This print-out should have 29 questions. A ball is thrown upward. At a height of 10 Multiple-choice questions may continue on meters above the ground, the ball has a po- the next column or page - find all choices tential energy of 50 Joules (with the potential before answering energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 001 10.0 points Joules. A rock of mass m is thrown horizontally off What is the maximum height & reached by a building from a height A. The speed of the the ball? Consider air friction to be negligible. rock as it leaves the thrower's hand at the edge of the building is up, as shown. 1. h = 30 m 2. h = 20 m 10 - 3. h = 40 m 4. h == 10m 5. he 50 m 004 10.0 points An ore car of mass 36000 kg starts from rest What is the kinetic energy of the rock just and rolls downhill on tracks from a mine. At before it hits the ground? the end of the tracks, 21 m lower vertically, is a horizontally situated spring with constant 1. K= = muj 3.3 x 10" N/m. 2. K= mgh The acceleration of gravity is 9.8 m/s?. 3. K = , muj - mgh Ignore friction. How much is the spring compressed in stop- A. K = mgh - = mu ping the ore car? 5. Kj = - muitmgh 1. 5.28132 2. 6.4796 002 10.0 points 3. 6.99391 A simple pendulum of mass m swings at the 4. 9.2481 end of a string of length &. The highest point 5. 4.53652 the pendulum reaches is a distance d above 6. 3.50175 its lowest point. 7. 4.96867 What is the kinetic energy of the pendulum 8. 7.61682 9. 6.70088 when it reaches a spot a distance - above its 10. 6.90355 lowest point? Answer in units of m. 1. my (( - d) 005 (part 1 of 3) 10.0 points 2. mge A 176 g arrow is shot straight up into the air 3. Zero with a speed of 17 m/s. It reaches a maximum 4. mga height of 13.7449 m. 2 Find the initial kinetic energy. The accel- 5. = mv2 2 eration of gravity is 9.8 m/s" . 6. mgd 7. mgh 1. 11.56 8. 2mgd 2. 20.736 3. 25.432 003 10.0 points 4. 26.1725kolsawala (afk472) - 3-PHY317K-Homework - lang - (57345) 2 5. 32.798 3. W = 0J 6. 16.7865 4. W = -300 J 7. 28.611 5. It cannot be determined without knowing 8. 24.624 the distance the block slides. 9. 11.3715 10. 18.15 009 (part 1 of 2) 10.0 points Answer in units of J. A block weighing 49 N is lifted from A to C. C 006 (part 2 of 3) 10.0 points Find the potential energy at its highest posi- 49 N tion. 1. 38.9795 6 m 2. 34.83 3. 23.7072 A 4. 17.8593 5. 32.2371 49 N 49 N 6. 30.564 7. 24.5187 - 14 m 8. 21.4926 Find the work done against the gravity (the 9. 15.0273 work done by the external force which lifted 10. 32.7402 the block). Answer in units of J. 1. 45 007 (part 3 of 3) 10.0 points 2. 294 Find the magnitude of the energy lost due to 3. 186 air resistance. 4. 222 5. 145 1. 1.5092 6. 160 2. 1.6954 7. 93 3. 0.7938 8. 240 4. 1.568 9. 57 5. 1.4308 10. 150 6. 1.7248 Answer in units of J. 7. 1.6268 8. 0.8428 010 (part 2 of 2) 10.0 points 9. 1.0486 Find the difference AU = VA - Up of the 10. 1.3132 potential energies. Answer in units of J. 1. -66 008 10.0 points 2. -54 When a block slides a certain distance down 3. -80 an incline, the work done by gravity is 300 J. 4. -64 What is the work done by gravity if this 5. -315 block slides the same distance up the incline? 6. -294 7. -259 1. W = 300 J 8. -210 2. It cannot be determined without knowing 9. -132 the coefficient of friction. 10. -270kolsawala (afk472) - 3-PHY317K-Homework - lang - (57345) 3 Answer in units of J. mass at the surface of the Earth (with ra- dius 6.4 x 10" m and mass 6 x 10" kg) with 011 10.0 points that on the surface of the Moon (with mass In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational 1 81.3 - Me and radius 0.27 RE). constant G uses lead spheres of masses 1.61kg and 9.3g whose centers are separated by 5.47cm. What is the force on the Earth? Calculate the gravitational force between these spheres, treating each as a point mass located at 1. 444.558 the center of the sphere. The value of the universal 2. 