This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: howard (cah3459) – extra credit 01 – turner – (56705) 1 This printout should have 9 questions. Multiplechoice 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 ....
View
Full
Document
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.
 Spring '10
 Turner

Click to edit the document details