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Final - Version 040 Final TSOI(58160 This print-out should...

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Version 040 – Final – TSOI – (58160) 1 This print-out should have 41 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 1 meter long hose of 8 cm diameter is con- nected to a faucet. The end of this hose is connected to a second hose which is 4 meters long and 4 cm in diameter. At the open end of the second hose water flows out at a rate of 8 cm 3 /s. What is the ratio of the speed of the water flowing in the second hose to the speed of the water flowing in the first hose? 1. v 2 v 1 = 1 3 2. v 2 v 1 = 8 3. v 2 v 1 = 1 8 4. v 2 v 1 = 2 5. v 2 v 1 = 1 4 6. v 2 v 1 = 1 7. Cannot be determined from the informa- tion given. 8. v 2 v 1 = 4 correct 9. v 2 v 1 = 3 10. v 2 v 1 = 1 2 Explanation: Let : r 1 = 4 cm and r 2 = 2 cm . According to the equation of continuity, v 2 v 1 = A 1 A 2 = π r 2 1 π r 2 2 = (4 cm) 2 (2 cm) 2 = 4 . 0 . 002 10.0 points A uniform horizontal rod of mass 1 . 1 kg and length 1 . 7 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I = m ℓ 2 12 . 1 . 7 m F 66 pivot 1 . 1 kg If a 6 N force at an angle of 66 to the hor- izontal acts on the rod as shown, what is the magnitude of the resulting angular accelera- tion about the pivot point? The acceleration of gravity is 9 . 8 m / s 2 . 1. 7.14862 2. 0.402982 3. 7.30469 4. 6.58537 5. 12.8831 6. 2.40796 7. 5.32443 8. 4.95164 9. 0.146427 10. 1.87803 Correct answer: 0 . 146427 rad / s 2 . Explanation: Let : = 1 . 7 m , m = 1 . 1 kg , θ = 66 , and F = 6 N . By the parallel axis theorem, the moment of inertia of a stick pivoted at the end is I = I cm + m d 2 = 1 12 m ℓ 2 + m parenleftbigg 2 parenrightbigg 2 = 1 3 m ℓ 2 .

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Version 040 – Final – TSOI – (58160) 2 The sum of the torques (counterclockwise ro- tation defined as positive) is summationdisplay τ = F ℓ sin θ m g 2 = I α F ℓ sin θ m g l 2 = parenleftbigg 1 3 m ℓ 2 parenrightbigg α 6 F sin θ 3 m g = 2 m ℓ α vectorα = 6 F sin θ 3 m g 2 m ℓ Since 6 F sin θ 3 m g = 6 (6 N) sin 66 3 (1 . 1 kg) (9 . 8 m / s 2 ) = 0 . 547636 N , then bardbl vectorα bardbl = vextendsingle vextendsingle vextendsingle 0 . 547636 N vextendsingle vextendsingle vextendsingle 2 (1 . 1 kg) (1 . 7 m) = 0 . 146427 rad / s 2 . 003 10.0 points Standing at a crosswalk, you hear a frequency of 555 Hz from the siren on an approaching police car. After the police car passes, the observed frequency of the siren is 475 Hz. Determine the car’s speed from these obser- vations. The speed of sound in air is 343 m / s. 1. 26.1394 2. 32.7622 3. 20.2939 4. 32.0744 5. 38.2577 6. 27.6199 7. 30.3597 8. 28.3635 9. 23.7723 10. 26.6408 Correct answer: 26 . 6408 m / s. Explanation: Let : f = 555 Hz , f ′′ = 475 Hz , and v = 343 m / s . From the Doppler effect, f = v ± v 0 v v s f . When the police car is approaching, the ob- served frequency is f = v v v s f and after the police car passes, the observed frequency is f ′′ = v v + v s , so f f ′′ = v v v s f v v + v s f = v + v s v v s f ( v v s ) = f ′′ ( v + v s ) v s = f f ′′ f + f ′′ v = 555 Hz 475 Hz 555 Hz + 475 Hz (343 m / s) = 26 . 6408 m / s .
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Final - Version 040 Final TSOI(58160 This print-out should...

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