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PHY303K-Spring07-Exam1-Moore

# PHY303K-Spring07-Exam1-Moore - midterm 01 SAENZ LORENZO Due...

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midterm 01 – SAENZ, LORENZO – Due: Feb 14 2007, 11:00 pm 1 Question 1 part 1 of 2 10 points A man pushing a mop across a floor causes the mop to undergo two displacements. The first has a magnitude of 206 cm and makes an angle of 148 with the positive x -axis. The resultant displacement has a magni- tude of 158 cm and is directed at an angle of 26 . 6 to the positive x -axis. Find the magnitude of the second displace- ment. 1. 289 . 999 cm 2. 299 . 061 cm 3. 308 . 532 cm 4. 318 . 301 cm correct 5. 328 . 172 cm 6. 338 . 35 cm 7. 348 . 842 cm 8. 359 . 675 cm 9. 371 . 104 cm 10. 382 . 593 cm Explanation: Let : A = 206 cm , θ A = 148 , R = 158 cm , and θ R = 26 . 6 . A = 206 cm B = 318 . 301 cm R = 158 cm Since vector A + vector B = vector R , vector B = vector R vector A . Thus, the vector vector B has components B x = R x A x = R cos θ R A cos θ A = (158 cm)(cos 26 . 6 ) (206 cm)(cos 148 ) = 315 . 974 cm and B y = R y A y = R sin θ R A sin θ A = (158 cm)(sin 26 . 6 ) (206 cm)(sin 148 ) = 38 . 4172 cm , which act at right angles to each other. From the Pythagorean Theorem, B = radicalBig B 2 x + B 2 y = radicalBig (315 . 974 cm) 2 + ( 38 . 4172 cm) 2 = 318 . 301 cm . Question 2 part 2 of 2 10 points Find the direction of the second displace- ment with respect to positive x axis. (Con- sider counterclockwise to be positive and an- swer within the limits of 180 to 180 .) 1. 7 . 97687 2. 7 . 71117 3. 7 . 42021 4. 7 . 14857 5. 6 . 93219 correct 6. 6 . 62341 7. 6 . 39429 8. 6 . 19765 9. 5 . 9778 10. 5 . 77275 Explanation: tan θ = B y B x θ = tan 1 parenleftbigg B y B x parenrightbigg = tan 1 parenleftbigg 38 . 4172 cm 315 . 974 cm parenrightbigg = 6 . 93219 , which is directed below the x axis.

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midterm 01 – SAENZ, LORENZO – Due: Feb 14 2007, 11:00 pm 2 Question 3 part 1 of 1 10 points A cat chases a mouse across a 0.79 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.7 m from the edge of the table. The acceleration of gravity is 9 . 81 m / s 2 . What was the cat’s speed when it slid off the table? 1. 5 . 37149 m / s 2. 5 . 54162 m / s 3. 5 . 7414 m / s 4. 5 . 9191 m / s 5. 6 . 11438 m / s 6. 6 . 30814 m / s 7. 6 . 51259 m / s 8. 6 . 72775 m / s correct 9. 6 . 95132 m / s 10. 7 . 18079 m / s Explanation: Basic Concept: Δ x = v x Δ t Δ y = 1 2 g t ) 2 Given: Δ y = 0 . 79 m Δ x = 2 . 7 m g = 9 . 81 m / s 2 . Solution: Δ t = radicalBigg 2 Δ y g = Δ x v x v x = radicalbigg g y Δ x = radicalBigg (9 . 81 m / s 2 ) 2 ( 0 . 79 m) (2 . 7 m) = 6 . 72775 m / s . Question 4 part 1 of 1 10 points A cannon fires a 0 . 355 kg shell with initial velocity v i = 9 . 4 m / s in the direction θ = 55 above the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . Δ x Δ h 9 . 4 m / s 55 Δ y y The shell’s trajectory curves downward be- cause of gravity, so at the time t = 0 . 794 s the shell is below the straight line by some vertical distance. Your task is to calculate this distance Δ h in the absence of air resistance. 1. 2 . 29249 m 2. 2 . 36682 m 3. 2 . 44926 m 4. 2 . 52607 m 5. 2 . 61121 m 6. 2 . 71233 m 7. 2 . 80053 m 8. 2 . 89014 m 9. 2 . 98881 m 10. 3 . 08914 m correct Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: ˆ x = t v i cos θ , ˆ y = t v i sin θ .
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PHY303K-Spring07-Exam1-Moore - midterm 01 SAENZ LORENZO Due...

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