Answer: 37.5 g
Explanation: you multiply 0.0375 by 1000 which equals 37.5
I think that the solar panel would work under a fluorescent or halogen light if the photons are being produced. These types of lights mimic sunlight so it would not work as good as the real thing but it could work. Just not be as powerful.
Answer:
<h2>
650W/m²</h2>
Explanation:
Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²
Given parameters
Power rating = 6.50Watts
Cross sectional area = 100cm²
Before we calculate the intensity, we need to convert the area to m² first.
100cm² = 10cm * 10cm
SInce 100cm = 1m
10cm = (10/100)m
10cm = 0.1m
100cm² = 0.1m * 0.1m = 0.01m²
Area (in m²) = 0.01m²
Required
Intensity of the sunlight I
I = P/A
I = 6.5/0.01
I = 650W/m²
Hence, the intensity of the sunlight in W/m² is 650W/m²
V^2 =U^2 +2AS
V^2 = 25 ^2 + 2x9.5x10
V^2 = 625 + 1900 = 2525
V = 50.25
Answer:
The answer is either 5 or I'm learning something different and I just can't read
Explanation:
I hope this helped...
