Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

Post by answerhappygod »

Ch 14 Hw Part 2 Problem 14 54 Part A What Is The Magnitude Of The Momentum Of An Electron In A Reference Frame In Which 1
Ch 14 Hw Part 2 Problem 14 54 Part A What Is The Magnitude Of The Momentum Of An Electron In A Reference Frame In Which 1 (40.38 KiB) Viewed 12 times
<Ch 14 HW Part 2 Problem 14.54 Part A What is the magnitude of the momentum of an electron in a reference frame in which it is moving at 0.540 co? p= IVF ΑΣΦ 6 1.47.10-24 Submit Previous Answers Request Answer * Incorrect; Try Again; 5 attempts remaining Part B V= At what speed must the electron move in a reference frame in which it has twice the momentum you calculated in part A? Submit IVI ΑΣΦΑ ^ ? Request Answer kg-m 6 ? C₁₁ 9 of 10 > Review
Problem 14.69 Antihydrogen is the only antimatter element that has been produced in the laboratory, albeit just a few atoms at a time. Each antihydrogen atom consists of a positron in orbit around an antiproton and has the same atomic mass as hydrogen. If an antihydrogen atom collides with a hydrogen atom, they annihilate each other and create gamma radiation. Part A What minimum amount of energy released in this process? Express your answer using three significant digits. | ΑΣΦ + E= Submit Part B Request Answer Value Submit If this energy could be harnessed in a matter-antimatter automobile engine, how far could a car travel on 10 mg each of antihydrogen and hydrogen? At highway speeds, a typical automobile expends about 2.5 x 10³ J per meter. Express your answer with the appropriate units. Request Answer Units ? ? 10 of 10 J
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply