Question

The intensity of solar radiation that falls on a detector on Earth is 1.00 kW/m2. The detector is a square that measures 4.21 m on a side and the normal to its surface makes an angle of 30.0° with respect to the Sun’s radiation. How long will it take for the detector to measure 448 kJ of energy?

Answer 1

Givens data & Intensity (I) Area of square = (side 2 (6.21m) 2 17.7241 m² Energy (E) = 448 Jouk. solution & Intensity power Area equalia C power [Energy T time equation from equation power (Intensity) - (Area) (103) [17.726,3 POST power generated by preen detector is pes per second. Joule 170724 x 103 power per second

detector The time required 648 x103 Joule measure E = Energy Docoer pocoer time (Power time) time [Energy power [ 648 x 103 [ 17.726 x 10 103 time 25:28 second Ansurog for detector to measure Energy (E) = 648X103 Joule is time required 25.28 second