solution32

# solution32 - fullwood(tf5349 – Homework 2 Vectors –...

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Unformatted text preview: fullwood (tf5349) – Homework 2 Vectors – Catala – (49033) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A person walks 27 . 4 ◦ north of east for 2 . 36 km. Another person walks due north, then due east to arrive at the same location. How far due north would this person walk? Correct answer: 1 . 08607 km. Explanation: Let : d = 2 . 36 km and θ = 27 . 4 ◦ . d x y θ The north component is sin θ = y d y = d sin θ = (2 . 36 km) sin27 . 4 ◦ = 1 . 08607 km . 002 (part 2 of 2) 10.0 points How far would this person walk due east? Correct answer: 2 . 09524 km. Explanation: The east component is cos θ = x d x = d cos θ = (2 . 36 km) cos 27 . 4 ◦ = 2 . 09524 km . 003 (part 1 of 2) 10.0 points WITHDRAWN 004 (part 2 of 2) 10.0 points WITHDRAWN 005 10.0 points WITHDRAWN 006 (part 1 of 2) 10.0 points A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 27 . 1 m / s. At the same time, it has a horizontal velocity of 52 . 3 m / s. At what speed does the vehicle move along its descent path? Correct answer: 58 . 9042 m / s. Explanation: Let : v v = 27 . 1 m / s , and v h = 52 . 3 m / s . v h v v v θ The speeds act at right angles to each other, so v = radicalBig v 2 h + v 2 v = radicalBig (52 . 3 m / s) 2 + (27 . 1 m / s) 2 = 58 . 9042 m / s . 007 (part 2 of 2) 10.0 points At what angle with the vertical is its path? Correct answer: 62 . 6084 ◦ . Explanation: v h is the side opposite and v v is the side adjacent to the angle, so tan θ = v h v v θ = arctan parenleftbigg v h v v parenrightbigg = arctan parenleftbigg 52 . 3 m / s 27 . 1 m / s parenrightbigg = 62 . 6084 ◦ . fullwood (tf5349) – Homework 2 Vectors – Catala – (49033) 2 008 10.0 points Consider three force vectors vector F 1 , vector F 2 , and vector F 3 . The vector vector F 1 has magnitude F 1 = 52 N and direction θ 1 = 27 ◦ ; the vector vector F 2 has magni- tude F 2 = 39 N and direction θ 2 = − 130 ◦ ; and the vector vector F 3 has magnitude F 3 = 19 N and direction θ 3 = 44 ◦ . All the direction an- gles θ are measured from the positive x axis: counter-clockwise for θ > 0 and clockwise for θ < 0. What is the magnitude F of the net force vector vector F = vector F 1 + vector F 2 + vector F 3 ? Correct answer: 35 . 6119 N. Explanation: In components, each force vector is F 1 ,x = F 1 × cos θ 1 = 46 . 3323 N , F 1 ,y = F 1 × sin θ 1 = 23 . 6075 N , F 2 ,x = F 2 × cos θ 2 = − 25 . 0687 N , F 2 ,y = F 2 × sin θ 2 = − 29 . 8757 N , F 3 ,x = F 3 × cos θ 3 = 13 . 6675 N , F 3 ,y = F 3 × sin θ 3 = 13 . 1985 N , and therefore the net force is F x = F 1 ,x + F 2 ,x + F 3 ,x = 34 . 9311 N , F y...
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solution32 - fullwood(tf5349 – Homework 2 Vectors –...

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