307.771 gravitational constant is 3. 185.64 6.67259 x 10 1N - m/kg. 4. 425.017 5. 190.525 6. 346.853 1. 4.706350-10 7. 175.869 2. 1.12266e-09 8. 39.082 3. 1.53595e-09 9. 87.9346 4. 9.765860-10 10. 122.131 5. 4.399280-09 Answer in units of N. 6. 2.45379e-09 7. 1.85660-09 8. 1.19809e-09 014 (part 2 of 2) 10.0 points What is it on the Moon? 9. 1.82771e-09 10. 3.33909e-10 1. 51.929 Answer in units of N. 2. 0.824269 3. 66.7658 012 10.0 points 4. 8.24269 A satellite moves in a circular orbit around 5. 32.1465 the Earth at a speed of 6.8 km/s. 6. 42.0377 Determine the satellite's altitude above 7. 64.293 the surface of the Earth. Assume the 8. 81.6026 Earth is a homogeneous sphere of radius 9. 65.9415 6370 km and mass 5.98 x 1024 kg. The 10. 65.1173 value of the universal gravitational constant Answer in units of N. is 6.67259 x 10-1 N . m'/kgz. 015 10.0 points 1. 2790.26 2. 3371.72 Energy is required to move a 924 kg mass from the Earth's surface to an altitude 2.48 3. 9590.84 4. 8386.69 times the Earth's radius Re. 5. 2259.35 What amount of energy is required to ac- 6. 2011.03 complish this move? The acceleration of grav- 7. 8971.06 ity near the Earth is9.8 m/s' , the mass of the 8. 4713.91 Earth is 5.98 x 10" kg , and the radius of the Earth is 6.37 x 105 m. 9. 5092.83 10. 4010.36 1. 2.61094 x 1010 Answer in units of km. 2. 4.11064 x 1010 3. 1.28511 x 1010 013 (part 1 of 2) 10.0 points 4. 3.16017 x 1010 Compare the gravitational force on a 31.5 kgkolsawala (afk472) - 3-PHY317K-Homework - lang - (57345) 5. 5.44275 x 1010 6. 2.64709 x 1010 018 10.0 points 7. 2.6068 x 1010 Tom has a mass of 72.1 kg and Sally has a 8. 3.87031 x 1010 mass of 50.1 kg. Tom and Sally are standing 9. 7.27302 x 1010 21.7 m apart on a massless dance floor. Sally 10. 6.0508 x 1010 looks up and she sees Tom. She feels an Answer in units of J. attraction. If the attraction is gravitation, find its mag- 016 10.0 points nitude. Assume both can be replaced by point A black hole is an object so heavy that neither masses and that the gravitational constant is matter nor even light can escape the influence 6.67259 x 10 N . m /kg of its gravitational field. Since no light can escape from it, it appears black. Suppose a 1. 1.9971 x 10-10 mass approximately the size of the Earth's 2. 8.53468 x 10-10 mass 6.73 x 1024 kg is packed into a small 3. 5.11856 x 10-10 uniform sphere of radius r. 4. 2.34807 x 10-10 Use: 5. 2.16097 x 10-10 The speed of light c = 2.99792 x 10 m/s. 6. 3.30703 x 10-10 The universal gravitational constant G = 7. 6.81781 x 10-10 6.67259 x 10 Nm /kg. 8. 1.77841 x 10-10 Hint: The escape speed must be the speed 9. 8.77328 x 10-10 of light. 10. 3.05985 x 10-10 Based on Newtonian mechanics, determine Answer in units of N. the limiting radius ro when this mass (approx- imately the size of the Earth's mass) becomes 019 10.0 points a black hole. When you drop a 0.38 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s to- 1. 0.00825577 ward the earth's surface. According to New- 2. 0.00631062 ton's third law, the apple must exert an equal 3. 0.00803305 but opposite force on Earth. 4. 0.00999305 If the mass of the earth 5.98 x 102 kg, what 5. 0.010097 is the magnitude of the earth's acceleration 6. 0.00841911 toward the apple? 7. 0.00862699 8. 0.00775092 1. 6.06355 x 10-25 9. 0.00761729 2. 6.3913 x 10-25 10. 0.0113146 3. 5.89967 x 10 25 Answer in units of m. 4. 7.37458 x 10-25 5. 5.73579 x 10-25 017 10.0 points 6. 6.71906 x 10-25 If the earth were of uniform density, what 7. 6.22742 x 10-25 would be the value of g inside the earth at 8. 6.55518 x 10-25 half its radius? (The value of g at the surface 9. 7.04682 x 10-25 of earth is 9.8 m/s'.) 10. 7.2107 x 10-25 Answer in units of m/s'. 1. 19.6 m/s 2. 39.2 m/s- 020 10.0 points 3. 9.8 m/s- A railroad car of mass m is moving at speed 4. 4.9 m/s- when it collides with a second railroad car