In large CRT televisions, electrons are accelerated from rest by a potential difference of 23.88 kV and shot onto a phosphorescent screen to produce an image. What is the speed of the electrons when they reach the screen? ( qe = 1.602 x 10 19C ;mel = 9.11 x 10 31 kg) %3D Answer: x10' (express your answers in tenths place or one decimal digit only)

In large CRT televisions, electrons are accelerated from rest by a potential difference of 23.88 kV and shot onto a phosphorescent screen to produce an image. What is the speed of the electrons when they reach the screen? ( qe = 1.602 x 10 19C ;mel = 9.11 x 10 31 kg) %3D Answer: x10' (express your answers in tenths place or one decimal digit only)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Lned weu/ student/Assignment-Responses/submit?dep=27380743&tags=autosave#Q5 15.0 V 10.0 V 30.0 V 35.0 V 45.0 V D 20.0 V 25.0 V Find the work W.n in electron volts done by the electric force on a proton that moves from point A to point B. Similarly, find Wac Wan, and WAE: (Assume the proton starts and stops at rest. Enter your answers AB in ev.) HINT (a) W AB The work done by any conservative force can be related to the change in an associated potential energy: W = -APE. eV (b) WAC 25 X ev (c) WAD 45 x ev (d) WAr 45 Xev 2:17 PM P Type here to search (? 9/13/2021
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2. The figure shows two parallel plates. The graph indicates the potential at points (X1= 0.300 m and X2= 0.400 m) between two plates, with V2 = 20 V and V1= 15 V. An electron is released from rest at x1, and that release point and the plate locations are aligned with the graph. What is the electrons speed (m/s) just as it runs into the plate? (You must decide which plate it hits) helpful info:: grap wo plate it hits.) V₂ V₁ 0 A | | X₁ X₂ e (Electron charge q = -1.6 x 10¹9 C, AK-q AV, electron mass = 9.1 x 10-³¹ kg, V=Joule/Coulomb, 1 eV = 1.6 x 10¹9 J, K-mv²/2)
How much potential difference is needed to stop an electron with a speed of 3x105 m/s.(m=9,11x10-31 kg; e=-1.6x10-19C)
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11. An electron in the picture tube of an old television set is accelerated from rest through a potential difference Vba = +5000V. a. What is the change in potential energy of the electron? (ans: -8.0 x 1010 J) b. What is the speed of the electron as a result of this acceleration? (ans: 4.2 x 10' m/s) c. What is the speed of a proton that is accelerated through a potential difference Vba = - 5000 V? (ans: 9.8 x 10° m/s) 020-12-11 Page 4 of 17
Suppose an electron (q = -e = -1.6 x 10¬9 C,m=9.1 x 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for %3D the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K = -U 1 Since K mv and using the formula for potential energy above, we arrive at an equation for speed: 2 v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=
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