# final 01 - Version 002/AAAAC – final 01 – Turner...

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Unformatted text preview: Version 002/AAAAC – final 01 – Turner – (59070) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Given: Assume the bar and rails have neg- ligible resistance and friction. In the arrangement shown in the figure, the resistor is 6 Ω and a 7 T magnetic field is directed into the paper. The separation between the rails is 2 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 5 m / s . m ≪ 1g 5 m / s 6Ω 7 T 7 T I 2m At what rate is energy dissipated in the resistor? 1. 2508.8 2. 5512.5 3. 648.0 4. 10035.2 5. 432.0 6. 400.0 7. 816.667 8. 6.25 9. 81.0 10. 8.0 Correct answer: 816 . 667 W. Explanation: Basic Concept: Motional E E = B ℓv . Ohm’s Law I = V R . Solution: The motional E induced in the circuit is E = B ℓv = (7 T) (2 m) (5 m / s) = 70 V . From Ohm’s law, the current flowing through the resistor is I = E R = B ℓv R = (7 T) (2 m) (5 m / s) R = 11 . 6667 A . The power dissipated in the resistor is P = I 2 R = B 2 ℓ 2 v 2 R 2 R = B 2 ℓ 2 v 2 R = (7 T) 2 (2 m) 2 (5 m / s) 2 (6 Ω) = 816 . 667 W . Note: Second of four versions. 002 10.0 points What is the longest wavelength that can be observed in the fifth-order spectrum using a diffraction grating with 4800 slits per cen- timeter? 1. 416.667 2. 540.541 3. 454.545 4. 408.163 5. 606.061 6. 512.821 7. 434.783 8. 666.667 9. 500.0 10. 487.805 Correct answer: 416 . 667 nm. Explanation: Version 002/AAAAC – final 01 – Turner – (59070) 2 Let : N = 4800cm − 1 = 4 . 8 × 10 5 m − 1 . The interference maxima are at angles given by d sin θ = mλ, and for a diffraction grating, d = 1 N . Therefore the longest wavelength that can be observed in the fifth-order spectrum is λ = d sin90 ◦ 5 = 1 5 N = 1 5 (4 . 8 × 10 5 m − 1 ) · 10 9 nm m = 416 . 667 nm . 003 10.0 points Consider this particular situation, which is spherically symmetric. We have a charge q 1 on a metallic ball at the center, inside of a conducting shell of inner radius R 2 and outer radius R 3 . There is a total charge of q 2 on the shell. O q 2 q 1 A B C a b c R 1 , q 1 R 2 , q ′ 2 R 3 , q ′′ 2 Find E at C, where OC = c . 1. E C = 0 2. E C = k 3 q 1 c 2 3. E C = k q 1 + q 2 b 2 4. E C = k q 1 − q 2 2 a 2 5. E C = k q 1 c 2 6. E C = k q 1 + q 2 c 2 correct 7. E C = k q 1 − q 2 c 2 8. E C = k q 1 2 b 2 9. E C = k q 1 a 2 10. E C = k 4 q 1 c 2 Explanation: Here the Gaussian surface is a sphere cen- tered at the point charge q 1 and of radius c . The enclosed charge in this sphere is all the charge, or q 1 + q 2 . The electric field at C is E C = k q 1 + q 2 c 2 ....
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final 01 - Version 002/AAAAC – final 01 – Turner...

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