Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337671729
Author: SERWAY
Publisher: Cengage
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Chapter 38, Problem 48AP
To determine
The reason for which the following situation is impossible.
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Chapter 38 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 38.1 - Which observer in Figure 38.1 sees the balls...Ch. 38.1 - Prob. 38.2QQCh. 38.4 - Suppose the observer O on the train in Figure 38.6...Ch. 38.4 - Prob. 38.4QQCh. 38.4 - Prob. 38.5QQCh. 38.4 - You are observing a spacecraft moving away from...Ch. 38.6 - You are driving on a freeway at a relativistic...Ch. 38.8 - Prob. 38.8QQCh. 38 - In a laboratory frame of reference, an observer...Ch. 38 - Prob. 2P
Ch. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - An astronaut is traveling in a space vehicle...Ch. 38 - Prob. 7PCh. 38 - You have been hired as an expert witness for an...Ch. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - A cube of steel has a volume of 1.00 cm3 and mass...Ch. 38 - Review. In 1963, astronaut Gordon Cooper orbited...Ch. 38 - You have an assistantship with a math professor in...Ch. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - A moving rod is observed to have a length of =...Ch. 38 - Prob. 18PCh. 38 - Prob. 19PCh. 38 - You have been hired as an expert witness in the...Ch. 38 - Figure P38.21 shows a jet of material (at the...Ch. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Prob. 27PCh. 38 - (a) Find the kinetic energy of a 78.0-kg...Ch. 38 - Prob. 29PCh. 38 - Prob. 30PCh. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Prob. 33PCh. 38 - Prob. 34PCh. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - An unstable particle with mass m = 3.34 1027 kg...Ch. 38 - Prob. 41PCh. 38 - Prob. 42APCh. 38 - Prob. 43APCh. 38 - Prob. 44APCh. 38 - Prob. 45APCh. 38 - Prob. 46APCh. 38 - Prob. 47APCh. 38 - Prob. 48APCh. 38 - Prob. 49APCh. 38 - Prob. 50APCh. 38 - Prob. 51APCh. 38 - Prob. 52APCh. 38 - The creation and study of new and very massive...Ch. 38 - Prob. 54CPCh. 38 - Prob. 55CP
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- Suppose an electron (q= -e = -1.6 x 10^-19 C, m = 9.1 x 10^-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. Use the template in the attached photo to solve for the problem.arrow_forwardSuppose an electron (g= - e= -1.6x 10-19 C ,m=9.1× 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for 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 Since K=mv and using the formula for potential energy above, we arrive at an equation for speed: v = ( 1/2 Plugging in values, the value of the electron's speed is: V= x 107 m/sarrow_forwardA proton is accelerated by a potential difference of 10 kV. How fast is the proton moving if it started from rest? A. 9.41 x 10-6 m/s B. 3.45 x 106 m/s C. 1.38 x 106 m/s D. 2.12 x 106 m/s Select one: О а. А O b. B О с. С d. D Clear my choicearrow_forward
- Starting from rest, a proton falls through a potential difference of 1600 V. What speed does it acquire? [mass for proton = 1.66 x 10-27 kg] а. 1.4 х 105 m/s b. 2.8 x 105 m/s с. 3.6 х 105 m/s d. 5.7 x 105 m/sarrow_forwardA proton is accelerated by a potential difference of 10 kV. How fast is the proton moving if it started from rest? A. 9.41 x 10-6 m/s В. 3.45 х 106 т/s С. 1.38 х 106 т/s D. 2.12 x 105 т/sarrow_forwardSuppose an electron (q= -e= -1.6 x 10-19 C,m=9.1x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for 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 Since K- and using the formula for potential energy above, we arrive at an equation for speed: v = ( 51/2 Plugging in values, the value of the electron's speed is: V= x 107 m/sarrow_forward
- A positron (a particle with a charge +e and a mass equal to that of electron) that is accelerated from rest between two points at a fixed potential difference acquires a speed of 9.0x10^7 m/s. What speed is achieved by a proton accelerated from rest between the same two points? (Disregard relativistic effects.) a) 2.5x10^6 m/s b) 2.1x10^6 m/s c) 2.8x10^7 m/s d) 4.9x10^7m/s e) None of the Abovearrow_forwardSuppose an electron (q = - e= - 1.6 x 10-19 C.m = 9.1 x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for 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: 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=arrow_forwardIn 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? (g. = 1.602 x 10-19C ;me = 9.11 x 10 -31 kg) Answer: x10' m (express your answers in tenths place or one decimal digit only)arrow_forward
- An electron is accelerated from rest through a potential difference of 3.00 kV. What is its final velocity? The mass of an electron is 9.109×10- 31 kg. a. 7.26x105 m/s b. 1.03×106 m/s c. 22.97x106 m/s d. 32.48×106 m/sarrow_forwardIn a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 17000 V. The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.arrow_forwardA cathode-ray tube accelerates electrons to a speed of 26500 kms−1V. What is the potential difference across the tube?arrow_forward
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