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Ch17-h1-solutions

# Ch17-h1-solutions - liu(ql744 Ch17-h1 chiu(56565 This...

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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 3575 m / s? Take the mass oF the proton to be 1 . 67 × 10 - 27 kg. Correct answer: 1 . 06718 × 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)(3575 m / s) 2 = 1 . 06718 × 10 - 20 J . 002 10.0 points IF the kinetic energy oF an electron is 5 . 3 × 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 . 41109 × 10 6 m / s. Explanation: The non-relativistic Formula For kinetic en- ergy is = 1 2 2 . Rearranging this expression to solve For the speed, we obtain v = ± 2( ) m = ² 2(5 . 3 × 10 - 18 J) 9 . 11 × 10 - 31 kg = 3 . 41109 × 10 6 m / s . 003 (part 1 of 4) 10.0 points Locations A , B ,a n d C are in a region oF uniForm electric feld, as shown in the diagram below. AB C % E Location A is at %r A = ±- 0 . 7 , 0 , 0 ² m . Location B is at B = ± 0 . 7 , 0 , 0 ² m . In the region the electric feld has the value % E = ± 740 , 0 , 0 ² N / C . ±or a path starting at B and ending at A , the displacement vector Δ % \$ will be oF the Form Δ % \$ = ± Δ x, 0 , 0 ² . ±ind Δ x . Correct answer: - 1 . 4m. Explanation: This is pretty straightForward. To fnd the vector pointing From B to A ,wejustsubtract: Δ % \$ = A - B = ±- 0 . 7 , 0 , 0 ² m 0 . 7 , 0 , 0 ² m = ±- 1 . 4 , 0 , 0 ² m . So Δ x is just - 1 . 4m . 004 (part 2 of 4) 10.0 points Now fnd the change in electric potential along this path. Correct answer: 1036 V. Explanation: We can just use the equation Δ V = - % E · Δ % \$.

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Δ V = - E x Δ x = - (740)( - 1 . 4) = 1036 V . 005 (part 3 of 4) 10.0 points Find the change in potential energy when ap r o t on( m p =1 . 7 × 10 - 27 kg and q p = 1 . 6 × 10 - 19 C) moves from B to A . Correct answer: 1 . 6576 × 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)(1036 V) = 1 . 6576 × 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 . 6576 × 10 - 16 J. Explanation: We can again use the expression Δ U = q Δ Δ U = q e Δ V =( - 1 . 6 × 10 - 19 C)(1036 V) = - 1 . 6576 × 10 - 16 J .
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Ch17-h1-solutions - liu(ql744 Ch17-h1 chiu(56565 This...

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