extra_credit 01-solutions

# extra_credit 01-solutions - howard (cah3459) – extra...

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Unformatted text preview: howard (cah3459) – extra credit 01 – turner – (56705) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points How much positive charge is in 0 . 5 kg of fluo- rine? The charge on a proton is 1 . 60218 × 10- 19 C and Avogadro’s number is 6 . 023 × 10 23 . The atomic weight (18 . 9984 g) of fluorine contains Avogadro’s number of atoms, with each atom having 9 protons and 9 electrons. Correct answer: 2 . 2857 × 10 7 C. Explanation: Let : e = 1 . 60218 × 10- 19 C , m = 0 . 5 kg = 500 g , N A = 6 . 023 × 10 23 , AW C = 18 . 9984 g , and n pr = 9 protons / atom . N pr = m m A n pr = mn pr N A AW = (500 g) (6 . 023 × 10 23 atoms) 18 . 9984 g × (9 protons / atom) = 1 . 42662 × 10 26 protons and Q = eN A = (1 . 60218 × 10- 19 C) (1 . 42662 × 10 26 ) = 2 . 2857 × 10 7 C . 002 10.0 points Two metal spheres that are initially un- charged are mounted on insulating stands, as shown. X Y---- A negatively charged rubber rod is brought close to but does not make contact with sphere X . Sphere Y is then brought close to X on the side opposite to the rubber rod. Y is allowed to touch X and then is removed some distance away. The rubber rod is then moved far away from X and Y . What are the final charges on the spheres? Sphere X Sphere Y 1. Zero Zero 2. Positive Negative correct 3. Negative Positive 4. Positive Positive 5. Negative Negative Explanation: The force is repulsive if the charges are of the same sign, so when the negatively charged rod moves close to the sphere X , the neg- atively charged electrons will be pushed to sphere Y . If X and Y are separated before the rod moves away, those charges will re- main on X and Y . Therefore, X is positively charged and Y is negatively charged. 003 10.0 points Three identical point charges, each of mass 160 g and charge + q , hang from three strings, as in the figure. 9 . 8 m / s 2 1 1 c m 160 g + q 1 1 c m 160 g + q 41 ◦ 160 g + q If the lengths of the left and right strings are each 11 cm, and each forms an an- gle of 41 ◦ 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 . howard (cah3459) – extra credit 01 – turner – (56705) 2 Correct answer: 0 . 794904 μ C. Explanation: Let : θ = 41 ◦ , m = 160 g = 0 . 16 kg , L = 11 cm = 0 . 11 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 ....
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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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extra_credit 01-solutions - howard (cah3459) – extra...

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