HW-02-solutions

HW-02-solutions - kuruvila (lk5992) HW-02 gokce (P111S11) 1...

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Unformatted text preview: kuruvila (lk5992) HW-02 gokce (P111S11) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The velocity of the transverse waves produced by an earthquake is 6 . 78 km / s, while that of the longitudinal waves is 11 . 187 km / s. A seis- mograph records the arrival of the transverse waves 71 . 9 s after that of the longitudinal waves. How far away was the earthquake? Correct answer: 1237 . 45 km. Explanation: Let : v t = 6 . 78 km / s , v = 11 . 187 km / s , and t = 71 . 9 s . If the distance to the earthquake is d , then the time lag is t = d v t- d v = d ( v - v t ) v v t d = t v v t v - v t = (71 . 9 s) (11 . 187 km / s) (6 . 78 km / s) 6 . 78 km / s- 11 . 187 km / s = 1237 . 45 km . Note: The longitudinal wave travels faster than the transverse waves. In fact, the ratio of the longitudinal velocity to the transverse wave velocity should be about 3 . 002 (part 1 of 2) 10.0 points A runner is jogging at a steady 7 . 3 km / hr. When the runner is 2 . 8 km from the finish line, a bird begins flying from the runner to the finish line at 14 . 6 km / hr (2 times as fast as the runner). When the bird reaches the finish line, it turns around and flies back to the runner. L v b v r finish line How far does the bird travel? Even though the bird is a dodo, assume that it occupies only one point in space (a zero length bird) and that it can turn without loss of speed. Correct answer: 3 . 73333 km. Explanation: Let : v r = 7 . 3 km / hr , L = 2 . 8 km , and v b = 2 v r . finish line L 1 dr 1 db 1 The runner travels a distance x until the encounter with the bird. In that time, the bird has traveled a distance L +( L- x ) = 2 L- x . The bird travel 2 times as fast as the runner during this time frame, so d b = 2 d r 2 L- x = 2 x 2 L = 3 x x = 2 3 L and the bird flies a distance of d b = 2 L- 2 3 L = 4 3 L = 4 3 (2 . 8 km) = 3 . 73333 km . 003 (part 2 of 2) 10.0 points After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back kuruvila (lk5992) HW-02 gokce (P111S11) 2 to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far does the bird travel from the be- ginning (including the distance traveled to the first encounter)? Correct answer: 5 . 6 km. Explanation: The distance remaining for the runner after the first encounter is L 1 = 1 3 L , and again the bird will fly 2 times as far as the runner until the next encounter. This pattern repeats over the entire original distance, so d b = 2 L = 2 (2 . 8 km) = 5 . 6 km ....
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This note was uploaded on 02/12/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.

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HW-02-solutions - kuruvila (lk5992) HW-02 gokce (P111S11) 1...

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