sol03 - Question 1, chap 3, sect 1. part 1 of 1 10 points A...

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Unformatted text preview: Question 1, chap 3, sect 1. part 1 of 1 10 points A vector of magnitude 3 CANNOT be added to a vector of magnitude 4 so that the magnitude of the resultant is 1. 3. 2. 7. 3. 1. 4. 5. 5. 0. correct Explanation: The smallest magnitude of the resultant occurs when the vectors are anti-parallel ( R = 1); the largest occurs when they are parallel ( R = 7). Therefore all listed values are possible except R = 0. Question 2, chap 3, sect 1. part 1 of 2 10 points Given two vectors vector F 1 , and vector F 2 . Where the magnitude of these vectors are F 1 = 77 N , and F 2 = 72 N . And where θ 1 = 21 ◦ , and θ 2 = 80 ◦ . The angles are measure from the positive x axis with the counter-clockwise angular direc- tion as positive. Draw the vectors to scale on a graph to determine the answer. What is the magnitude of the resultant vec- tor bardbl vector F bardbl , where vector F = vector F 1 + vector F 2 ? 1. 109 . 458 N 2. 78 . 4916 N 3. 129 . 706 N correct 4. 87 . 5005 N 5. 121 . 553 N 6. 50 . 9967 N 7. 124 . 715 N 8. 136 . 289 N Explanation: Drawing a diagram to scale of the vectors, we have θ F 1 F 2 F Scale: 10 N where : F 1 = 77 N , θ 1 = 21 ◦ , F 1 x = 71 . 8857 N , F 1 y = 27 . 5943 N , F 2 = 72 N , θ 2 = 80 ◦ , F 2 x = 12 . 5027 N , F 2 y = 70 . 9062 N , and F = 129 . 706 N , θ = 49 . 4123 ◦ , F x = 84 . 3884 N , F y = 98 . 5005 N . The x components of the forces vector F 1 and vector F 2 are F 1 x = F 1 cos(21 ◦ ) = 71 . 8857 N F 2 x = F 2 cos(80 ◦ ) = 12 . 5027 N . And the y components are F 1 y = F 1 sin(21 ◦ ) = 27 . 5943 N F 2 y = F 2 sin(80 ◦ ) = 70 . 9062 N . The x and y components of the resultant vec- tor vector F are F x = F 1 x + F 2 x = (71 . 8857 N) + (12 . 5027 N) = 84 . 3884 N F y = F 1 y + F 2 y = (27 . 5943 N) + (70 . 9062 N) = 98 . 5005 N . Hence the magnitude of the resultant vector bardbl vector F bardbl is bardbl vector F bardbl = radicalBig F 2 x + F 2 y = radicalBig (84 . 3884 N) 2 + (98 . 5005 N) 2 = 129 . 706 N . Question 3, chap 3, sect 1. part 2 of 2 10 points Note: Give the angle in degrees, use coun- terclockwise as the positive angular direction, between the limits of − 180 ◦ and +180 ◦ from the positive x axis. What is the direction of this resultant vec- tor vector F ? 1. 49 . 4123 ◦ correct 2. − 76 . 3995 ◦ 3. 104 . 994 ◦ 4. 131 . 792 ◦ 5. 89 . 2517 ◦ 6. − 134 . 449 ◦ 7. − 177 . 375 ◦ 8. − 65 . 1028 ◦ Explanation: The angle is given by θ = arctan parenleftbigg F y F x parenrightbigg = tan − 1 parenleftbigg 98 . 5005 N 84 . 3884 N parenrightbigg = 49 . 4123 ◦ . The “arctan” function is defined between lim- its − 90 ◦ and +90 ◦ . Thus you must check to see in which quadrant your resultant vector lies. You should use the “atan2(y,x)” in your calculator, if available, since it does not have this ambiguity....
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This note was uploaded on 12/04/2008 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas.

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sol03 - Question 1, chap 3, sect 1. part 1 of 1 10 points A...

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