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Unformatted text preview: pan (zp695) – Ch17h1 – chiu – (56565) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What is the kinetic energy of a proton that is traveling at a speed of 3850 m / s? Take the mass of the proton to be 1 . 67 × 10 27 kg. Correct answer: 1 . 23768 × 10 20 J. Explanation: Since this speed is much less than the speed of light, we just use the approximate formula for kinetic energy: KE = 1 2 mv 2 = 1 2 (1 . 67 × 10 27 kg)(3850 m / s) 2 = 1 . 23768 × 10 20 J . 002 10.0 points If the kinetic energy of an electron is 4 . 7 × 10 18 J, what is the speed of the elec tron? You can use the approximate (non relativistic) formula here. Take the mass of the electron to be 9 . 11 × 10 31 kg. Correct answer: 3 . 21222 × 10 6 m / s. Explanation: The nonrelativistic formula for kinetic en ergy is KE = 1 2 mv 2 . Rearranging this expression to solve for the speed, we obtain v = 2( KE ) m = 2(4 . 7 × 10 18 J) 9 . 11 × 10 31 kg = 3 . 21222 × 10 6 m / s . 003 (part 1 of 4) 10.0 points Locations A , B , and C are in a region of uniform electric field, as shown in the diagram below. A B C $ E Location A is at $r A = . 7 , , m . Location B is at $r B = . 8 , , m . In the region the electric field has the value $ E = 780 , , N / C . For a path starting at B and ending at A , the displacement vector Δ $ # will be of the form Δ $ # = Δ x, , . Find Δ x . Correct answer: 1 . 5 m. Explanation: This is pretty straightforward. To find the vector pointing from B to A , we just subtract: Δ $ # = $r A $r B = . 7 , , m . 8 , , m = 1 . 5 , , m . So Δ x is just 1 . 5 m . 004 (part 2 of 4) 10.0 points Now find the change in electric potential along this path. Correct answer: 1170 V. Explanation: We can just use the equation Δ V = $ E · Δ $ #. pan (zp695) – Ch17h1 – chiu – (56565) 2 Δ V = E x Δ x = (780)( 1 . 5) = 1170 V . 005 (part 3 of 4) 10.0 points Find the change in potential energy when a proton ( m p = 1 . 7 × 10 27 kg and q p = 1 . 6 × 10 19 C) moves from B to A . Correct answer: 1 . 872 × 10 16 J. Explanation: Here we can use the expression Δ U = q Δ V. Δ U = q p Δ V = (1 . 6 × 10 19 C)(1170 V) = 1 . 872 × 10 16 J . 006 (part 4 of 4) 10.0 points Find the change in potential energy when an electron ( m e = 9 . 1 × 10 31 kg and q e = 1 . 6 × 10 19 C) moves from B to A ....
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This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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