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Unformatted text preview: brown (twb493) – 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 3800 m / s? Take the mass of the proton to be 1 . 67 × 10 27 kg. Correct answer: 1 . 20574 × 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)(3800 m / s) 2 = 1 . 20574 × 10 20 J . 002 10.0 points If the kinetic energy of an electron is 4 . 2 × 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 . 03655 × 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 = radicalbigg 2( KE ) m = radicalBigg 2(4 . 2 × 10 18 J) 9 . 11 × 10 31 kg = 3 . 03655 × 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. b b b A B C vector E Location A is at vectorr A = (− . 5 , , ) m . Location B is at vectorr B = ( . 4 , , ) m . In the region the electric field has the value vector E = ( 790 , , ) N / C . For a path starting at B and ending at A , the displacement vector Δ vector ℓ will be of the form Δ vector ℓ = ( Δ x, , ) . Find Δ x . Correct answer: − . 9 m. Explanation: This is pretty straightforward. To find the vector pointing from B to A , we just subtract: Δ vector ℓ = vectorr A − vectorr B = (− . 5 , , ) m − ( . 4 , , ) m = (− . 9 , , ) m . So Δ x is just − . 9 m . 004 (part 2 of 4) 10.0 points Now find the change in electric potential along this path. Correct answer: 711 V. Explanation: We can just use the equation Δ V = − vector E · Δ vector ℓ. brown (twb493) – Ch17h1 – chiu – (56565) 2 Δ V = − E x Δ x = − (790)( − . 9) = 711 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 . 1376 × 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)(711 V) = 1 . 1376 × 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 12/01/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Energy, Kinetic Energy

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