promotion image of download ymail app
Promoted
Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 months ago

Physics help!?

In a wire,  8.45x10^20 electrons flow past any point during 3.89s. What is the magnitude of the current

(I)

in the wire?

1 Answer

Relevance
  • Anonymous
    4 months ago
    Favorite Answer

    charge Q = 8.45*10^20 el / (6.24*10^18) el/Co = 135 Co (A*sec)

    current I = Q/t = 135 A*sec / 3.89 sec = 34.8 A

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.