$\frac{\mathbf{8}\mathbf{}\mathbf{hr}}{\mathbf{1}\mathbf{}\mathbf{day}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{kW}}{\mathbf{1}\mathbf{}{\mathbf{m}}^{\mathbf{2}}}\mathbf{\times}\frac{\mathbf{16}\mathbf{}}{\mathbf{100}}$** = 1.28 kWh/m ^{2 }per day**

You may want to reference (Pages 279 - 282) section 6.10 while completing this problem.

In a sunny location, sunlight has a power density of about 1 kW/m^{2}. Photovoltaic solar cells can convert this power into electricity with 16% efficiency.

If a typical home uses 390 kWh of electricity per month, how many square meters of solar cells would be required to meet its energy requirements? Assume that electricity can be generated from sunlight for 8 hours per day.

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