Ch17-h1-solutions - yoon (jey283) Ch17-h1 turner (56545)...

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yoon (jey283) – Ch17-h1 – turner – (56545) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points What is the kinetic energy oF a proton that is traveling at a speed oF 3650 m / s? Take the mass oF the proton to be 1 . 67 × 10 - 27 kg. Correct answer: 1 . 11243 × 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)(3650 m / s) 2 = 1 . 11243 × 10 - 20 J . 002 10.0 points IF the kinetic energy oF an electron is 4 . 1 × 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 . 00018 × 10 6 m / s. Explanation: The non-relativistic Formula For kinetic en- ergy is KE = 1 2 mv 2 . Rearranging this expression to solve For the speed, we obtain v = r 2( KE ) m = R 2(4 . 1 × 10 - 18 J) 9 . 11 × 10 - 31 kg = 3 . 00018 × 10 6 m / s . 003 (part 1 of 4) 10.0 points Locations A , B , and C are in a region oF uniForm electric feld, as shown in the diagram below. b b b A B C V E Location A is at Vr A = a− 0 . 7 , 0 , 0 A m . Location B is at Vr B = a 0 . 8 , 0 , 0 A m . In the region the electric feld has the value V E = a 730 , 0 , 0 A N / C . ±or a path starting at B and ending at A , the displacement vector Δ V will be oF the Form Δ V = a Δ x, 0 , 0 A . ±ind Δ x . Correct answer: 1 . 5 m. Explanation: This is pretty straightForward. To fnd the vector pointing From B to A , we just subtract: Δ V = Vr A Vr B = a− 0 . 7 , 0 , 0 A m − a 0 . 8 , 0 , 0 A m = a− 1 . 5 , 0 , 0 A m . So Δ x is just 1 . 5 m . 004 (part 2 of 4) 10.0 points Now fnd the change in electric potential along this path. Correct answer: 1095 V. Explanation: We can just use the equation Δ ± = V E · Δ V ℓ.
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yoon (jey283) – Ch17-h1 – turner – (56545) 2 Δ V = E x Δ x = (730)( 1 . 5) = 1095 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 . 752 × 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)(1095 V) = 1 . 752 × 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 . Correct answer: 1 . 752 × 10 - 16 J.
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Ch17-h1-solutions - yoon (jey283) Ch17-h1 turner (56545)...

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