apr 22 2008, 4:00 am Explanation: Question 1, chap 16, sect 4. part 1 of 2 10 points A standing wave of frequency 5 hertz is set up on a string 2 meters long with nodes at both ends and in the center, as shown. 2 m
University of Texas - PHY - 58235
homework 35 PAPAGEORGE, MATT Due: Apr 24 2008, 4:00 am Question 1, chap 16, sect 2. part 1 of 3 10 points Two waves in one string are described by the relationships so y1 = A1 cos(k1 x 1 t) where A1 = 3.4 cm, A2 = 4.9 cm, k1 = 6 cm1 , k2 = 4 cm1 ,
University of Texas - PHY - 58235
homework 36 PAPAGEORGE, MATT Due: Apr 26 2008, 4:00 am The sound level at L2 is Question 1, chap 17, sect 2. part 1 of 1 10 points An explosive charge is detonated at a height of several kilometers in the atmosphere. At a distance of 256 m from the
University of Texas - PHY - 58235
Morby, Grant Homework 27 Due: Apr 14 2006, noon Inst: Drummond This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - 58235
Morby, Grant Homework 28 Due: Apr 14 2006, noon Inst: Drummond This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - 58235
Morby, Grant Homework 29 Due: Apr 17 2006, noon Inst: Drummond This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - 58235
Morby, Grant Homework 30 Due: Apr 19 2006, noon Inst: Drummond This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of
University of Texas - PHY - 58235
Morby, Grant Homework 31 Due: Apr 21 2006, noon Inst: Drummond This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - 58235
Morby, Grant Homework 32 Due: Apr 24 2006, noon Inst: Drummond This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - 58235
Morby, Grant Homework 33 Due: Apr 26 2006, noon Inst: Drummond This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - 58235
Morby, Grant Homework 34 Due: Apr 28 2006, noon Inst: Drummond This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 o
University of Texas - PHY - 58235
Morby, Grant Homework 35 Due: May 1 2006, noon Inst: Drummond This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of
University of Texas - PHY - 58235
Morby, Grant Homework 36 Due: May 3 2006, noon Inst: Drummond This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of
University of Texas - PHY - 58235
midterm 04 PAPAGEORGE, MATT Due: Apr 30 2008, 11:00 pm1Mechanics - Basic Physical ConceptsMath: Circle: 2 r, r2 ; Sphere: 4 r2 , (4/3) r3 b2 Quadratic Eq.: a x2 + b x + c = 0, x = b 2 a4 a cCartesian and polar coordinates: y x = r cos ,
University of Texas - PHY - 58235
midterm 04 KELLERMANN, MARC Due: Dec 6 2006, 11:00 pmGravity F21 = G m12m2 r12 ,r121for r R,g(r) = G M r2G = 6.67259 1011 N m2 /kg2 Rearth = 6370 km, Mearth = 5.98 1024 kg2 Circular orbit: ac = v = 2 r = r2 2 r = g(r) T M U = G m
University of Texas - PHY - 58235
oldhomewk 28 PAPAGEORGE, MATT Due: Apr 6 2008, 4:00 am Question 1, chap 14, sect 3. part 1 of 2 10 points A solid bar of length L has a mass m1 . The bar is fastened by a pivot at one end to a wall which is at an angle with respect to the horizont
University of Texas - PHY - 58235
oldhomewk 29 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am Question 1, chap 14, sect 1. part 1 of 1 10 points A ladder is leaning against a smooth wall. There is friction between the ladder and the oor, which may hold the ladder in place; the 1 ladder
University of Texas - PHY - 58235
oldhomewk 30 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am and the period of oscillation is Question 1, chap 15, sect 1. part 1 of 2 10 points A 1.83 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose forc
University of Texas - PHY - 58235
oldhomewk 31 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am Question 1, chap 15, sect 1. part 1 of 2 10 points Two massless springs with spring constants 744 N/m and 5022 N/m are hung from a horizontal support. A block of mass 2 kg is suspended from th
University of Texas - PHY - 58235
oldhomewk 32 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am Question 1, chap 15, sect 3. part 1 of 3 10 points A mass of 357 g connected to a light spring of force constant 12.7 N/m oscillates on a horizontal, frictionless track. The amplitude of the m
University of Texas - PHY - 58235
oldhomewk 33 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am Question 1, chap 16, sect 1. part 1 of 1 10 points When a particular wire is vibrating with a frequency of 3.8 Hz, a transverse wave of wavelength 76.5 cm is produced. Determine the speed of w
University of Texas - PHY - 58235
oldhomewk 34 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am Question 1, chap 16, sect 2. part 1 of 1 10 points The distance between two successive maxima of a certain transverse wave is 1.52 m. Eight crests, or maxima, pass a given point along the dire
