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Hw 1 Solutions

# Hw 1 Solutions - Bell Evan Homework 1 Due 11:00 pm Inst...

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Bell, Evan – Homework 1 – Due: Jan 19 2005, 11:00 pm – Inst: Furlan 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Assume the Sun is mostly composed of hy- drogen atoms. Assume the mass of the Sun is 1 . 54 × 10 30 kg, and the mass of a hydrogen atom is 1 . 672 × 10 - 27 kg. How many atoms are there in the Sun? Correct answer: 9 . 21053 × 10 56 atoms. Explanation: Basic Concepts: Elementary Arithmetic N atoms = M sun M atom = 1 . 54 × 10 30 kg 1 . 672 × 10 - 27 kg = 9 . 21053 × 10 56 atoms . 002 (part 1 of 1) 10 points The base of a pyramid covers an area of 19 . 7 acres (1 acre = 43 , 560 ft 2 ) and has a height of 69 ft. If the volume of a pyramid is given by the expression V = 1 3 b h , where b is the area of the base and h is the height, find the volume of this pyramid in cubic meters. Correct answer: 558891 m 3 . Explanation: Given : b = 19 . 7 acres and h = 69 ft . V = 1 3 b h = 1 3 (19 . 7 acres) (69 ft) · 43560 ft 2 acre 0 . 3048 m ft · 3 = 558891 m 3 . 003 (part 1 of 3) 10 points Consider the triangle shown in the figure. 7 . 3 m 13 m φ θ What is the length of the unknown side? Correct answer: 10 . 7569 m. Explanation: Given : c = 13 m and a = 7 . 3 m . Applying the Pythagorean Theorem, b = p c 2 - a 2 = q (13 m) 2 - (7 . 3 m) 2 = 10 . 7569 m 004 (part 2 of 3) 10 points Find the tangent of θ . Correct answer: 0 . 678637 . Explanation: tan θ = opposite adjacent = 7 . 3 m 10 . 7569 m = 0 . 678637 . 005 (part 3 of 3) 10 points Find the sin of φ . Correct answer: 0 . 827451 . Explanation:

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Bell, Evan – Homework 1 – Due: Jan 19 2005, 11:00 pm – Inst: Furlan 2 sin φ = opposite hypotenuse = 10 . 7569 m 13 m = 0 . 827451 . 006 (part 1 of 1) 10 points A car travels along a straight stretch of road. It proceeds for 12 . 4 mi at 52 mi / h, then 24 . 2 mi at 43 mi / h, and finally 41 mi at 35 . 7 mi / h. What is the car’s average velocity during the entire trip? Correct answer: 39 . 8008 mi / h. Explanation: The total distance the car traveled is Δ d = d A + d B + d C = 77 . 6 mi The total time the car spent on the road is Δ t = d A v A + d B v B + d C v C = 12 . 4 mi 52 mi / h + 24 . 2 mi 43 mi / h + 41 mi 35 . 7 mi / h = 1 . 94971 h . Hence the average velocity is v = Δ d Δ t = 77 . 6 mi 1 . 94971 h = 39 . 8008 mi / h . 007 (part 1 of 1) 10 points Assume: 1 . 28 km / lap. A race car is one lap behind the lead race car when the lead car has 52 laps to go in a race. If the speed of the lead car is 55 . 9 m / s, what must be the average speed of the second car to catch the lead car just before the end of the race ( i. e., right at the finish line)? Correct answer: 56 . 975 m / s. Explanation: Basic Concepts Constant Velocity Solution The time needed for the lead car to end the race is t = n v = (52 laps) (55 . 9 m / s) = (52 laps) (1 . 28 km / lap) (55 . 9 m / s) 1000 m 1 km = 1190 . 7 s . In the same period of time, the trailing car must go 1 lap more than the lead car to catch up; thus its average speed must be v avg = m t = (53 laps) (1190 . 7 s) = (53 laps) (1 . 28 km / lap) (1190 . 7 s) 1000 m 1 km = 56 . 975 m / s .
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Hw 1 Solutions - Bell Evan Homework 1 Due 11:00 pm Inst...

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