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Unformatted text preview: Nguyen, Thanh – Homework 19 – Due: Apr 11 2008, 7:00 pm – Inst: Weathers 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A plane electromagnetic wave has an energy flux of 331 W / m 2 . A flat, rectangular surface of dimensions 35 cm × 29 cm is placed per pendicular to the direction of the wave. The surface absorbs half of the energy and reflects half. The speed of light is 2 . 99792 × 10 8 m / s. Calculate the total energy absorbed by the surface in 0 . 8 min. Correct answer: 806 . 316 J. Explanation: Let : I = 331 W / m 2 , a = 35 cm = 0 . 35 m , b = 29 cm = 0 . 29 m , and t = 0 . 8 min . The total power incident on the surface is equal to the intensity times the area. Since this is energy per unit time, the total energy hitting the surface is U tot = I abt. From this, only half the energy is absorbed, so U abs = U tot 2 = 1 2 (331 W / m 2 )(0 . 35 m) × (0 . 29 m)(0 . 8 min) = 806 . 316 J . 002 (part 2 of 2) 10 points Calculate the momentum absorbed in this time. Correct answer: 8 . 06874 × 10 6 kg · m / s. Explanation: The momentum of the incident wave is p inc = U abs c = 2 U abs c . Since the surface reflects only 50% of the wave, the momentum upon reflection is p refl = 1 2 U tot c = U abs c . By conservation of momentum p inc = p abs + p refl p abs = p inc p refl = 2 U abs c µ U abs c ¶ = 3 U abs c = 3(806 . 316 J) 2 . 99792 × 10 8 m / s = 8 . 06874 × 10 6 kg · m / s ....
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This note was uploaded on 10/24/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
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