Optimal voltage distribution. The physical meaning of best-matched load is that at this resistance value, the internal resistance of the solar cell and the load resistance form the best-matched voltage distribution, so that the electrical energy generated inside the solar cell can be transferred to the load with maximum. efficiency. In this state, an optimal match is achieved between the voltage and current of the solar cell, thereby maximizing the power output.
Use of solar energy from middle school physics questions
Solution: Q absorption=CM(t-t0)=225kg*4200*(40-20)= 18900000J
Solar energy The solar energy absorbed by the water heater in 8 hours W=1400*3*8*3600=120960000J
The efficiency of the solar water heater= 18900000/120960000=15.6%< /p>
Physical problem of solar temperature
1: Motor power=U*I=50*300=15 kilowatts.
2: P=F*v, whens of a movement at constant speed, the traction force is equal to the resistance, resistance = 15000/10=1500 N
3: The surface solar radiation power/square meter is 4* 10 ^26/(4*3.14*(1.5*10^11)^2)*70%
Solar radiation power surface/square meter*15%*panel surface solar = 15,000, can be calculated Minimum area.
Formula for calculating solar energy High school physics
If it's just water, under ideal conditions it has nothing to do with the volume.
Suppose your water heater contains 100 kg of water. When full it is 100*70*4.2*10^3, pour half out, leaving 50*70*4.2*10^3. heat. Adding 50 kg of water, the current temperature is 50*70*4.2*10^3=100*X*4.2*10^3, so X is 35 degrees Celsius.
Similarly, assuming it was 70kg of water, the calories when full were 70*70*4.2*10^3, pour half that out and the remaining calories would beof 35*70*. 4.2*10^3. Add 35 kg of water again, then the temperature is 35*70*4.2*10^3=70*X*4.2*10^3. Calculated this way, X always equals 35 degrees Celsius.
To calculate it this way, the prerequisite is that the water you add each time is at 0 degrees Celsius.
If the injected water is not at the same temperature each time, it has something to do with the capacity
The power p to raise the water in the cylinder is p=Q/ t=cmΔt/ t=(4.2*10 to the third power×0.6×1)/(2×60)=21w
The radiant power p′ reaching the surface of 3×10-2m2 of the upper layer of the earth's atmosphere is p`=p Total×p/η=p×4лR cube/s=4.4×10de21 power w