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# phyfinal - Version 116 – Final – florin –(58140 1...

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Unformatted text preview: Version 116 – Final – florin – (58140) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A cannon fires a 0 . 732 kg shell with initial velocity v i = 10 m / s in the direction θ = 42 ◦ above the horizontal. Δ x Δ h 1 m / s 4 2 ◦ Δ y y The shell’s trajectory curves downward be- cause of gravity, so at the time t = 0 . 664 s the shell is below the straight line by some vertical distance Δ h . Find this distance Δ h in the absence of air resistance. The acceleration of gravity is 9 . 8 m / s 2 . 1. 0.230736 2. 1.18611 3. 2.16039 4. 0.700132 5. 1.55315 6. 0.57648 7. 1.53116 8. 1.64836 9. 2.89767 10. 2.58979 Correct answer: 2 . 16039 m. Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: ˆ x = tv i cos θ , ˆ y = tv i sin θ . The gravity does not affect the x coordinate of the shell, but it does pull its y coordinate downware at a constant acceleration a y = − g , so x = tv i cos θ, y = tv i sin θ − g t 2 2 . Thus, x = ˆ x but y = ˆ y − 1 2 gt 2 ; in other words, the shell deviates from the straight-line path by the vertical distance Δ h = ˆ y − y = g t 2 2 . Note: This result is completely indepen- dent of the initial velocity v i or angle θ of the shell. It is a simple function of the flight time t . Δ h = g t 2 2 = (9 . 8 m / s 2 ) (0 . 664 s) 2 2 = 2 . 16039 m . 002 10.0 points A car is moving at constant speed on the freeway. The driver sees a patrol car at time t 1 and rapidly slows down by around 10 miles per hour. After continuing at this speed for a few minutes, the driver at time t 2 returns to the earlier constant speed. Which graph correctly describes the car’s acceleration a ( t ) as a function of time? Take the forward direction of motion as positive. 1. t 1 time a t 2 2. t 1 time a t 2 correct Version 116 – Final – florin – (58140) 2 3. t 1 time a t 2 4. t 1 time a t 2 5. t 1 time a t 2 6. t 1 time a t 2 7. t 1 time a t 2 8. t 1 time a t 2 Explanation: Analyze the acceleration over each part of the trip: 1) Moves at constant speed: a = 0 2) Rapidly slows down: a < 0 briefly 3) Continues at this speed: a = 0 4) Returns to earlier speed: a > 0 briefly 5) Original constant speed: a = 0 003 10.0 points A mouse is running against the motion of a conveyor belt. He discovers that if he aims into the belt’s motion at an angle 11 ◦ away from the motion of the belt, he ends up run- ning directly across. If the mouse can maintain a speed of 42 cm / s relative to the belt, what is the speed of the belt? 1. 39.2043 2. 57.3212 3. 36.133 4. 38.3133 5. 49.2622 6. 45.8385 7. 50.2295 8. 35.4639 9. 34.1847 10. 41.2283 Correct answer: 41 . 2283 cm / s....
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phyfinal - Version 116 – Final – florin –(58140 1...

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