GR8677 #25



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harmonxjim33 20191001 14:06:03  It has been simply incredibly generous with you to provide openly what exactly many individuals wouldâ€™ve marketed for an ebook to end up making some cash for their end, primarily given that you could have tried it in the event you wanted. subway surfers unblocked   fredluis 20190809 04:26:58  Yosun has the answer as 1 ohm, that 1/2 you\'re seeing is just a 1 with capital omega. pressure cleaning   joshuaprice153 20190809 02:27:41  I feel that may be a captivating point, it made me suppose a bit. Thanks for sparking my pondering cap. Now and again I am getting such a lot in a rut that I just really feel like a record. makeup artist Tampa   UNKNOWNUMBER 20181014 22:55:11  This is how I got the correct answer, can someone confirm if it\'s viable method or if I just got lucky? \r\n\r\nWe\'re given a constant magnetic field, this means that according at Faraday\'s law, the curl of the electric field = 0. Since the electric field is what propels the charge, it must move in a straight path because there is no curl; there is no rotational aspect so it can\'t be any of the other answers. \r\n   FutureDrSteve 20111029 15:02:56  I did this one qualitatively using the righthand rule. To find the force the Bfield exerts on the particle, I first reasoned that the Efield would propel it on a vector parallel to the Efield. Pointing my fingers in that direction, it's impossible to curl them towards the Bfield, since they're parallel. Therefore, the Bfield doesn't act on the particle, and it should continue on its straightline path.   ME 20091011 16:16:13  Could it be that on GR9677: #86 you meant cycloid instead of helical as in choice B instead of C? Thank you so much for your site. I love it!
LAStew 20110912 17:11:19 
You mean answer (D)

  a19grey2 20081102 22:23:12  Note that if the fields were perpendicular to each other the electron would drift in a helical pattern as in choice (C). It seems ETS likes questions of this type (see: GR9677: #86)
 

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