University of Texas - PHY - 58235
oldhomewk 35 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am Question 1, chap 16, sect 4. part 1 of 1 10 points Two guitar strings, of equal length and linear density, are tuned such that the second harmonic of the rst string has the same frequency as t
University of Texas - PHY - 58235
oldhomewk 36 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am From Eq. (4), we have Question 1, chap 17, sect 3. part 1 of 1 10 points Given: The speed of sound in air is 344 m/s. A student uses an audio oscillator of adjustable frequency to measure the
University of Texas - PHY - 58235
oldhomewk 37 PAPAGEORGE, MATT Due: Apr 28 2008, 4:00 am Question 1, chap 18, sect 3. part 1 of 1 10 points In testing a new material for shielding spacecraft, 149 small ball bearings, each moving at a supersonic speed of 415.7 m/s, collide head-on
University of Texas - PHY - 58235
homework 15 MOHAJER, CHRISTOPHER Due: May 6 2008, 4:00 am Question 1, chap 18, sect 2. part 1 of 1 10 points A cowboy at a dude ranch lls a horse trough that is 1.6 m long, 64 cm wide, and 47 cm deep. He uses a 2.1 cm diameter hose from which water
University of Texas - PHY - 58235
quizb 01 MOHAJER, CHRISTOPHER Due: Mar 5 2008, 11:00 pm1Mechanics - Basic Physical ConceptsMath: Circle: 2 r, r2 ; Sphere: 4 r2 , (4/3) r3 b2 Quadratic Eq.: a x2 + b x + c = 0, x = b 2 a4 a cCartesian and polar coordinates: y x = r cos ,
University of Texas - PHY - 58235
quizb 02 MOHAJER, CHRISTOPHER Due: Apr 2 2008, 11:00 pm1Mechanics - Basic Physical ConceptsMath: Circle: 2 r, r2 ; Sphere: 4 r2 , (4/3) r3 b2 Quadratic Eq.: a x2 + b x + c = 0, x = b 2 a4 a cCartesian and polar coordinates: y x = r cos ,
University of Texas - PHY - 58235
quizb 03 MOHAJER, CHRISTOPHER Due: Apr 30 2008, 11:00 pm1Mechanics - Basic Physical ConceptsMath: Circle: 2 r, r2 ; Sphere: 4 r2 , (4/3) r3 b2 Quadratic Eq.: a x2 + b x + c = 0, x = b 2 a4 a cCartesian and polar coordinates: y x = r cos
University of Texas - PHY - 58235
nal 01 MOHAJER, CHRISTOPHER Due: May 8 2008, 1:00 pm1Mechanics - Basic Physical ConceptsMath: Circle: 2 r, r2 ; Sphere: 4 r2 , (4/3) r3 b2 Quadratic Eq.: a x2 + b x + c = 0, x = b 2 a4 a cCartesian and polar coordinates: y x = r cos , y
UNC - ECON - 101
Econ 101 Class 2 Questions 1. It is hot at the football game and Joe is thirsty. He would pay up to \$5.00 for a 20 oz cola. He goes to the concession stand and finds that he can buy a 20 oz cola for \$3.00. His economic surplus from the transaction is
UNC - ECON - 101
Econ 101 Class 3 Questions 1. On Saturday morning, you have decided to wash, wax, and detail your car. The improvements to your cars appearance are worth \$50 to you. You have decided to do the work yourself because the next best use of your time is w
UNC - ECON - 101
Econ 101 Class 4 Questions 1. Which of the following best explains the connection between the principle of comparative advantage and Ari and Sams Production Possibilities Frontier (PPF)? A. B. C. D. Sam and Ari are inside the PPF unless Sam spends al
UNC - ECON - 101
Econ 101 Class 5 Questions1. The quantity demanded of a slice of pepperoni pizza falls as the price of the slice rises because A. The opportunity cost of the slice grows smaller. B. The opportunity cost of the slice grows larger. * C. Buyers have s
UNC - ECON - 101
Econ 101 Class 6 Questions 1. Given the displayed demand and supply data, equilibrium price and quantity are: A. \$3.00, 200 B. \$1.50, 200 C. \$2.25, 125 D. \$2.25, 75 E. None of the Above *2. Which of the following correctly describes the effects of
UNC - ECON - 101
Econ 101 Class 7 Questions 1. An increase in the price of gasoline will cause a (an) _ in the equilibrium price and a (an) _ in the equilibrium quantity of new SUVs relative to small engine cars. A. B. C. D. Decrease, decrease * Decrease, increase In
UNC - ECON - 101
Econ 101 Class 8 Questions 1. Over the next several years, it is reasonable to predict that the ratio of larger to smaller vehicles made in the U.S. will A. Rise as the market price of larger cars falls. B. Rise as a result of the increased cost of g
UNC - ECON - 101
Econ 101 Class 9 Questions 1. Which is correct? The supply schedule for pizza is upward sloping because A. B. C. D. As price rises, pizzeria owners hire more labor to supply pizza. As price rises, business people open new pizzerias. To increase pizza
UNC - ECON - 101
Econ 101 Class 10 Questions 1. Agostino released some of his catch because he believed that, at the quantity demanded associated with his total catch of Tuna, the demand schedule for Tuna was _ and releasing some of his catch would _ his revenue. A.
