Physics, asked by kartheekrayadurgam3, 10 months ago

The operating temperature of the filament of a bulb is 3000 K. If its surface area is 0.25 cm2 and emissivity is 0.35, the wattage of the bulb is (o
5.67x10 -8 Wm m2 K-4) (in W)

Answers

Answered by fa006453

0

Answer:

The 3000.K. and 0.25 cm2 this all answers 300002.0.25 this is a answer

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