The average intensity of the sunlight in Miami,Florida,is 1.04 kW/m².For surfaces on which the light is all absorbed,what is the average value of the radiation pressure due to this sunlight in Miami? (c = 3.00 × 10⁸ m/s,μ₀ = 4π × 10^-7 T ∙ m/A,ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)2.28 µPa B)1.73 µPa C)6.93 µPa D)3.47 µPa E)9.78 µPa

Question

Quizplus · Accepted Answer

The radiation pressure $P$ due to absorbed light is given by $P = \frac{I}{c}$, where $I$ is the intensity of the sunlight, and $c$ is the speed of light. Substituting the given values, $I = 1.04 \, \text{kW/m}^2 = 1.04 \times 10^3 \, \text{W/m}^2$ and $c = 3.00 \times 10^8 \, \text{m/s}$, we get $P = \frac{1.04 \times 10^3}{3.00 \times 10^8} = 3.47 \times 10^{-6} \, \text{Pa} = 3.47 \, \mu\text{Pa}$.

The Average Intensity of the Sunlight in Miami,Florida,is 1