UNC - ECON - 101
Econ 101 Class 12 Questions1. Reducing the Supply of Drugs is not likely to lower property crime because demand for drugs is _ and lowering the supply will _ the amount that users spend on drugs.A. B. C. D. Elastic, Raise Elastic, Lower Inelastic,
UNC - ECON - 101
Econ 101 Class 13 Questions 1. Durham gets water from three sources. The table shows cost and capacity of each source. Source Spring Lake Michie Kerr Lake Capacity (millions of gallons per day 2 2 2 Cost per Gallon \$0.01 \$0.015 \$0.03If Durham resid
UNC - ECON - 101
Econ 101 Class 14 Questions 1. Sellers are more likely to bear the burden of a tax imposed on _ because _. A. B. C. D. Tobacco products, because demand is highly inelastic. Tobacco products, because farmers can easily switch to other crops. Imported
UNC - ECON - 101
Econ 101 Class 15 Questions1. Scott Adams loves to draw and has created a well liked cartoon series called Dilbert. When he first started out, Scott earned \$3,000 per month drawing Dilbert for subscribing newspapers. Dilbert became popular and now
UNC - ECON - 101
Econ 101 Class 16 Questions1. A. B. C. D. 2. A. B. C. D.Which of the following soft drink sellers is likely to have the most market power? Concession Area at the Smith Center. * Franklin Street Snack Shop. Lenoir Dining Hall. Food Court at a Shop
UNC - ECON - 101
Econ 101 Class 17 Questions 1. Which of the following correctly explains the effect of imperfect competition on economic efficiency? Imperfect Competition A. Enhances efficiency by providing consumers with specific brands. B. Reduces efficiency becau
UNC - ECON - 101
Econ 101 Class 18 Questions 1. The following table gives the value of a mill and the value of a fishery as a function of the mills pollution decision Pollution Decision High Discharge Rate Low Discharge Rate Value of Mill \$5000 \$1000 Value of Fishery
UNC - ECON - 101
Econ 101 Class 19 Questions1.Given the following cost data and a SO2 reduction target of 1200 tons, firm _ will buy permits to pollute and firm _ will sell them.SO2 Reduction Marginal Cost per Ton of SO2 Reduction Firm A 0-200 201-400 401-600 60
UNC - ECON - 101
Econ 101 Class 22 Questions 1. The U. S. economy grows primarily because A. The inflation rate remains low. B. U.S. exports are an engine of growth. C. Labor productivity continues to increase. * D. Recessions have been infrequent and mild.2. In 20
UNC - ECON - 101
Econ 101 Class 23 Questions 1. If the CPI of Simple Land is 1.00 in 2005, 1.07 in 2006, and 1.05 in 2007 then A. B. C. D. Simple Land experienced inflation in 2006 and deflation in 2007. * Prices in Simple Land rose by about 12 percent between 2005 a
UNC - ECON - 101
Econ 101 Class 24 Questions1.In accounting for growth in US output per capita A. Growth in labor productivity is more important. * B. Growth in the fraction of the population is more important. C. Neither labor productivity nor changes in the wil