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Unformatted text preview: magalhaes (bam2734) – Homework #2 – Erskine – (58200) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A physics book is moved once around the perimeter of a table of dimensions 1 m by 3 m. If the book ends up at its initial position, what is the magnitude of its displacement? 1. 2 m 2. 8 m 3. 5 m 4. 3 m 5. 1 m 6. 6 m 7. None of these 8. 7 m 9. 0 m correct 10. 4 m Explanation: Since its final position is the same as its initial position, the displacement is 0. 002 (part 2 of 2) 10.0 points What is the distance traveled? 1. 6 m 2. None of these 3. 2 m 4. 3 m 5. 7 m 6. 1 m 7. 0 m 8. 4 m 9. 5 m 10. 8 m correct Explanation: The distance traveled is the perimeter of the table: d = 2 ℓ + 2 w = 2 (1 m) + 2 (3 m) = 8 m . 003 (part 1 of 2) 10.0 points While John is traveling along a straight inter state highway, he notices that the mile marker reads 236 km. John travels until he reaches the 130 km marker and then retraces his path to the 169 km marker. What is John’s resultant displacement from the 236 km marker? Correct answer: 67 km. Explanation: Let : s = 236 km and s f = 169 km . Δ s = s f s = 169 km 236 km = 67 km . 004 (part 2 of 2) 10.0 points How far has he traveled? Correct answer: 145 km. Explanation: Let : s 2 = 130 km . magalhaes (bam2734) – Homework #2 – Erskine – (58200) 2 The distance traveled is given by d =  s 2 s  +  s f s 2  =  130 km 236 km  +  169 km 130 km  = 145 km . 005 10.0 points Two ants race across a table 68 cm long. One travels at 5 . 03 cm / s and the other at 2 cm / s. When the first one crosses the finish line, how far behind is the second one? Correct answer: 40 . 9622 cm. Explanation: Let : ℓ = 68 cm , v 1 = 5 . 03 cm / s , and v 2 = 2 cm / s . The time it takes the first (faster) ant to cross the finish line is t = ℓ v 1 , and the distance the slower ant covers in that time is s 2 = v 2 t = v 2 ℓ v 1 . The slower ant is ℓ s 2 = ℓ v 2 ℓ v 1 = 68 cm (2 cm / s) (68 cm) 5 . 03 cm / s = 40 . 9622 cm from the finish line when the faster one crosses it. 006 (part 1 of 2) 10.0 points A bug completes one lap along the edge of a circular planter of radius 10 . 9 cm in 17 . 9 s. How fast was it traveling? Correct answer: 3 . 82607 cm / s. Explanation: Let : r = 10 . 9 cm and t = 17 . 9 s . The distance around the circular planter s = 2 π r , so the velocity v = s t = 2 π r t = 2 π (10 . 9 cm) 17 . 9 s = 3 . 82607 cm / s . 007 (part 2 of 2) 10.0 points How fast in m/s? Correct answer: 0 . 0382607 m / s. Explanation: v = (3 . 82607 cm / s) 1 m 100 cm = . 0382607 m / s ....
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This note was uploaded on 04/13/2010 for the course PHY phy303k taught by Professor Erskine during the Spring '09 term at University of TexasTyler.
 Spring '09
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