kerryjenkins300
05.05.2020 •
Physics
A certain light bulb consumes 200J of electrical energy per second, but only emits 25J of light energy per second. Calculate the efficiency of this bulb.
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Ответ:
B. 35.9°
Step-by-step explanation:
sin P = opposite/hypotenuse = 33.8/57.6
∡ P = arcsin(33.8/57.6)=35.9°