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ZanzabumX [31]
2 years ago
8

When creating a story in Einstein Discovery, do all potential collinear fields need to be removed before executing the build sto

ry'5 A. No. Einstein Discovery is impervious to collinearity, so the story and subsequent model will be fine. B. No. Although it is ideal to eliminate collinearity as soon as possible, Einstein will give a warning post-build and the ridge regression will prevent collinearity from over-fitting. C. yes. If all collinear variables are not excluded, the model will over-fit and not make any sense. D. Yes. If the collinear variables are not removed, the Einstein Discovery model build will fail.
Physics
1 answer:
Blizzard [7]2 years ago
3 0

Answer:

The correct option is B i.e. <em>No. Although it is ideal to eliminate collinearity as soon as possible, Einstein will give a warning post-build and the ridge regression will prevent collinearity from over-fitting.</em>

Explanation:

With the help of creating a story in Einstein Discovery, the collinears are not needed to be removed because the ride regression, methodology used will automatically prevent overfitting.

Option A is incorrect as the story in Einstein Discovery is not impervious to the collinearity.

Option C  is incorrect as on one hand it does explain the effect of collinearity but as this is catered automatically in the ridge regression thus the need to do this is not satisfied.

Option D is incorrect because it does not depict the correct outcome of the collinear variables. Collinear variables do not fail the model. It just overfit it and make it difficult to be generalized.

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Answer:

T₂ = 5646 K

Explanation:

Let's start by finding the power received by the first disc, for this we use Stefan's law

          P = σ. A e T⁴

Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of ​​the disk, T the absolute temperature and e the emissivity that for a black body is  1

The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is

           I = P / A

Writing this expression for both discs

          I₁ A₁ = I₂ A₂

          I₂ = I₁ A₁ / A₂

The area of ​​a sphere is

          A = 4π r²

           I₂ = I₁ (r₁ / r₂)²

          r₂ = r₁ ± 5

          I₁ = I₂ ( (r₁ ± 5)/r₁)²

.

        Let's write the Stefan equation

         P / A = σ e T⁴

          I = σ e T⁴

This is the intensity that affects the disk, substitute in the intensity equation

         σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²

The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values ​​of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white

        e₁ T₁⁴ = T₂⁴   (r1 + 5)²/r₁²

       T₁ = T₂  {(e₂/e₁)}^{1/4}  √(1 ± 1/ r₁)  

If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term

           (1 ±x) n = 1 ± n x

Where x = 5 / r₁ << 1

We replace

          T₁ = T₂ {(e₂/e₁)}^{1/4}  (1 ± ½   5/r1)

           T₁ = T₂ {(e₂)}^{1/4}   (1 ± 5/2 1/r1)

If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach

              T₁ = T₂  {(e₂)}^{1/4} ¼

              T<u>₁</u> = T₂  0.90.9^{1/4}

               5500 = T₂  0.974

               T₂ = 5646 K

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