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uysha [10]
3 years ago
5

Please help quick

Physics
1 answer:
Tatiana [17]3 years ago
6 0

Answer:

C is the right answer for the question

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find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4​
Dima020 [189]

The rate of energy radiated by the man is 3.86 x 10^{-8}  J/s. m^{2}.

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = AeσT^{4}

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

\frac{Q(t)}{T^{4} } = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x 10^{-8} J/s.

So that;

\frac{Q(t)}{T^{4} } = 1.7 * 0.4 * 5.67 x 10^{-8}

     = 3.8556 x 10^{-8}

     = 3.86 x 10^{-8}  J/s. m^{2}

Thus, the rate of energy radiated by the man is 3.86 x 10^{-8}  J/s. m^{2}.

Learn more on energy radiation of objects by visiting: brainly.com/question/12550129

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