Respuesta :
(a) 423 J
The power of the battery is the ratio between the total energy stored (E) and the time elapsed (t):
[tex]P=\frac{E}{t}[/tex]
However, the power is also the product of the voltage (V) and the current (I):
[tex]P=VI[/tex]
Linking the two equations together,
[tex]\frac{E}{t}=VI\\E=VIt[/tex]
Since we know:
V = 9.0 V
[tex]I \cdot t = 47.0 A\cdot h[/tex]
We can calculate the total energy:
[tex]E=(9.0 V)(47 A \cdot h)=423 J[/tex]
(b) [tex]7.79\cdot 10^{-6}[/tex] dollars
The battery has a total energy of E = 423 J. (2)
1 Watt (W) is equal to 1 Joule (J) per second (s):
[tex]1 W = \frac{1 J}{1 s}[/tex]
so 1 kW corresponds to 1000 J/s:
[tex]1 kW = \frac{1000 J}{1 s}[/tex]
Multiplying both side by 1 hour (1 h):
[tex]1 kW \cdot h = \frac{1000 J}{1 s} 1 h[/tex]
and [tex]1 h = 3600 s[/tex], so
[tex]1 kWh = \frac{1000 J}{1 s}\cdot 3600 s =3.6\cdot 10^6 J[/tex]
So we find the conversion between kWh and Joules. So now we can convert the energy from Joules (2) into kWh:
[tex]1 kWh = 3.6\cdot 10^6 J = x : 423 J\\x=\frac{1 kWh \cdot 423 J}{3.6\cdot 10^6 J}=1.18\cdot 10^{-4}kWh[/tex]
And since the cost is $0.0660 per kilowatt-hour, the total cost will be
[tex]C=$0.0660\cdot 1.18\cdot 10^{-4} kWh=7.79\cdot 10^{-6}[/tex] dollars