homework assignment 4 1 judging space in seconds answers

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Seminar assignments - homework assignment 4 - questions + answers (a)

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rash (tgr353) Homework 4 (review for test) fischler (58920) This should have 50 questions. questions may continue on the next column or page find all choices before answering. 001 10 points A car rounds a curve while maintaining a constant speed. Is there a net force on the car as it rounds the curve? 1. Yes. correct 2. It depends on the sharpness of the curve and speed of the car. 3. No its speed is constant. Explanation: Acceleration is a change in the speed direction of an object. Thus, because its direction has changed, the car has accelerated and a force must have been exerted on it. 002 10 points If an object is moving in a circle at constant speed, what does Second Law inform us about the motion? X 1. F 0 since v for the object is constant. X 2. F points toward the center of the circle. correct X 3. F points in the direction the object is moving. X 4. F points in the opposite direction to the velocity. X 5. F points directly outward, away from the center of the circle. Explanation: An object moving in a circle at constant speed is accelerating toward the center of the X F circle. second law, a , M X so F must point toward the center of the 1 circle. 003 (part 1 of 2) 10 points Consider the dimension content of the centripetal force. The force should depend on the mass m, the tangential speed v and the radius r. We write it in the form F mx v y r z . Based on dimensional analysis, determine the powers x, y and z. 1. x y z 1. 2. x 1, y 2, z correct 3. x y z 4. x y 2, z 5. x 1, y z 6. x 1, y z 1. 7. x y 2, z 1. 8. x 1, y 2, z 1. Explanation: In order that F mx v y r z to be dimensionally correct: M 1 L1 T L Lz v y r z M x T M x T . equating the powers of M , L, and T x 1, y z 1, and Therefore, x 1, y 2, and z 004 (part 2 of 2) 10 points Consider the following set of equations, where s, s0 , x and r have units of length, t has units of time, v has units of velocity, g and a have units of acceleration, and k is dimensionless. Which one is dimensionally incorrect? r s a 1. t k correct g v rash (tgr353) Homework 4 (review for test) fischler (58920) 2 L2 T v2 2 L L T T s v t 0 v a L T 2. s s0 v t a v x 3. t a v is also consistent. r ksv s a 2 4. v 2 a s So only t k is dimensionally t g v incorrect. k v v2 5. a g t s0 005 (part 1 of 2) 10 points Explanation: In the Bohr model of the hydrogen atom, For an equation to be dimensionally corthe speed of the electron is approximately rect, all its terms must have the same units. v x 2 106 (1): t Find the central force acting on the electron a v as it revolves in a circular orbit of radius T 4 m. h v i h x i L T L a v L T L T Correct answer: 1 N. T Explanation: is consistent. 2 The centripetal force for the electron is kv v (2): a g t s0 v2 Fc m L T r k v v2 L T L2 T (9 kg)(2 106 LT t s0 T L 4 m LT 1 N is also consistent. r s a (3): t k g v T r r s a L L T k g v L T L T T T is not dimensionally consistent. ksv (4): v 2 2 a s t 2 L2 T ksv L T L L L T T t L2 T is also consistent. v2 a L (5): s s0 v t 006 (part 2 of 2) 10 points Find the centripetal acceleration of the electron. Correct answer: 1 1023 . Explanation: The centripetal acceleration for the electron is v2 r (2 106 4 m 1 1023 007 10 points A small ball of mass 37 g is suspended from a rash (tgr353) Homework 4 (review for test) fischler (58920) v2 r 4 π2 r m2 g m1 T2 2 4 π m1 r m2 g r m1 r T 2π m2 g s (8 kg) (6 m) 2π (50 kg) (9 ) frictionless hole in the center of the table. A second 50 kg mass is attached to the other end of the string. The acceleration of gravity is 9 . v 8 kg m2 g m1 6m 50 kg 1 s . Determine the period (the time for one revolution). Correct answer: 1 s. Explanation: Let : m1 m2 r g 8 kg , 50 kg , 6 m , and 9 . The linear velocity v can be expressed in terms of the distance it travels each revolution: 2πr . T Consider the forces