Physics Solution Manual for 1100 and 2101

# 55 10 n 180 10 m f r2 q k 899 109 n m 2 c2 n 8 e e 160

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Unformatted text preview: sion of the wire, L0 is its length at 25.0 °C, and ∆T is the amount by which the temperature drops. Since the wire is prevented from contracting, there must be a stretching force exerted at each end of the wire. According to Equation 10.17, the magnitude of this force is ∆L ∆F = Y A L 0 where Y is the Young's modulus of the wire, and A is its cross-sectional area. Combining this relation with Equation 12.2, we have α L ∆T ∆F = Y 0 L 0 A = α ( ∆T )Y A Thus, the frequency f at the lower temperature is f= ( F0 + ∆F ) /(m / L) v = = 2 L0 2 L0 F0 + α ( ∆T )Y A /(m / L) 2 L0 (2) Using Equations (1) and (2), we find that the frequency f is f = f0 f = ( 2093 Hz ) F0 + α ( ∆T )Y A /(m / L) F + α ( ∆T )Y A = f0 0 F0 F0 /(m / L) 818.0 N + (12 × 10 –6 /C°)(5.0 C°)(2.0 × 1011 N/m 2 ) (7.85 × 10 –7 m 2 ) 818.0 N = 2105 Hz Therefore, the beat frequency is 2105 Hz − 2093 Hz = 12 Hz . CHAPTER 18 ELECTRIC FORCES AND ELECTRIC FIELDS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. 1.9 × 1013 2. (b) Suppose that A is positive and B is negative. Since C and A also attract each other, C must be negative. Thus, B and C repel each other, because they have like charges (both negative). Suppose, however, that A is negative and B is positive. Since C and A also attract each other, C must be positive. Again we conclude that B and C repeal each other, because they have like charges (both positive). 3. (a) The ball is electrically neutral (net charge equals zero). However, it is made from a conducting material, so it contains electrons that are free to move. The rod attracts some of these (negative) electrons to the side of the ball nearest the rod, leaving the opposite side of the ball positively charged. Since the negative side of the ball is closer to the positive rod than the positive side, a net attractive force arises. 4. (d) The fact that the positive rod repels one object indicates that that object carries a net positive charge. The fact that the rod repels the other object indicates that that object carries a net negative charge. Since both objects are identical and made from conducting material, they share the combined net charges equally after they are touched together. Since the rod repels each object after they are touched, each object must then carry a net positive charge. But the net electric charge of any isolated system is conserved, so the total net charge initially must also have been positive. This means that the initial positive charge had the greater magnitude. 5. (c) This distribution is not possible because of the law of conservation of electric charge. The total charge on the three objects here is 9 q , whereas only q was present initially. 8 6. (c) This is an example of charging by induction. The negatively charged rod repels free electrons in the metal. These electrons move through the point of contact and into the sphere farthest away from the rod, giving it an induced charge of −q. The sphere nearest the rod acquires an induced charge of +q. As long as the rod is kept in place while the spheres are separated, these induced charges cannot recombine and remain on the spheres. 7. (b) Coulomb’s law states that the magnitude of the force is given by F = k q1 q2 . Doubling r2 the magnitude of each charge as in A would increase the numerator by a factor of four, but this is offset by the change in separation, which increases the denominator by a factor of 22 = 4. Doubling the magnitude of only one charge as in D would increase the numerator by Chapter 18 Answers to Focus on Concepts Questions 943 a factor of two, but this is offset by the change in separation, which increases the denominator by a factor of ( 2) 2 = 2. 8. (e) Coulomb’s law states that the magnitude of the force is given by F = k q1 q2 . The force r2 is directed along the line between the charges and is an attraction for unlike charges and a repulsion for like charges. Charge B is attracted by charge A with a force of magnitude qq k 2 and repelled by charge C with a force of the same magnitude. Since both forces d qq point to the left, the net force acting on B has a magnitude of 2k 2 . Charge A is attracted d qq qq by charge B with a force of k 2 and also by charge C with a force of k . Since both 2 d ( 2d ) forces point to the right, the net force acting on A has a magnitude of (1.25 ) k C is pushed to the right by B with a force of k of k qq ( 2d ) 2 qq . Charge d2 qq and pulled to the left by A with a force d2 . Since these two forces have different directions, the net force acting on C has a magnitude of ( 0.75 ) k qq . d2 9. (b) According to Coulomb’s law, the magnitude of the force that any one of the point charges qq exerts on another point charge is given by F = k 2 , where d is the length of each side of d the triangle. The charge at B experiences a repulsive force from the charge at A and an attractive force from the charge at C. Both forces have vertical components, but one points in the +y direction and the other in the −y direction. These vertical components have eq...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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