The deuterium nucleus starts out with a kinetic energy of 1.88e-13 joules, and the proton starts out with a kinetic energy of 3.77e-13 joules. The radius of a proton is 0.9e-15 m; assume that if the particles touch, the distance between their centers will be twice that. What will be the total kinetic energy of both particles an instant before they touch?
Remember that the q values of a deuterium and proton is 1.6e-19. You can apply this to the (9e9)*(q1q2/r) formula where Ki + Kf = (9e9)*(q1q2/r)(1.88e-13 joules + 3.77e-13 joules). Advice for my son who's planning his curriculum :: Here is the breakdown of requirements for Physics: AP Physics requires that he have either . Please read carefully the Google Answers Terms of Service. http://answers.google.com/answers/threadview?id=210144HOME | maths, physics, heat transfer q. regarding pond:: Latent heat's working in our favour to help prevent freezing. Trying to calculate the rate of Please read carefully the Google Answers Terms of Service. http://answers.google.com/answers/threadview?id=140171HOME | Physics problems:: I have the book Physics sixth edtion by cutnell and johnson. Please read carefully the Google Answers Terms of Service. http://answers.google.com/answers/threadview/id/508144.htmlHOME |
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