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First Exam-solutions

First Exam-solutions - Version 003 – First Exam –...

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Unformatted text preview: Version 003 – First Exam – Hoffmann – (56745) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A vertical electric field of magnitude 18900 N / C exists above Earth’s surface on a day when a thunderstorm is brewing. A car with a rectangular size of 2 . 84 m by 6 . 38 m is traveling along a roadway sloping downward at 7 ◦ . Determine the electric flux through the bot- tom of the car. 1. 381784.0 2. 264972.0 3. 750151.0 4. 1115550.0 5. 973672.0 6. 339903.0 7. 285036.0 8. 472045.0 9. 491798.0 10. 700047.0 Correct answer: 3 . 39903 × 10 5 N · m 2 / C. Explanation: Let : a = 2 . 84 m , b = 6 . 38 m , E = 18900 N / C , and θ = 7 ◦ . Flux is Φ = E A cos θ = E ab cos θ = (18900 N / C) (2 . 84 m) (6 . 38 m) cos 7 ◦ = 3 . 39903 × 10 5 N · m 2 / C 002 10.0 points A charged particle is connected to a string that is is tied to the pivot point P . The particle, string, and pivot point all lie on a horizontal table (consequently the figure below is viewed from above the table). The particle is initially released from rest when the string makes an angle 47 ◦ with a uniform electric field in the horizontal plane (shown in the figure). 47 ◦ 2 . 1 m 323 V / m P . 012 kg 1 μ C initial release ω parallel P Determine the speed of the particle when the string is parallel to the electric field. 1. 0.463588 2. 0.239069 3. 0.243348 4. 0.189605 5. 0.235129 6. 0.587406 7. 0.207405 8. 0.161848 9. 0.195513 10. 0.204166 Correct answer: 0 . 189605 m / s. Explanation: Let : L = 2 . 1 m , q = 1 μ C = 1 × 10 − 6 C , m = 0 . 012 kg , and θ = 47 ◦ . θ L E P m q Let V = 0 at point P . The potential at the initial position is V i = − vector E · vectors = − E L cos θ , Version 003 – First Exam – Hoffmann – (56745) 2 and the potential at the final position is V f = − E L. The potential energy U is given by U = qV , so by conservation of energy ( K + U ) i = ( K + U ) f − q E L cos θ = 1 2 mv 2 − q E L. v = radicalbigg 2 q E L (1 − cos θ ) m . Since q E L = (1 × 10 − 6 C) (323 V / m) (2 . 1 m) = 0 . 0006783 m 2 · kg / s 2 , then v = radicalBig 2 (0 . 0006783 m 2 · kg / s 2 ) × radicalbigg 1 − cos 47 ◦ . 012 kg = . 189605 m / s . 003 10.0 points Three identical point charges, each of mass 60 g and charge + q , hang from three strings, as in the figure. 9 . 8 m / s 2 1 . 4 c m 60 g + q 1 . 4 c m 60 g + q 53 ◦ 60 g + q If the lengths of the left and right strings are each 10 . 4 cm, and each forms an an- gle of 53 ◦ with the vertical, determine the value of q . The acceleration of gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 ....
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First Exam-solutions - Version 003 – First Exam –...

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