# Cha1Cha2 - HWK Set One Joy Zhou October 6 2009 1 Chapter...

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Unformatted text preview: HWK Set One Joy Zhou October 6, 2009 1 Chapter One 1.1, 1.2, 1.3, 1.5, 1.10, 1.13 Remarks: 1. Read the problems carefully. Dr. Conroy says a lot of students lost points on exams simply because they did not understand the problem in the right way, as opposed to making math mistakes. 2. You are not required to remember how many feet is in a mile, etc. You can put them all in the sheet of paper that you can bring into the exams. 3. The ”multiplication by one” trick is very important. Problem 1.1 SOL: (b) 1 mile = 5280 ft. 100 mph = 100 mile 1 hour = 1000 mile 1 hour × 1 hour 3600 seconds × 5280 ft 1 mile = 146 . 67 ft / s So 150 ft/s is faster. You can also try convert ft/s into mph. (c) Total Change = Rate × Time. You can either calculate how much does Gina earn in 40 hours, and com- pare it with \$1400 which is the amound Tiare earns in 40 hours; or you can calculate the rate of Tiare’s earnings in terms of cent/second and compare it 1 with 1 cent/second which is the rate of earning of Gina. Gina’s salary in 40 hours: 40 hours = 40 hours × 3600 seconds 1 hour = 144000 seconds. Total salary = rate × time = 1 cent/second × 144000 seconds = 144000 cents = 144000 cents × 1 dollar 100 cents = 1440 dollars. Compared with \$1400, Gina earns more. (d) Modeling: Let T be the hours to invest in each quarter credit, T l be the hours for lecture for each quarter credit, and T s be the hours to study for each quarter credit. Then T = T l + T s . T l = 1 classroom hour/week × 10 weeks = 10 classroom hours. T s = 10 classroom hours × 2 . 5 hourstostudy 1 classroomhour = 25 hours to study. T = T l + T s = 35 hours. So you need to invest 35 hours in 1 quarter credit. Therefore, 180 quarter credits = 180 × 35 hours to invest = 6300 hours to invest. Problem 1.2 SOL: Note that 2hr 40 min = 160 min. 6 min less than that is 154 min. Also, 1 km/hr = 1000 60 m / min = 50 3 m / min. Suppose Sarah’s speed before the increase was s m/min. Since the loop length doesn’t change, we have the equation s m / min × 160 min = (s + 50 3 ) m / min × 154 min . Solving this equation, we find that the speed before increase is s = 427 . 7 m / min. Therefore the loop length L is : L = 427.7 m/min × 160 min = 68.44 km. 2 Problem 1.3 SOL: Note: It is given in the Appendix A that Volume = 4 3 π Radius 3 , So 3 × Volume = 4 π Radius 3 , and Radius 3 = 3 × Volume 4 π . Therefore Radius = ( 3 × Volume 4 × π ) 1 / 3 . Now, 50 kg = 50000g 50 kg of lead: Volume = Mass Density = 50000g 11 . 34 g / cm 3 = 4409 . 1711 cm 3 Radius = ( 3 × Volume 4 × π ) 1 / 3 = 10 . 1724 cm 50 kg of aluminum: Volume = Mass Density = 50000g 2 . 69 g / cm 3 = 18587 . 3606 cm 3 Radius = ( 3 × Volume 4 × π ) 1 / 3 = 16 . 4327 cm Problem 1.5 SOL: I’ll only mention the (i) part.I’ll only mention the (i) part....
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Cha1Cha2 - HWK Set One Joy Zhou October 6 2009 1 Chapter...

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