Final answer:
The angle at which the sun hits the 6ft x 4ft solar panel, given an efficiency of 0.25 and the sun's intensity of 1360 W/m², is approximately 7°.
Step-by-step explanation:
To determine the angle at which the sun hits the solar panel, we need to use the concept of power and intensity. The power of the solar panel is given as 80W, and the intensity of the sun is given as 1360 W/m². We can calculate the area of the solar panel by converting the dimensions from feet to meters (6ft = 1.83m and 4ft = 1.22m) and multiplying them together (1.83m * 1.22m = 2.23m²).
Now, we can calculate the incident power on the solar panel by multiplying the intensity by the area (1360 W/m² * 2.23m² = 3036.8 W). The efficiency of the solar panel is given as 0.25, so the actual power output of the solar panel would be 3036.8 W * 0.25 = 759.2 W.
To find the angle at which the sun hits the solar panel, we need to use the trigonometric relationship between power and intensity. The power output of the solar panel is proportional to the cosine of the angle between the sun's rays and the normal to the solar panel's surface. Taking the inverse cosine of the ratio of the actual power output to the incident power, we can find the angle at which the sun hits the solar panel. In this case, the angle is approximately 7°.