CH18_Sol - homework 13 BROWNING, ADAM Due: Nov 30 2007,...

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Unformatted text preview: homework 13 BROWNING, ADAM Due: Nov 30 2007, 1:00 am 3 What is the wavelength of the sound of the ambulances siren if you are standing at the position of the car? Correct answer: 0 . 413604 m (tolerance 1 %). Explanation: Let : v car = 33 . 3 m / s , v amb = 64 . 4 m / s , v sound = 343 m / s , and f = 985 Hz . The wavelength of the sound emitted be- hind the ambulance is = v sound v amb f = v sound + v amb f = 343 m / s + 64 . 4 m / s 985 Hz = . 413604 m . The positive sign arises because the ambu- lance driver is traveling in the opposite direc- tion as the sound waves. Here the wave form is stretched out, which results in a longer wavelength. Question 7, chap 17, sect 4. part 2 of 2 10 points At what frequency does the driver of the car hear the ambulances siren? Correct answer: 909 . 807 Hz (tolerance 1 %). Explanation: The observed frequency is f = v . The car driver sees the sound wave overtaking him with a velocity of v = v sound + v car , so the car driver hears a sound with a frequency f = v sound + v car . In terms of the original frequency, f = parenleftbigg v sound v car v sound v amb parenrightbigg f = parenleftbigg v sound + v car v sound + v amb parenrightbigg f = parenleftbigg 343 m / s + 33 . 3 m / s 343 m / s + 64 . 4 m / s parenrightbigg (985 Hz) = 909 . 807 Hz . Question 8, chap 17, sect 4. part 1 of 1 10 points A jet fighter plane travels in horizontal flight at Mach 1 . 63. At the instant an observer on the ground hears the shock wave, what is the angle her line of sight makes with the horizontal as she looks at the plane? Correct answer: 37 . 8428 (tolerance 1 %). Explanation: Let : v = 1 . 63 c. sin = v sound v jet = c 1 . 63 c = arcsin parenleftbigg 1 1 . 63 parenrightbigg = 37 . 8428 . Question 9, chap 18, sect 4. part 1 of 1 10 points The small piston of a hydraulic lift has a cross-sectional area of 8 cm 2 and the large piston has an area of 32 cm 2 , as in the figure below. homework 13 BROWNING, ADAM Due: Nov 30 2007, 1:00 am 4 34 kN F 8 cm 2 area 32 cm 2 What force F must be applied to the small piston to raise a load of 34 kN? Correct answer: 8500 N (tolerance 1 %). Explanation: Let : A 1 = 8 cm 2 , A 2 = 32 cm 2 , W = 34 kN , and F = F . According to Pascals law, the pressure ex- erted on A 1 must be equal to the one exerted on A 2 . The pressure P 1 = F A 1 must be equal to the pressure P 2 = W A 2 due to the load. F A 1 = W A 2 , so F = A 1 A 2 W = (8 cm 2 ) (32 cm 2 ) (34000 N) = 8500 N . Question 10, chap 18, sect 2. part 1 of 1 10 points A constriction in a pipe reduces its diameter from 7 . 1 cm to 3 . 3 cm. Where the pipe is wider, the fluid velocity is 8.0 m/s....
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CH18_Sol - homework 13 BROWNING, ADAM Due: Nov 30 2007,...

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