A defibrillator is used during a heart attack to restore the heart to its normal beating pattern. A defibrillator passes 18 A of current

through the torso of a person in 2.0 ms.

a) How much charge moves during this time?

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

a) 0.036 C

b) 2.25E17

Q = I*t, which means

(charge,coulombs)
= [current,amps)*(time,seconds)

Divide the Charge Q by the electron charge, e, to get the number of electrons.

Look up the electron charge if you don't know it. It is a number you should memorize. You should also memorize the formula.

122

in the mean time (nine) minus (one) (eight)

They're gonna want his milk money next poor Pluto
Bring back Pluto
This is Russian dolls that getting smaller and smaller still, jesus pleebus man now tell wtf am i supposed to do
we were busy putting bars on a big iron gate

To find the answers to these questions, we can use the formula:

Q = I * t

where Q is the charge, I is the current, and t is the time.

a) To find the amount of charge that moves during this time, we need to calculate Q using the given values. Given that the current is 18 A and the time is 2.0 ms (which can be converted to seconds by dividing by 1000), we can calculate Q as follows:

Q = 18 A * 2.0 ms / 1000

First, we convert 2.0 ms to seconds:

2.0 ms = 2.0 ms / 1000 = 0.002 s

Plugging in the values, we have:

Q = 18 A * 0.002 s = 0.036 C

Therefore, during this time, 0.036 C (coulombs) of charge moves.

b) To determine how many electrons pass through the wires connected to the patient, we need to know the charge of a single electron. The elementary charge, denoted by e, is approximately equal to 1.6 x 10^-19 coulombs.

To calculate the number of electrons, we can use the formula:

Number of electrons = Charge (Q) / Elementary charge (e)

Given that Q is 0.036 C and e is 1.6 x 10^-19 C, we can substitute these values into the formula:

Number of electrons = 0.036 C / (1.6 x 10^-19 C)

Simplifying,

Number of electrons = 2.25 x 10^17 electrons

Therefore, approximately 2.25 x 10^17 electrons pass through the wires connected to the patient during this time.