Asked by Mary

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.
Answered by drwls
You posted this before as "Physics please clarify". See my previous answer.
Answered by Mary
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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