A defibrillator is used during a heart attack to restore the heart to its normal beating pattern. A defibrillator passes 21 A of current through the torso of a person in 1.8 ms. (a) How much charge moves during this time? C (b) How many electrons pass through the wires connected to the patient? electrons
I = 21 A , t=1.8 ms = 1.8e(-3) s
a) charge (q) = I*t = 21*1.8e(-3)
q = 3.78e(-2) coulombs.
That is three point seven eight e(-2)
b) number of electrons (n) = q/e here e is charge of electron which is 1.6e(-19) C
Therefore n = 3.78e(-2)/(1.6e(-19))
n = 2.36e17 electrons
That is two point three six e17
Hope this helps...
A defibrillator is used during a heart attack to restore the heart to its normal beating...
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