An electron (m = 9.11x10-31 kg) is accelerated from rest to speed v by a conservative force. In this process, its potential energy decreases by 7.30x10-14 J. Determine the electron's speed, v. V = xa — ΑΣΦ Xb √x vx X ↑ X ? |X| x 10n •C

An electron (m = 9.11x10-31 kg) is accelerated from rest to speed v by a conservative force. In this process, its potential energy decreases by 7.30x10-14 J. Determine the electron's speed, v. V = xa — ΑΣΦ Xb √x vx X ↑ X ? |X| x 10n •C

Name: An electron (m = 9.11×10-31 kg ) is acceleratedfrom rest to speed v b
Uploaded: 2024-03-20T06:57:25.000Z
Channel: Bartleby
Description: An electron (m = 9.11×10-31 kg ) is acceleratedfrom rest to speed v by a conservative force. In thisprocess, its potential energy decreases by7.30×10-14 J.Part ADetermine the electron's speed, v.V =ΓΙ ΑΣΦxaXb01618√x vx x<४?|X| X.10n•C

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter5: Energy

Section: Chapter Questions

Problem 4CQ: (a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a...

See similar textbooks

Similar questions

A particle of mass 2.0 kg moves under the influence of the force F(x)=(3/x)N. If its speed at x=2.0 m is v=6.0 m/s, what is its speed at x = 7.0 m?
A system consists of five particles. How many terms appear in the expression for the total gravitational potential energy of the system? (a) 4 (b) 5 (c) 10 (d) 20 (e) 25
(a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.
mes New... BIUA 田回, 三三 12 + 三| 1三 2. 3 2. Ryan having mass 60 kg possesses a kinetic energy of 1.3 kJ. What is his speed? 3. Steven having mass 52 kg jumps to the surface of the table which is at the height of 45 cm. Find his gravitational potential energy.
64. A particle of mass 'm' moves under central force field 'F (r)= 7.Iftötal energy of the particle is E, what is speed of the particle when it is at a distance 'a' from the center of force. k (a) ma? 2E 2k 2E k (d) 2ma? 2E (b) V 2ma? (c) ma Solve 65 k 65. In the previous question if the force law were F(F) = -, then speed will be 2E k (a) V ma? 2k (c) k 2E (b) V 2ma? (d) 2ma? та m |E一m
II. A) What are the gravitational potential energies for Peter (mass is 80.0 kg), David (mass is 60.0 kg), and Mark (mass is 75.0 kg) from the graph. Choose the ground as the zero potential energy level, take g 10.0 m/s. B) How much work the gravitational force has to do to bring Mark and David to the ground (think about positive and negative work)? Daviel 4.00m Peter Ground help! 3 Mark 5.00m
A body with mass 0.40 kg moves in one dimension according to a potential of theform U(x) = A/x3-B/x , where A = 4.0 Jm3 abd B = 3.0 Jm, and we take x > 0. (a) If the speed of the body at x = 4.0m is v = 3.9 ms−1, what is the speed at x = 1.5 m? (b) Explain briefly what type(s) of motion are possible in this potential if the totalenergy E = 3.0 J.
An electron (mass = 9.11 × 10−31 kg) is accelerated along a straight line at 3.9 × 1014 ms−2.The electron has an initial speed of 3.2 × 106 ms−1 and travels a distance of 3.0 cm. Calculatethe electron’s: a) final speed.b) change in kinetic energy.
k: 64. A particle of mass 'm' moves under central fórce field F(r)=-.Iftötal energy of the particle is E, what is speed of the particle when it is at a distance 'a' from the center of force. 2E k E 2k 2E 2E (a) ma? (b) 2ma (c) ma (d) 2ma? +. m m m
A linear accelerator has a bundle of electrons travelling North. The bundle experiences acceleration of 276.6 m/s^2 North. The bundle has 7.49e23 electrons. Each electron has mass 9.10938e-31kg. What is the work done on the bundle due to gravity if the bundle travels 8.904 m?
A moving electron has a Kinetic Energy K1. After a net amount of work is done on it, the electron is moving one-quarter as fast in the opposite direction. What is the work done W in terms of K1? Solution To solve for the work done, first we must determine what is the final kinetic energy of the electron. By concept, we know that K1=(1/2)mv21 K2=(1/2)mv2 But it was mentioned that: v2=( V1 so, K2 in terms of v1 is K2=( mv²1 Substituting the expression for K1 results to K2=( K1 Since work done is W=AK=K -K Evaluating results to W=( K1
A moving electron has a Kinetic Energy Kq. After a net amount of work is done on it, the electron is moving one-quarter as fast in the opposite direction. What is the work done W in terms of Kq? Solution To solve for the work done, first we must determine what is the final kinetic energy of the electron. By concept, we know that K1=(1/2)mv²1 K2=(1/2)mv2 But it was mentioned that: v2=( so, K2 in terms of v1 is K2=( )mv21 Substituting the expression for K1 results to K2=( )K1 Since work done is W=AK=K -K Evaluating results to W=( )K1