acting on each mass: T N T m1 m2 The tension T in the string provides the centripetal force required to keep m1 moving in X a circular path. Applying Fr m ar to m1 , we obtain T m1 ac m1 X v2 r Fy m ay 0 to m2 , so m2 g T 0 4 and 011 10 points A satellite moves in a circular orbit around the Earth at a speed of 6 Determine the altitude above the surface of the Earth. Assume the Earth is a homogeneous sphere of radius 6370 km and mass 5 1024 kg. The value of the universal gravitational constant is 6 N m2 . Correct answer: 2790 km. Explanation: Let : v 6 , Re 6370 km , Me 5 1024 kg , and G 6 N m2 . The gravitational force provides the centripetal acceleration, so G m Me m v2 r2 r G Me v2 (6 N m2 ) 5 1024 kg 1 km (6 m 9160 km , rash (tgr353) Homework 4 (review for test) fischler (58920) G m ME 1 2 (0)2(81) RE FE (0)2(81) 254 N (0)2(81) and the height of the satellite above the surface is h r Re 9160 km 6370 km 2790 km . 012 (part 1 of 2) 10 points Compare the gravitational force on a 26 kg mass at the surface of the Earth (with radius 6 106 m and mass 6 kg) with that on the surface of the Moon with mass 1 ME and radius 0 RE . 81 What is the force on the Earth? Correct answer: 254 N. Explanation: Let : FE m 26 kg , ME 6 1024 kg , RE 6 106 m , and G 6 N m2 . G m ME 2 RE (6 N m2 ) (26 kg) (6 106 m)2 (6 1024 kg) 254 N . 013 (part 2 of 2) 10 points What is it on the Moon? Correct answer: 42 N. Explanation: Let : F MM 81 ME and RM 0 RE . Gm ME G m MM 2 (0 )2 81 RM 5 42 N . 014 10 points In another solar system is planet Driff, which has 5 times the mass of the earth and also 5 times the radius. How does the gravitational acceleration on the surface of Driff compare to the gravitational acceleration on the surface of the earth? 1 1. th as much. correct 5 2. the same, 10 . 3. 1 th as much. 25 4. 25 times as great. 5. There is no gravity on Driff because 5 times miles (the radius of the earth), is miles, far beyond the pull of gravity. 6. 5 times as much. Explanation: Let : MD 5 Me and RD 5 Re . Gravitational force is Mm r2 GM M g 2 2 , so r r F mg G mD r2 (5 me ) re2 MD re2 1 gD D 2 2 me ge me (5 re ) 5 me rD 2 re 1 gD ge . 5 rash (tgr353) Homework 4 (review for test) fischler (58920) universal law of gravitation, m1 m2 d2 (6 N m2 ) (7 kg) (0 kg) 0 m2 F 8 N . 018 (part 1 of 2) 10 points The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbit around the sun is at a distance 560 million km from the sun. What is the period of the orbit? Explanation: Te 1 year , re 150 million km , ra 560 million km . and From laws, Te2 Ta2 re3 ra3 ra Ta Te re 560 million km (1 year) 150 million km 7 year . 019 (part 2 of 2) 10 points What is the orbital velocity of the asteroid? Assume there are 365 days in one year. Correct answer: 15467 Explanation: va 2 π ra Ta 2 π (5 1011 m) 1 y 7 year 365 d 1d 1h 24 h 3600 s 15467 . 020 10 points A distant star has a single planet circling it in a circular orbit of radius 2 1011 m. The period of the motion about the star is 800 days. What is the mass of the star? The value of the universal gravitational constant is 6 N m2 . Correct answer: 2 1030 kg. Explanation: Correct answer: 7 year. Let : 7 Let : G 6 N m2 , RB 2 1011 m , and TB 800 day . 24 h TB (800 day) 1 day 7 6 10 s . 3600 s 1h According to explanation of third law 3 RB G Ms const. 2 4 π2 TB The mass of the star is thus Ms 3 4 π 2 RB G TB2 4 π2 6 N m2 (2 1011 m)3 (6 107 s)2 2 1030 kg . 021 10 points rash (tgr353) Homework 4 (review for test) fischler (58920) It takes a satellite 1 h to revolve around an unknown asteroid in a circular orbit of radius 510 km. How long it will take an object without initial speed to reach surface of the asteroid from this orbit? The radius of the asteroid is very much smaller than radius of the orbit and can be considered negligible. Correct answer: 0 h. Explanation: Trajectory of object when it falls from some height without initial velocity to the asteroid could be considered as extreme case of ellipse. The axis in this case is distance from initial point to the center of the asteroid and period is twice time of fall t. Therefore third law could be written as (2 r)3 r3 T2 (2 t)2 T 1 h t 0 h . 