solutions 19 - Patel, Kinal Homework 19 Due: Apr 11 2008,...

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Patel, Kinal – Homework 19 – Due: Apr 11 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 2) 10 points A plane electromagnetic wave has an energy ±ux oF 445 W / m 2 . A ±at, rectangular surFace oF dimensions 57 cm × 65 cm is placed per- pendicular to the direction oF the wave. The surFace absorbs halF oF the energy and re±ects halF. The speed oF light is 2 . 99792 × 10 8 m / s. Calculate the total energy absorbed by the surFace in 1 . 4 min. Correct answer: 6924 . 64 J. Explanation: Let : I = 445 W / m 2 , a = 57 cm = 0 . 57 m , b = 65 cm = 0 . 65 m , and t = 1 . 4 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. ²rom this, only halF the energy is absorbed, so U abs = U tot 2 = 1 2 (445 W / m 2 )(0 . 57 m) × (0 . 65 m)(1 . 4 min) = 6924 . 64 J . 002 (part 2 oF 2) 10 points Calculate the momentum absorbed in this time. Correct answer: 6 . 92944 × 10 - 5 kg · m / s. Explanation: The momentum oF the incident wave is p inc = U abs c = 2 U abs c . Since the surFace re±ects only 50% oF the wave, the momentum upon re±ection 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(6924 . 64 J) 2 . 99792 × 10 8 m / s = 6 . 92944 × 10 - 5 kg · m / s . keywords: 003 (part 1 oF 2) 10 points Consider a light bulb S emitting light isotrop- ically, ( i.e. uniFormly in all directions), with a power oF 106 W. The paper is 1 m away and has an area oF 0 .
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This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.

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solutions 19 - Patel, Kinal Homework 19 Due: Apr 11 2008,...

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