[Solutions Manual] [Instructors] Physics by Resnick Halliday Krane, 5th Ed. Vol 2

[Solutions Manual] [Instructors] Physics by Resnick Halliday Krane, 5th Ed. Vol 2

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Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College Volume 2

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A Note To The Instructor... The solutions here are somewhat brief, as they are designed for the instructor, not for the student. Check with the publishers before electronically posting any part of these solutions; website, ftp, or server access must be restricted to your students. I have been somewhat casual about subscripts whenever it is obvious that a problem is one dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution. Although this does not change the validity of the answer, it will sometimes obfuscate the approach if viewed by a novice. There are some traditional formula, such as v 2 x = v 2 0 x + 2 a x x, which are not used in the text. The worked solutions use only material from the text, so there may be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an easier approach existed. But if it was not in the text, I did not use it here. I also tried to avoid reinventing the wheel. There are some exercises and problems in the text which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer you to the previous solution. I adopt a different approach for rounding of significant figures than previous authors; in partic- ular, I usually round intermediate answers. As such, some of my answers will differ from those in the back of the book. Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual with considerably more detail and, when appropriate, include discussion on any physical implications of the answer. These student solutions carefully discuss the steps required for solving problems, point out the relevant equation numbers, or even specify where in the text additional information can be found. When two almost equivalent methods of solution exist, often both are presented. You are encouraged to refer students to the Student’s Solution Manual for these exercises and problems. However, the material from the Student’s Solution Manual must not be copied. Paul Stanley Beloit College [email protected] 1
E25-1 The charge transferred is Q = (2 . 5 × 10 4 C / s)(20 × 10 - 6 s) = 5 . 0 × 10 - 1 C . E25-2 Use Eq. 25-4: r = s (8 . 99 × 10 9 N · m 2 / C 2 )(26 . 3 × 10 - 6 C)(47 . 1 × 10 - 6 C) (5 . 66 N) = 1 . 40 m E25-3 Use Eq. 25-4: F = (8 . 99 × 10 9 N · m 2 / C 2 )(3 . 12 × 10 - 6 C)(1 . 48 × 10 - 6 C) (0 . 123 m) 2 = 2 . 74 N . E25-4 (a) The forces are equal, so m 1 a 1 = m 2 a 2 , or m 2 = (6 . 31 × 10 - 7 kg)(7 . 22 m / s 2 ) / (9 . 16 m / s 2 ) = 4 . 97 × 10 - 7 kg . (b) Use Eq. 25-4: q = s (6 . 31 × 10 - 7 kg)(7 . 22 m / s 2 )(3 . 20 × 10 - 3 m) 2 (8 . 99 × 10 9 N · m 2 / C 2 ) = 7 . 20 × 10 - 11 C E25-5 (a) Use Eq. 25-4, F = 1 4 π 0 q 1 q 2 r 2 12 = 1 4 π (8 . 85 × 10 - 12 C 2 / N · m 2 ) (21 . 3 μ C)(21 . 3 μ C) (1 . 52 m) 2 = 1 . 77 N (b) In part (a) we found F 12 ; to solve part (b) we need to first find F 13 . Since q 3 = q 2 and r 13 = r 12 , we can immediately conclude that F 13 = F 12 .

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