Question: At time t = 0 , an electron with kinetic energy 9 . 7 keV moves through x = 0 in the positive direction of

At time t =0, an electron with kinetic energy 9.7 keV moves through x =0 in the positive direction of an x axis that is parallel to the horizontal component of Earth's magnetic field . The field's vertical component is downward and has magnitude 45.1T.
(a) What is the magnitude of the electron's acceleration due to ?
m/s2
(b) What is the electron's distance from the x axis itself, when the electron reaches coordinate x =23.0 cm?

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