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Unformatted text preview: miller (zdm77) – Homework03 – Chiu – (60180) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A ball rolling up a hill has vector velocities vectorv 1 and vectorv 2 at times t 1 and t 2 , respectively, as shown in the figure. v 1 initial v 2 final Which vector diagram below most accu rately depicts the direction of the ball’s aver age acceleration over the interval? 1. 2. 3. 4. 5. 6. 7. correct 8. 9. Zero vector. Explanation: In fact the two forces exerting on the ball, the gravitational force from the earth and the force from the incline remain unchanged during the interval, so the total acceleration should be downward to the left and won’t change either. 002 (part 1 of 2) 10.0 points A commuter airplane starts from an airport and takes the route shown in the figure. It first flies to city A located at 178 km in a direction 40 ◦ north of east. Next, it flies 154 km 17 ◦ west of north to city B . Finally, it flies 182 km due west to city C . 1 7 8 k m 40 ◦ A 1 5 4 k m 17 ◦ B 182 km C R C x (km) y (km) 50 100 150 200 50 100 150 200 250 W E S N How far away from the starting point is city C ? Correct answer: 276 . 95 km. Explanation: Let : a = 178 km , α = 40 ◦ , b = 154 km , β = 17 ◦ , and c = 182 km . a b α β The xcomponent of the resultant is r x = a x + b x + c x = a cos α b sin β c = (178 km) cos40 ◦ (154 km) sin17 ◦ 182 km = 90 . 6693 km and the ycomponent of the resultant is r y = a y + b y + c y miller (zdm77) – Homework03 – Chiu – (60180) 2 = a sin α + b cos β + 0 = (178 km) sin40 ◦ + (154 km) cos17 ◦ = 261 . 687 km , so the resultant is R = radicalBig r x 2 + r y 2 = radicalBig ( 90 . 6693 km) 2 + (261 . 687 km) 2 = 276 . 95 km . 003 (part 2 of 2) 10.0 points What is the direction of the final position vector r , measured from North? Use coun terclockwise as the positive angular direction, between the limits of 180 ◦ and +180 ◦ . Correct answer: 19 . 1102 ◦ . Explanation: r γ r y r x Since γ is the angle between r and the y axis, for r in the second quadrant tan γ =  r x   r y  . Thus γ = tan − 1 parenleftbigg  r x   r y  parenrightbigg = tan − 1 parenleftbigg 90 . 6693 km 261 . 687 km parenrightbigg = 19 . 1102 ◦ . 004 10.0 points In a TV set, an electron beam moves with horizontal velocity of 4 . 6 × 10 7 m / s across the cathode ray tube and strikes the screen, 41 cm away. The acceleration of gravity is 9 . 8 m / s 2 . How far does the electron beam fall while traversing this distance? Correct answer: 3 . 89267 × 10 − 16 m. Explanation: Since we are looking for vertical drop, we use the horizontal displacement to find the time. Horizontally, x = 0 and a x = 0, so x = v x t t = x v x = 41 cm 4 . 6 × 10 7 m / s · 1 m 100 cm = 8 . 91304 × 10 − 9 s Vertically, v y = 0, y = 0 and all motion is downward, so y = 1 2 g t 2 = 1 2 ( 9 . 8 m / s...
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This note was uploaded on 09/30/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

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