Physics Problems - thwing(jrt2329 – Homework 01 – Yao...

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Unformatted text preview: thwing (jrt2329) – 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. They are attracted to the five thou- sand billion positively charged protons in the atomic nuclei of atoms in the penny. correct 2. They cause a jam when they try to fly away. 3. They don’t have enough speed. 4. The electrons attract each other. 5. The shell of the penny prevents the elec- trons from flying. Explanation: The electrons are bound to the nuclei. 002 10.0 points Two uncharged metal balls, Z and X , stand on insulating glass rods. A third ball, carrying a positive charge, is brought near the ball X as shown in the figure. A conducting wire is then run between Z and X and then removed. Finally the third ball is removed. X Z + conducting wire When all this is finished 1. ball Z is negative and ball X is neutral. 2. balls Z and X are still uncharged. 3. ball Z is neutral and ball X is positive. 4. balls Z and X are both negative, but ball X carries more charge than ball Z . 5. ball Z is positive and ball X is negative. correct 6. balls Z and X are both negative, but ball Z carries more charge than ball X . 7. ball Z is neutral and ball X is negative. 8. balls Z and X are both positive. 9. ball Z is positive and ball X is neutral. 10. ball Z is negative and ball X is positive. Explanation: When the conducting wire is run between Z and X , some positive charge flows from X to Z under the influence of the positive charge of the third ball. Therefore, after the wire is removed, Z is charged positive and X is charged negative. 003 10.0 points A particle of mass 67 g and charge 22 μ C is released from rest when it is 38 cm from a second particle of charge − 26 μ C. Determine the magnitude of the initial ac- celeration of the 67 g particle. Correct answer: 531 . 365 m / s 2 . Explanation: Let : m = 67 g , q = 22 μ C = 2 . 2 × 10 − 5 C , d = 38 cm = 0 . 38 m , Q = − 26 μ C = − 2 . 6 × 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 = ma bardbl vectora bardbl = k e bardbl vectorq bardblbardbl vector Q bardbl md 2 = k e vextendsingle vextendsingle 2 . 2 × 10 − 5 C vextendsingle vextendsingle vextendsingle vextendsingle − 2 . 6 × 10 − 5 C vextendsingle vextendsingle (0 . 067 kg) (0 . 38 m 2 ) = 531 . 365 m / s 2 . thwing (jrt2329) – 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....
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This note was uploaded on 11/11/2009 for the course PHY 59110 taught by Professor Yao during the Spring '09 term at University of Texas.

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Physics Problems - thwing(jrt2329 – Homework 01 – Yao...

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