(a) ![8.13\cdot 10^{-21}](https://tex.z-dn.net/?f=8.13%5Ccdot%2010%5E%7B-21%7D)
The magnitude of the charge of one electron is
![q=1.6\cdot 10^{-19}C](https://tex.z-dn.net/?f=q%3D1.6%5Ccdot%2010%5E%7B-19%7DC)
Here the total amount of charge that passed through the battery pack is
Q = 1300 C
So this total charge is given by
Q = Nq
where
N is the number of electrons that has moved through the battery
Solving for N,
![N=\frac{Q}{q}=\frac{1300 C}{1.6\cdot 10^{-19} C}=8.13\cdot 10^{-21}](https://tex.z-dn.net/?f=N%3D%5Cfrac%7BQ%7D%7Bq%7D%3D%5Cfrac%7B1300%20C%7D%7B1.6%5Ccdot%2010%5E%7B-19%7D%20C%7D%3D8.13%5Ccdot%2010%5E%7B-21%7D)
(b) ![4.16\cdot 10^5 J](https://tex.z-dn.net/?f=4.16%5Ccdot%2010%5E5%20J)
First, we can find the current through the battery, which is given by the ratio between the total charge (Q = 1300 C) and the time interval (t = 8.0 s):
![I=\frac{Q}{t}=\frac{1300 C}{8.0 s}=162.5 A](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BQ%7D%7Bt%7D%3D%5Cfrac%7B1300%20C%7D%7B8.0%20s%7D%3D162.5%20A)
Now we can find the power, which is given by:
![P=VI](https://tex.z-dn.net/?f=P%3DVI)
where
V = 320 V is the voltage
I = 162.5 A is the current
Subsituting,
![P=(320 V)(162.5 A)=52,000 W](https://tex.z-dn.net/?f=P%3D%28320%20V%29%28162.5%20A%29%3D52%2C000%20W)
And now we can find the total energy transferred, which is the product between the power and the time:
![E=Pt = (52,000 W)(8.0 s)=4.16\cdot 10^5 J](https://tex.z-dn.net/?f=E%3DPt%20%3D%20%2852%2C000%20W%29%288.0%20s%29%3D4.16%5Ccdot%2010%5E5%20J)
(c) 69.7 hp
Now we have to convert the power from Watt to horsepower.
We know that
1 hp = 746 W
So we can set up the following proportion:
1 hp : 746 W = x : 52,000 W
And by solving for x, we find the power in horsepower:
![x=\frac{1 hp \cdot 52,000 W}{746 W}=69.7 hp](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%20hp%20%5Ccdot%2052%2C000%20W%7D%7B746%20W%7D%3D69.7%20hp)