OH-04-solutions - mohammed(omm269 – OH-04 – bradley...

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Unformatted text preview: mohammed (omm269) – OH-04 – bradley – (19101) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No. 4 (OH-04). All answers should be submitted online before 2300 on Tuesday 22 February (UAE time). Computers and network connections can give trouble, so do not wait until the last minute! 001 10.0 points A mass slides with negligible friction on a circular track of 1 m radius oriented vertically. Its speed at the position shown in the figure is 3.13 m/s. v At the position shown in the figure, which arrow best represents the direction of the ac- celeration of the mass? The acceleration of gravity is 9.8 m/s 2 and no external forces act on the system. 1. 2. correct 3. 4. 5. 6. 7. 8. 9. The mass is traveling at a constant veloc- ity, so it has no acceleration. Explanation: The magnitude of the centripetal accelera- tion is a r = v 2 r = (3 . 13 m / s) 2 1 m ≈ 9 . 8 m / s 2 , acting inward ( − ˆ r ) with gravity acting down- ward ( − ˆ k ). a a r g 002 10.0 points A particle is moving at constant speed in a counterclockwise circle of radius r about the point x = y = 0. At a certain instant it is at x = − r , y = 0. What is the direction of its acceleration at that instant? 1. It is along ˆ ı . correct 2. It is along − ˆ ı . 3. It is along ˆ  . 4. It is along − ˆ  . Explanation: The particle must have an acceleration to- ward the center of the circle, which is at x = 0, y = 0. If the particle is at point x = − r , y = 0, the acceleration is toward the origin of coordinates, along the x axis, in the direction of ˆ ı . 003 (part 1 of 2) 10.0 points The orbit of a Moon about its planet is ap- proximately circular, with a mean radius of 2 . 48 × 10 8 m. It takes 41 days for the Moon to complete one revolution about the planet. Find the mean orbital speed of the Moon. Correct answer: 439 . 88 m / s. Explanation: mohammed (omm269) – OH-04 – bradley – (19101) 2 Dividing the length C = 2 πr of the trajectory of the Moon by the time T = 41 days = 3 . 5424 × 10 6 s of one revolution (in seconds!), we obtain that the mean orbital speed of the Moon is v = C T = 2 π r T = 2 π (2 . 48 × 10 8 m ) 3 . 5424 × 10 6 s = 439 . 88 m / s . 004 (part 2 of 2) 10.0 points Find the Moon’s centripetal acceleration. Correct answer: 0 . 000780218 m / s 2 . Explanation: Since the magnitude of the velocity is con- stant, the tangential acceleration of the Moon is zero. The centripetal acceleration is a c = v 2 r = (439 . 88 m / s ) 2 2 . 48 × 10 8 m = 0 . 000780218 m / s 2 . 005 (part 1 of 2) 10.0 points A jet airliner moving initially at 961 mph to the east where there is no wind moves into a region where the wind is blowing at 542 mph in a direction 26 ◦ north of east....
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This note was uploaded on 03/22/2011 for the course PHYSICS 19101 taught by Professor Walschap,g during the Spring '11 term at The Petroleum Institute.

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OH-04-solutions - mohammed(omm269 – OH-04 – bradley...

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