2 2 2 2 022 (part 1 of 2) 10 points On the way to the moon, the Apollo astronauts reach a point where the gravitational pull is stronger than that of Find the distance of this point from the center of the Earth. The masses of the Earth and the Moon are 5 1024 kg and 7 1022 kg, respectively, and the distance from the Earth to the Moon is 3 108 m. Explanation: Me 5 1024 kg , mm 7 1022 kg , Re 3 108 m . 2 rm Mm re2 Me r Mm rm . re Me It is also true that R re rm r Mm re re Me R r re Mm Me 3 108 m s 7 1022 kg 5 1024 kg 3 108 m . 023 (part 2 of 2) 10 points What would the acceleration of the astronaut be due to the gravity at this point if the moon was not there? The value of the universal gravitational constant is 6 N m2 . Correct answer: . Explanation: F G Me mm re2 (6 N m2 ) 5 1024 kg (3 108 m)2 Correct answer: 3 108 m. Let : 8 and Consider the point where the two forces are equal. If re is the distance from this point to the center of the Earth and rm is the distance from this point to the center of the Moon, then G m Me G m Mm 2 2 re rm . 024 10 points A runner is jogging at a steady 1 When the runner is 0 km from the finish line a bird begins flying from the runner to the finish line at 3 (3 times as fast as the runner). When the bird reaches the rash (tgr353) Homework 4 (review for test) fischler (58920) What will be his position from point A after that time? Correct answer: 48 Explanation: The initial speed vo 0, so the speed of the particle after t seconds is Correct answer: 573 m. Explanation: Let : 10 v v0 a t a t (6 )(7 s) d0 242 m . 48 . The position is d d0 242 m 331 m 573 m . 027 (part 1 of 2) 10 points A glacier advances at 3 How far will it move in s? Explanation: v 3 t s . Correct answer: 435 m. Explanation: Correct answer: 0 cm. Let : 030 10 points A sound wave, traveling at 339 is emitted the foghorn of a tugboat. An echo is heard 2 s later. How far away is the reflecting object? Let : and s v t (3 s) 0 cm . 028 (part 2 of 2) 10 points How far will it move in 3 y? Correct answer: 385 cm. Explanation: v 339 t 2 s . and If d is the distance to the reflecting object, the total distance traveled the wave is 2 d, so 2d t (339 s) vt 2 2 435 m . 031 10 points A vehicle moves in a straight line with an acceleration of 4 . how much does the speed change each second? Correct answer: s vt (3 (3 y) 365 d 24 h 3600 s 1y 1d 1h 385 cm . 029 10 points A particle accelerates from rest at 6 . What is its speed 7 s after the particle starts moving? Explanation: Let : a 4 1 s . and When a vehicle is moving in a straight line, 1h a (4 )(1 s) 3600 s . rash (tgr353) Homework 4 (review for test) fischler (58920) 032 11 a 10 points Henry hits a hockey puck in the positive at time, t t0 . The puck is then stopped a net starting at time, t t1 . Which of the following curves could describe the acceleration of the hockey puck if we ignore any effects of friction? 8. t1 t t0 a t0 t t1 9. a 1. t1 t t1 t Explanation: at t0 , sudden acceleration, v increases t0 t t1 , constant v , a 0 at t1 , sudden deceleration, v decreases to 0 The contact with the hockey stick involves a short time duration, while the puck slows down for a longer duration of time in the that is, the acceleration peak is narrower at t0 and wider at t1 . t0 a 2. t0 a 3. t1 t t0 correct 033 10 points The graph shows position as a function of time for two trains running on parallel tracks. At time t 0 the position of both trains is 0 (at the origin). position 4. None of these graphs are correct. A B 5. a t0 t t1 time a t0 tB t t1 6. Which is true? 1. Both trains speed up all the time. a 7. t0 t t1 2. Somewhere on the graph, both trains have the same acceleration. 3. At time tB , both trains have the same velocity. rash (tgr353) Homework 4 (review for test) fischler (58920) Explanation: The acceleration is the slope of the velocity vs time i., the acceleration is zero. 039 10 points A body moving with uniform acceleration has a velocity of 6 when its x coordinate is 3 cm. If its x coordinate 2 s later is cm, what is the of its acceleration? Correct answer: . 