Test #1-solutions

# Test #1-solutions - Version 097/ABCAB Test #1 Antoniewicz...

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Unformatted text preview: Version 097/ABCAB Test #1 Antoniewicz (57380) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Vector vector A has components A x = 6 . 6 , A y = 8 . 7 , A z = 6 . 4 , while vector vector B has components B x = 7 . 2 , B y = 1 . 6 , B z = 5 . 9 . What is the angle AB between these vec- tors? (Answer between 0 and 180 .) 1. 142.444 2. 137.806 3. 126.051 4. 119.909 5. 93.3259 6. 131.142 7. 127.534 8. 98.74 9. 101.424 10. 105.88 Correct answer: 101 . 424 . Explanation: bardbl vector A bardbl = radicalBig A 2 x + A 2 y + A 2 z = radicalBig ( 6 . 6) 2 + 8 . 7 2 + 6 . 4 2 = 12 . 6574 and bardbl vector B bardbl = radicalBig B 2 x + B 2 y + B 2 z = radicalBig 7 . 2 2 + ( 1 . 6) 2 + 5 . 9 2 = 9 . 4451 , so using vector A vector B = A x B x + A y B y + A z B z = ( 6 . 6) 7 . 2 + 8 . 7( 1 . 6) + 6 . 4(5 . 9) = 23 . 68 , cos AB = vector A vector B bardbl vector A bardbl bardbl vector B bardbl = 23 . 68 (12 . 6574) (9 . 4451) = . 198075 AB = arccos( . 198075) = 101 . 424 . Two vectors can define a plane. When these two vectors are plotted in this plane, we have A B 1 1 . 4 2 4 002 (part 1 of 3) 4.0 points A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 2 m, y = 6 . 5 m, and has velocity vectorv o = (2 . 5 m / s) + ( 8 . 5 m / s) . The acceleration is given by vectora = (3 . 5 m / s 2 ) + (5 m / s 2 ) . What is the x component of velocity after 1 . 5 s? 1. 15.5 2. 44.0 3. 60.0 4. 18.0 5. 27.0 6. 10.5 7. 26.5 8. 20.0 9. 41.0 10. 7.75 Correct answer: 7 . 75 m / s. Explanation: Let : a x = 3 . 5 m / s 2 , v xo = 2 . 5 m / s , and t = 1 . 5 s . After 1 . 5 s, vectorv x = vectorv xo + vectora x t = (2 . 5 m / s) + (3 . 5 m / s 2 ) (1 . 5 s) = (7 . 75 m / s) . Version 097/ABCAB Test #1 Antoniewicz (57380) 2 003 (part 2 of 3) 3.0 points What is the y component of velocity after 1 . 5 s? 1. 1.0 2. 21.5 3. 15.0 4. 15.5 5. -0.5 6. 9.5 7. 57.5 8. -1.0 9. 36.5 10. 5.75 Correct answer: 1 m / s. Explanation: Let : a y = 5 m / s 2 and v yo = 8 . 5 m / s . vectorv y = vectorv yo + vectora y t = ( 8 . 5 m / s) + (5 m / s 2 ) (1 . 5 s) = ( 1 m / s) ....
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## This note was uploaded on 04/02/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Test #1-solutions - Version 097/ABCAB Test #1 Antoniewicz...

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