(a) [tex]8.13\cdot 10^{-21}[/tex]
The magnitude of the charge of one electron is
[tex]q=1.6\cdot 10^{-19}C[/tex]
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,
[tex]N=\frac{Q}{q}=\frac{1300 C}{1.6\cdot 10^{-19} C}=8.13\cdot 10^{-21}[/tex]
(b) [tex]4.16\cdot 10^5 J[/tex]
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):
[tex]I=\frac{Q}{t}=\frac{1300 C}{8.0 s}=162.5 A[/tex]
Now we can find the power, which is given by:
[tex]P=VI[/tex]
where
V = 320 V is the voltage
I = 162.5 A is the current
Subsituting,
[tex]P=(320 V)(162.5 A)=52,000 W[/tex]
And now we can find the total energy transferred, which is the product between the power and the time:
[tex]E=Pt = (52,000 W)(8.0 s)=4.16\cdot 10^5 J[/tex]
(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:
[tex]x=\frac{1 hp \cdot 52,000 W}{746 W}=69.7 hp[/tex]
