(a) A defibrillator passes 11.0 A of current through the torso of a person for 0.0300 s. How much charge moves?

C
(b) How many electrons pass through the wires connected to the patient?
electrons

To find the amount of charge that moves through the torso of a person when a defibrillator passes a current of 11.0 A for 0.0300s, we can use the formula:

Q = I * t

where Q represents the charge in coulombs (C), I represents the current in amperes (A), and t represents the time in seconds (s).

(a) Calculate the charge moved:
Q = 11.0 A * 0.0300 s
Q = 0.33 C

Therefore, the amount of charge that moves through the torso of the person is 0.33 C.

(b) To find the number of electrons that pass through the wires, we need to know the charge of a single electron and use the formula:

Q = n * e

where Q represents the charge in coulombs (C), n represents the number of electrons, and e represents the charge of a single electron.

For this calculation, we need to know the charge of a single electron, which is approximately -1.602 x 10^-19 C.

To find the number of electrons, rearrange the formula:

n = Q / e

Plugging in the values:
n = 0.33 C / (-1.602 x 10^-19 C)
n ≈ -2.1 x 10^18 electrons

Therefore, approximately 2.1 x 10^18 electrons pass through the wires connected to the patient. Note that the negative sign is due to the convention that electrons carry a negative charge.