physics solutions

# physics solutions - sanne(as42476 Homework 01 Yao(59110...

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sanne (as42476) – Homework 01 – Yao – (59110) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The five thousand billion freely moving elec- trons in a penny repel one another. Why don’t they fly off the penny? 1. The electrons attract each other. 2. They are attracted to the five thou- sand billion positively charged protons in the atomic nuclei of atoms in the penny. correct 3. They don’t have enough speed. 4. The shell of the penny prevents the elec- trons from flying. 5. They cause a jam when they try to fly away. Explanation: The electrons are bound to the nuclei. 002 10.0 points Two uncharged metal balls, X and Z , stand on insulating glass rods. A third ball, carrying a negative charge, is brought near the ball Z as shown in the figure. A conducting wire is then run between X and Z and then removed. Finally the third ball is removed. Z X conducting wire When all this is finished 1. ball X is neutral and ball Z is negative. 2. ball X is negative and ball Z is positive. correct 3. balls X and Z are both negative. 4. ball X is positive and ball Z is neutral. 5. balls X and Z are still uncharged. 6. balls X and Z are both positive, but ball Z carries more charge than ball X . 7. balls X and Z are both positive, but ball X carries more charge than ball Z . 8. ball X is neutral and ball Z is positive. 9. ball X is negative and ball Z is neutral. 10. ball X is positive and ball Z is negative. Explanation: When the conducting wire is run between X and Z , some negative charge flows from Z to X under the influence of the negative charge of the third ball. Therefore, after the wire is removed, X is charged negative and Z is charged positive. 003 10.0 points A particle of mass 66 g and charge 31 μ C is released from rest when it is 65 cm from a second particle of charge 28 μ C. Determine the magnitude of the initial ac- celeration of the 66 g particle. Correct answer: 279 . 762 m / s 2 . Explanation: Let : m = 66 g , q = 31 μ C = 3 . 1 × 10 5 C , d = 65 cm = 0 . 65 m , Q = 28 μ C = 2 . 8 × 10 5 C , and k e = 8 . 9875 × 10 9 . The force exerted on the particle is F = k e | q 1 | | q 2 | r 2 = m a bardbl vectora bardbl = k e bardbl vectorq bardbl bardbl vector Q bardbl m d 2 = k e vextendsingle vextendsingle 3 . 1 × 10 5 C vextendsingle vextendsingle vextendsingle vextendsingle 2 . 8 × 10 5 C vextendsingle vextendsingle (0 . 066 kg) (0 . 65 m 2 ) = 279 . 762 m / s 2 .

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sanne (as42476) – Homework 01 – Yao – (59110) 2 004 10.0 points Two charges q 1 and q 2 are separated by a distance d and exert a force F on each other. What is the new force F , if charge 1 is increased to q 1 = 5 q , charge 2 is decreased to q 2 = q 2 / 2, and the distance is decreased to d = d/ 2? Choose one 1. F = 5 / 2 F 2. F = 20 F 3. F = 50 F 4. F = 25 / 4 F 5. F = 25 F 6. F = 5 / 4 F 7. F = 25 / 2 F 8. F = 5 F 9. F = 100 F 10. F = 10 F correct Explanation: F = k q 1 q 2 r 2 = k (5 q 1 ) ( q 2 2 ) ( d 2 ) 2 = 10 k q 1 q 2 d 2 = 10 F F = 10 F .

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