sol03 - 6 50.9967 N Question 1 chap 3 sect 1 part 1 of 1 10...

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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
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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. Question 4, chap 3, sect 1. part 1 of 1 10 points A person walks 25 m west and then 45 m at an angle of 60 north of east. What is the magnitude of the total dis- placement? 1. 31.0 m 2. 67.8 m 3. 45.7 m 4. 26.5 m 5. 39.1 m correct 6. 61.4 m 7. 40.9 m 8. 43.6 m 9. 56.3 m 10. 60.2 m
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Explanation: In a triangle with three sides of a, b, and c , if the angle between a and b is θ , then c 2 = a 2 + b 2 2 ab cos θ .
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