hw1 solutions - homework 01 BOBBITT, JAMES Due: Jan 21...

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Unformatted text preview: homework 01 BOBBITT, JAMES Due: Jan 21 2008, 4:00 am 1 Question 1, chap 22, sect 1. part 1 of 1 10 points You have 1 . 4 kg of water. One mole of water has a mass of 18 . 1 g / mol and each molecule of water contains 10 electrons since water is H 2 O. What is the total electron charge contained in this volume of water? Correct answer: 7 . 45281 10 7 C (tolerance 1 %). Explanation: Let : N A = 6 . 02214 10 23 molec / mol , q e = 1 . 6 10 19 C / electron , M = 18 . 1 g / mol = 0 . 0181 kg / mol , m = 1 . 4 kg , and Z = 10 electrons / molec . The mass is proportional to the number of molecules, so for m grams in n molecules and M grams in N A molecules, m M = n N A n = m M N A . Since 10 electrons are in each molecule of water, then the total number of electrons n e in the coin is n e = Z n = Z m M N A and the total charge q for the n e electrons is q = n e q e = Z mN A q e M = (10 electrons / molec) 1 . 4 kg . 0181 kg / mol (6 . 02214 10 23 molec / mol) ( 1 . 6 10 19 C / electron) = 7 . 45281 10 7 C . Question 2, chap 22, sect 1. part 1 of 1 10 points What happens to the mass of an object when it acquires a positive net charge by the transfer of electrons? 1. Decreases correct 2. Cannot be determined 3. Increases 4. Doesnt change Explanation: When an object acquires a positive charge it loses electrons, so its mass decreases. Question 3, chap 22, sect 2. part 1 of 1 10 points A particle of mass 79 g and charge 14 C is released from rest when it is 72 cm from a second particle of charge 14 C. Determine the magnitude of the initial ac- celeration of the 79 g particle. Correct answer: 43 . 0133 m / s 2 (tolerance 1 %). Explanation: Let : m = 79 g = 0 . 079 kg , q = 14 C = 1 . 4 10 5 C , r = 72 cm = 0 . 72 m , Q = 14 C = 1 . 4 10 5 C , and k e = 8 . 9875 10 9 N m 2 / C 2 . The force exerted on the particle is F = k e | q || Q | r 2 = m | a | | a | = k e | q || Q | mr 2 = (8 . 9875 10 9 N m 2 / C 2 ) vextendsingle vextendsingle 1 . 4 10 5 C vextendsingle vextendsingle vextendsingle vextendsingle 1 . 4 10 5 C vextendsingle vextendsingle (0 . 079 kg) (0 . 72 m) 2 = 43 . 0133 m / s 2 . homework 01 BOBBITT, JAMES Due: Jan 21 2008, 4:00 am 2 Question 4, chap 22, sect 2. part 1 of 2 10 points What can we conclude from the attrac- tive nature of the force between a positively charged rod and an object? + + + F F ? 1. The object is a conductor. 2. Cannot be determined correct 3. The object is an insulator. Explanation: The attractive nature of the force will be the same whether the object is a conductor or insulator. Question 5, chap 22, sect 2....
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This note was uploaded on 09/18/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw1 solutions - homework 01 BOBBITT, JAMES Due: Jan 21...

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