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Oksanka [162]
3 years ago
8

What of the following does NOT influence resistance? *

Physics
1 answer:
Marina86 [1]3 years ago
4 0

Answer:

D. Number of components

Explanation:

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Explain why a black and white aeronautical map would be difficult to use
Aneli [31]

Aeronautical maps are usually meant to be used by pilots and air aviation professionals in other to navigate or traverse though the sky. With various elements such as vegetation, hills, valleys being depicted by color coded keys or legend. Hence, the absence of color on an aeronautical map make the <em>representation of elements very difficult</em>.

Visual map interpretation is usually aided by the use of legends. The legend hold the key to the elements which are represented on the map. Usually, a combination of colors and shapes makes up the legend and makes map interpretation easy.

Therefore, the absence of various color palletes for representation on a black and white aeronautical map will make it difficult to use.

Learn more : brainly.com/question/25323763

5 0
2 years ago
In which one or more of the following is the earth assumed to be the center of the universe?
sladkih [1.3K]
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Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission
bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

       P₁ = σ A T₁⁴

T = 12000K

       P₂ = σ A T₂⁴

       P₂ / P₁ = T₂⁴ / T₁⁴

       P₂ / P₁ = (12000/3000)⁴

       P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

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The potential energy Wood be 1000
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Yessir it sure does
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