A) What is the kinetic energy of a proton that is traveling at a speed of 2350 m

Question

A) What is the kinetic energy of a proton that is traveling at a speed of 2350 m/s?
K = _____ J

B) If the kinetic energy of an electron is 3.1e-18 J, what is the speed of the electron? (You can use the approximate (nonrelativistic) formula here.)
v = ______ m/s

C) You move from location i at < 5, 3, 5 > m to location f at < 7, 5, 11 > m. All along this path there is a nearly uniform electric field = < 1100, 170, -520 > N/C. Calculate ?V = Vf - Vi, including sign and units.
= ________V

Explanation / Answer

a ) speed of the proton is v = 2350 m/s     kinetic energy of the proton is K = 1/2 mv^2                                                         = 1/2 * 1.67 *10^-27kg * ( 2350 m/s)^2                                                         = 4.61*10^-21 J b ) kinetic energy of the electron is K = 3.1*10^-18 J                                                        K = 1/2 m v^2                                      3.1*10^-18 J = 1/ 2 * 9.31 *10^-31 kg * v^2                                              v = 2.58*10^6 m/s

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