Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 38, Problem 48AP
To determine
The reason for which the following situation is impossible.
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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.
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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 (q = - e = - 1.6 x 10¬19 C,m=9.1 × 10¬3' 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 mv and using the formula for potential energy above, we arrive at an equation for speed: 2 Since K= v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=arrow_forwardAn electron is accelerating under a potential difference of 105 volts. (m,=9,108x10 31) a) What is the kinetic energy of the electron?arrow_forward
- Suppose an electron (q = - e= -1.6 × 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= 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: x 107 m/s V=arrow_forward(a) An positron (electron with a positive charge) starts at rest and accelerates through an electric field established by a set of parallel plates with a potential difference of 35 V. What is the speed of the positron the instant before it hits the negative plate?(e = 1.6 × 10-19 C, melectron = 9.1 × 10-31 kg) (b) Instead of hitting the negative plate, the positron, travelling East, escapes the parallel plates through a small hole and enters a magnetic field of 0.75 T directed downward. What will be the magnetic force (magnitude and direction) on the charge?(c) Once the positron has entered the magnetic field, it is in circular motion. What is the radius of the positron's circular path?arrow_forwardA proton is accelerated from rest through a potential difference of 370 V. What is its final speed?arrow_forward
- Suppose 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_forwardA 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_forward
- Starting from rest, a proton falls through a potential difference of 1 MV. Find its final velocity.arrow_forwardIf an electron is accelerated from rest through a potential difference of 9.9kV what speedarrow_forwardAn electron is accelerated from rest through a potential difference of 181 V. What is the electron's final speed? (e = 1.60×10-¹9 C, Melectron 9.11x10-31 kg) =arrow_forward
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