Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 39, Problem 82AP
To determine
The reason for which the following situation is impossible.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
This transmission electron microscope (TEM) image of coronavirus can be taken using a beam of
electrons accelerated from rest through a potential difference of 25 kV. What is the final speed of
the electrons?
Provide the answer: .
x 108 m/s
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 proton is accelerated through a potential difference of 6 MV. (megavolts)
from rest. Calc. the final velocity in m/s (a) 1.4 X 10' (b) 2.4 X 10' (c) 3.4 X 10'
(d) 4.4 X 10'.
Chapter 39 Solutions
Physics for Scientists and Engineers With Modern Physics
Ch. 39.1 - Which observer in Figure 38.1 sees the balls...Ch. 39.1 - Prob. 39.2QQCh. 39.4 - Suppose the observer O on the train in Figure 38.6...Ch. 39.4 - Prob. 39.4QQCh. 39.4 - Prob. 39.5QQCh. 39.4 - Prob. 39.6QQCh. 39.4 - You are observing a spacecraft moving away from...Ch. 39.6 - You are driving on a freeway at a relativistic...Ch. 39.8 - Prob. 39.9QQCh. 39 - Prob. 1OQ
Ch. 39 - A spacecraft zooms past the Earth with a constant...Ch. 39 - Prob. 3OQCh. 39 - Prob. 4OQCh. 39 - Prob. 5OQCh. 39 - Prob. 6OQCh. 39 - Prob. 7OQCh. 39 - Prob. 8OQCh. 39 - Prob. 9OQCh. 39 - Prob. 10OQCh. 39 - Prob. 1CQCh. 39 - Prob. 2CQCh. 39 - Prob. 3CQCh. 39 - Prob. 4CQCh. 39 - Prob. 5CQCh. 39 - Prob. 6CQCh. 39 - Prob. 7CQCh. 39 - Prob. 8CQCh. 39 - Prob. 9CQCh. 39 - Prob. 10CQCh. 39 - Prob. 11CQCh. 39 - Prob. 12CQCh. 39 - Prob. 13CQCh. 39 - Prob. 14CQCh. 39 - Prob. 1PCh. 39 - In a laboratory frame of reference, an observer...Ch. 39 - The speed of the Earth in its orbit is 29.8 km/s....Ch. 39 - Prob. 4PCh. 39 - A star is 5.00 ly from the Earth. At what speed...Ch. 39 - Prob. 6PCh. 39 - Prob. 7PCh. 39 - Prob. 8PCh. 39 - Prob. 9PCh. 39 - An astronaut is traveling in a space vehicle...Ch. 39 - Prob. 11PCh. 39 - Prob. 12PCh. 39 - Prob. 13PCh. 39 - Prob. 14PCh. 39 - Prob. 15PCh. 39 - Prob. 16PCh. 39 - Prob. 17PCh. 39 - A cube of steel has a volume of 1.00 cm3 and mass...Ch. 39 - Prob. 19PCh. 39 - Prob. 20PCh. 39 - Prob. 21PCh. 39 - Review. In 1963, astronaut Gordon Cooper orbited...Ch. 39 - Prob. 23PCh. 39 - Prob. 24PCh. 39 - Prob. 25PCh. 39 - Prob. 26PCh. 39 - Prob. 27PCh. 39 - Prob. 28PCh. 39 - Prob. 29PCh. 39 - Prob. 30PCh. 39 - Prob. 31PCh. 39 - Prob. 32PCh. 39 - Prob. 33PCh. 39 - Prob. 34PCh. 39 - Prob. 35PCh. 39 - Prob. 36PCh. 39 - Prob. 37PCh. 39 - Prob. 38PCh. 39 - Prob. 39PCh. 39 - Prob. 40PCh. 39 - Prob. 41PCh. 39 - Prob. 42PCh. 39 - Prob. 43PCh. 39 - Prob. 44PCh. 39 - Prob. 45PCh. 39 - Prob. 46PCh. 39 - Prob. 47PCh. 39 - (a) Find the kinetic energy of a 78.0-kg...Ch. 39 - Prob. 49PCh. 39 - Prob. 50PCh. 39 - Prob. 51PCh. 39 - Consider electrons accelerated to a total energy...Ch. 39 - Prob. 53PCh. 39 - Prob. 54PCh. 39 - Prob. 55PCh. 39 - Prob. 56PCh. 39 - Prob. 57PCh. 39 - Prob. 58PCh. 39 - Prob. 59PCh. 39 - Prob. 60PCh. 39 - Prob. 61PCh. 39 - An unstable particle with mass m = 3.34 1027 kg...Ch. 39 - Prob. 63PCh. 39 - Prob. 64PCh. 39 - Prob. 65PCh. 39 - Prob. 66APCh. 39 - Prob. 67APCh. 39 - Prob. 68APCh. 39 - Prob. 69APCh. 39 - Prob. 70APCh. 39 - Prob. 71APCh. 39 - Prob. 72APCh. 39 - Prob. 73APCh. 39 - Prob. 74APCh. 39 - Prob. 75APCh. 39 - Prob. 76APCh. 39 - Prob. 77APCh. 39 - Prob. 78APCh. 39 - Prob. 79APCh. 39 - Prob. 80APCh. 39 - Prob. 81APCh. 39 - Prob. 82APCh. 39 - An alien spaceship traveling at 0.600c toward the...Ch. 39 - Prob. 84APCh. 39 - Prob. 85APCh. 39 - Prob. 86APCh. 39 - Prob. 87APCh. 39 - Prob. 88CPCh. 39 - The creation and study of new and very massive...Ch. 39 - Prob. 90CPCh. 39 - Owen and Dina are at rest in frame S, which is...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- 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
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Electric Fields: Crash Course Physics #26; Author: CrashCourse;https://www.youtube.com/watch?v=mdulzEfQXDE;License: Standard YouTube License, CC-BY