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hw1 solutions - homework 01 BOBBITT JAMES Due 4:00 am...

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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 m N A q e M = (10 electrons / molec) 1 . 4 kg 0 . 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. Doesn’t 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 μ C is released from rest when it is 72 cm from a second particle of charge 14 μ 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 μ C = 1 . 4 × 10 5 C , r = 72 cm = 0 . 72 m , Q = 14 μ 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 | m r 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 .
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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. part 2 of 2 10 points What can we conclude from the attrac- tive nature of the force between a positively charged rod and an object? 1. The object has no net charge. 2. The object has an appreciable positive charge. 3. The object does not have an appreciable negative charge. 4. Cannot be determined. 5. The object has an appreciable negative charge. 6. The object does not have an appreciable positive charge. correct Explanation: Because of the electric induction caused by the (positively or negatively) charged rod, there is always attractive force between the rod and the object unless the object has some net charge with the same sign as the charge of
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