13 d 1 at t 2 1 1 km (0 ) (24 s) 24 s 2 39 . v0 041 (part 1 of 3) 10 points In deep space (no gravity), the bolt (arrow) of a crossbow accelerates at 159 and attains a speed of 135 when it leaves the bow. For how long is it accelerated? Explanation: Correct answer: 0 s. Let : v0 6 , x cm , x0 3 cm , and t 2 s . The distance is 1 x x0 v0 t a t2 2 2 (x x0 v0 t) t2 2 cm 3 cm (6 (2 (2 s)2 . Explanation: Let : a 159 and v 135 . v vo a t a t v 135 0 s . a 159 042 (part 2 of 3) 10 points What speed will the bolt have attained 2 s after leaving the crossbow? Correct answer: 135 040 10 points A car, moving along a straight stretch of highway, begins to accelerate at 0 . It takes the car 24 s to cover 1 km. How fast was the car going when it first began to accelerate? Correct answer: 39 Explanation: Explanation: It has left the crossbow, so a 0 and the velocity of the bolt after 2 s will remain 135 043 (part 3 of 3) 10 points How far will the bolt have traveled during the 2 s? Correct answer: 378 m. Let : a 0 , t 24 s , and d 1 km . d v0 t 1 2 at 2 Explanation: 1 2 a t v t2 2 (135 s) 378 m . s so vo t rash (tgr353) Homework 4 (review for test) fischler (58920) 044 10 points Objects fall near the surface of the earth with a constant downward acceleration of 10 . Suppose a falling object is moving downward at 10 at a certain instant. How fast is it falling 2 sec later? 14 046 10 points A cannon fires a 0 kg shell with initial velocity vi 9 in the direction θ 56 above the horizontal. 1. None of these s 2. 50 9. 7 3. 20 4. 40 5. 5 56 y 6. 30 correct 7. Still 10 Explanation: The downward velocity increases at 10 every second, so after 2 sec the object is moving downward at 30 045 10 points A drag racer starts her car from rest and accelerates at 12 for the entire distance of 226 m. How long did it take the car to travel this distance? Correct answer: 5 s. Explanation: Let : 226 m and a 12 . The trajectory curves downward because of gravity, so at the time t 0 s the shell is below the straight line some vertical distance Find this distance in the absence of air resistance. The acceleration of gravity is 9 . Correct answer: 1 m. Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: x ˆ t vi cos θ , yˆ t vi sin θ . The gravity does not affect the x coordinate of the shell, but it does pull its y coordinate downware at a constant acceleration ay so x t vi cos θ, Under constant acceleration, 1 1 vi t a t2 a t2 2 s 2 r 2 2(226 m) a 12 5 s . y t vi sin θ g t2 . 2 1 Thus, x x ˆ but y yˆ in other words, 2 the shell deviates from the path the vertical distance g t2 . yˆ y 2 rash (tgr353) Homework 4 (review for test) fischler (58920) Choose the origin at the base of the building. The initial position of the brick is y0 h, the vertical component of the initial velocity is v0y v0 sin θ directed upward, and y 0 when the brick reaches the ground, so 0 h v0y t 1 2 gt 2 1 2 gt 2 (2 s) 1 9 (2 s)2 2 23 m . h t 050 10 points A football is thrown at an angle of above the horizontal. Assume the ball is throw and received at the same height. To throw a 51 m pass, what must be the initial speed of the ball? The acceleration due to gravity is 9 . Correct answer: 22 Explanation: Let : θ , d 51 m , and g 9 . The horizontal distance (range) is R (v0 cos θ) t R v0 cos θ with a corresponding vertical height of 0 (v0 sin θ) t 1 2 gt 2 1 gt 2 g R 0 v0 sin θ 2 v0 cos θ 2 0 v0 (2 sin θ cos θ) g R 0 v02 sin(2 θ) g R 0 v0 sin θ gR sin(2 θ) s r gR (9 )(51 m) v0 sin(2 θ) sin v02 22 . 16

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Course : Elements of Physics (PHY 309K)

University : university of texas at austin.

homework assignment 4 1 judging space in seconds answers

Discover more from: Elements of Physics PHY 309K University of Texas at Austin 41 Documents Go to course
More from: Elements of Physics PHY 309K University of Texas at Austin 41 Documents Go to course