If one star has a temperature of 5200 k and another star has a temperature of 7900 k, how much more energy per second will the hotter star radiate from each square meter of its surface?
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Ответ:
5.327
Explanation:
Stefan-Boltzmann law states that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature.
W = σT⁴
Where,
W is the total radiant heat power emitted from a surface
σ is constant of proportionality, called the Stefan–Boltzmann constant = 5.67 × 10⁻⁸ Wm⁻²K⁻⁴
T is absolute temperature in kelvin
For the first star, T = 5200 K
∴ W₁ = σ(5200)⁴
For the second star, T = 7900 K
∴ W₂ = σ(7900)⁴
The amount of energy radiated by the hotter star W₂, with respect to the other star W₁ is,
W₂ / W₁ = σ(7900)⁴ / σ(5200)⁴
Ответ:
the potential energy is higher as the object when it falls will have more room to accelerate using kinetic energy