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Unformatted text preview: homework 03 HELD, ETHAN Due: Feb 6 2007, 4:00 am 1 Question 1 part 1 of 2 10 points A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 33 m / s. At the same time, it has a horizontal velocity of 55 m / s. At what speed does the vehicle move along its descent path? Correct answer: 64 . 1405 m / s (tolerance 1 %). Explanation: The speeds act at right angles to each other, so the actual speed is the hypotenuse of the right triangle formed, and v = radicalBig v 2 h + v 2 v Question 2 part 2 of 2 10 points At what angle with the vertical is its path? Answer between 180 and +180 . Correct answer: 59 . 0363 (tolerance 1 %). Explanation: v h is the side opposite and v v is the side adjacent, and = arctan parenleftbigg v h v v parenrightbigg Question 3 part 1 of 1 10 points A pair of 340 . 4 N vectors are perpendicular. What is the magnitude of their resultant? Correct answer: 481 . 398 N (tolerance 1 %). Explanation: The resultant of two perpendicular vectors is the hypotenuse of a right triangle, so R = radicalbig F 2 + F 2 = 2 F 2 = radicalBig 2(340 . 4 N) 2 = 481 . 398 N Question 4 part 1 of 3 10 points Consider vectors vector A and vector B with coordinate components shown in the illustration below. A B An isometric drawing: Each coordinate has a length of 5 units. To indicate the coordinates of each vector, a line is projected to the hor izontal plane then two lines are pro jected to the horizontal coordinates. As well, a line is directly projected to the vertical coordinate. Find the vertical component of vector A vector B . Correct answer: 14 units 2 (tolerance 1 %). Explanation: See Eq. 2 in the next part. ( vector A vector B ) z = bracketleftBig A x B y A y B x bracketrightBig k = bracketleftBig (5 units) (4 units) ( 3 units) ( 2 units) bracketrightBig k = 14 units 2 k . Question 5 part 2 of 3 10 points What is the magnitude of the vector prod uct of these two vectors? Correct answer: 24 . 1661 units 2 (tolerance 1 %). Explanation: Basic Concept: vector A vector B = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle k A x A y A z B x B y B z vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle (1) homework 03 HELD, ETHAN Due: Feb 6 2007, 4:00 am 2 = vextendsingle vextendsingle vextendsingle vextendsingle A y A z B y B z vextendsingle vextendsingle vextendsingle vextendsingle + vextendsingle vextendsingle vextendsingle vextendsingle A z A x B z B x vextendsingle vextendsingle vextendsingle vextendsingle + vextendsingle vextendsingle vextendsingle vextendsingle A x A y B x B y vextendsingle vextendsingle vextendsingle vextendsingle k = ( A y B z A z B y ) + ( A z B x A x B z ) + ( A x B y A y B x ) k Let : A x = 5 units , A y = 3 units , A z = 1 units , B x = 2 units , B y = 4 units ,...
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
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