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Unformatted text preview: Su (ycs73) – HW01 – Tsoi – (58020) 1 This printout should have 19 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 person travels by car from one city to an other. She drives for 27 . 8 min at 89 . 1 km / h, 13 . 4 min at 110 km / h, 38 . 6 min at 45 . 5 km / h, and spends 10 . 5 min along the way eating lunch and buying gas. Determine the distance between the cities along this route. Correct answer: 95 . 1213 km. Explanation: Distances traveled are x 1 = v 1 t 1 x 2 = v 2 t 2 x 3 = v 3 t 3 Then the total distance traveled is x = x 1 + x 2 + x 3 = (41 . 283 km) + (24 . 5667 km) + (29 . 2717 km) = 95 . 1213 km . 002 (part 2 of 2) 10.0 points Determine the average speed for the trip. Correct answer: 63 . 2035 km / h. Explanation: And, time spent is t = t 1 + t 2 + t 3 + t other = (27 . 8 min) + (13 . 4 min) + (38 . 6 min) + (10 . 5 min) = 1 . 505 h . Hence v av = x t = 95 . 1213 km 1 . 505 h = 63 . 2035 km / h . 003 (part 1 of 2) 10.0 points Two particles moving along parallel paths in the same direction pass the same point at the same time. Particle A has an initial velocity of 6 . 5 m / s and an acceleration of 2 . 7 m / s 2 . Particle B has an initial velocity of 2 m / s and an acceleration of 7 . 2 m / s 2 . a) At what time will B pass A? Correct answer: 2 s. Explanation: Basic Concept: s = v t + 1 2 at 2 for each particle. Solution: Particle B will pass particle A when the positions are the same, so equate the positions and solve the resulting equation for t : v A t + 1 2 a A t 2 = v B t + 1 2 a B t 2 2 v A t + a A t 2 = 2 v B t + a B t 2 a A t 2 − a B t 2 + 2 v A t − 2 v B t = 0 t ( a A t − a B t + 2 v A − 2 v B ) = 0 t parenleftBig ( a A − a B ) t + 2( v A − v B ) parenrightBig = 0 t = 2( v B − v A ) ( a A − a B ) The trivial solution t = 0 must be rejected. 004 (part 2 of 2) 10.0 points b) At what position will the faster particle pass the slower? Correct answer: 18 . 4 m. Explanation: the time found in Part 1 into the following position equation: s = v A t + 1 2 a A t 2 = v B t + 1 2 a B t 2 . 005 (part 1 of 2) 10.0 points The radius of the planet Jupiter is 7 . 14 × 10 7 m, and its mass is 1 . 9 × 10 30 kg. Su (ycs73) – HW01 – Tsoi – (58020) 2 Find the average density of Jupiter (its mass divided by its volume) if the volume of a sphere is given by 4 3 π r 3 . Correct answer: 1246 . 15 g / cm 3 . Explanation: Let : r = 7 . 14 × 10 7 m = 7 . 14 × 10 9 cm , and m = 1 . 9 × 10 30 kg = 1 . 9 × 10 33 g . Density is ρ = m V = m 4 3 π r 3 = 1 . 9 × 10 33 g 4 3 π (7 . 14 × 10 9 cm) 3 = 1246 . 15 g / cm 3 . 006 (part 2 of 2) 10.0 points Find the surface area of Jupiter if the surface area of a sphere is given by 4 π r 2 ....
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This note was uploaded on 04/15/2009 for the course PHY 58020 taught by Professor Tsoi during the Spring '09 term at University of Texas at Austin.
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