This leads (using Eq. 4-23) to 61126005sin(2.00 10 m/s)sin40.0(8.78 10 m/s )(1.96 10s)4.34 10 m/s .yvvatθ−=−=×°−×××Since the xcomponent of velocity does not change, then the final velocity is v→= (1.53×106m/s) i^−(4.34×105m/s) j^. 50. Due to the fact that the electron is negatively charged, then (as a consequence of Eq. 22-28 and Newton’s second law) the field
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This note was uploaded on 07/07/2010 for the course PHYSICS 244 taught by Professor Ds during the Fall '10 term at The Petroleum Institute.