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MArishka [77]
3 years ago
10

A python can detect thermal radiation from objects that differ in temperature from their environment as long as the received int

ensity of thermal radiation is greater than 0.60 W/m2 . Your body emits a good deal of thermal radiation. Assume that your body has a surface area of 1.6 m2 , a surface temperature of 30°C, and an emissivity e = 0.97 at infrared wavelengths. As we’ve seen, the intensity of a source of radiation decreases with the distance from the source. If you are outside on a cool, dark night, what is the maximum distance from which a python could detect your presence?
Physics
1 answer:
elena55 [62]3 years ago
5 0

Answer:

 R = 9.92 m

Explanation:

For this exercise let's use the Stefa radiation equation

      P = σ A e T⁴

Where the Stefan-Boltzmann constant is worth 5,670 10⁻⁸ W/m²K⁴, A is the area of ​​the body and the emissivity and T is the absolute temperature

Let's calculate irradiated

     P = 5,670 10⁻⁸ 1.6 0.97 (273 +30) 4

     P = 7.417 10² W

The emitted power is distributed on a spherical surface as it progresses, whereby the intensity detected by the python I = 0.6 W / m2

        P = I A

The area of ​​the sphere is

        A = 4π R²

       

        P = I 4π R²

        R² = P / 4π I

        R = √ [7.417 102 / (4π 0.6)]

        R = √ 98.37

        R = 9.92 m

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An event occurs in system K' at x' = 2 m, y' = 3.7 m, z' = 3.7 m, and t' = 0. System K' and K have their axes coincident at t =
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Answer:

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

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x=\frac{x^{|}+vt^{|} }{\sqrt{1-\frac{v^{2} }{c{2} } } } \\x=\frac{2m+0.92c(0) }{\sqrt{1-\frac{(0.92c)^{2} }{c{2} } } }\\x=5.103m\\y=y^{|}\\ y=3.7m\\z=z^{|}\\ z=3.7m

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r=\frac{1}{\sqrt{1-v^{2} } } \\r=\frac{1}{\sqrt{1-(0.92)^{2} } } \\r=2.551\\t=r(t^{|}+vx^{|}/c^{2}   )\\t=2.551(0s+(0.92c)(2)/c^{2} )\\t=1.57*10^{-8}s

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