The potential at location A is 379 V. A positively charged particle is released there from rest and arrives at location B with a speed vB. The potential at location C is 789 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2vB. Find the potential at B.
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You posted this before as "Physics please clarify". See my previous answer.
Sorry I mistakenly posted the wrong question. I will post the correct question. I really appreciate all of your help! You make physics same less foreign to me. Again thanks!
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