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maxonik [38]
2 years ago
15

Properties that could be used in developing a temperature scale?​

Physics
2 answers:
satela [25.4K]2 years ago
8 0

Answer:

Explanation:

Y=Kx

Andru [333]2 years ago
5 0

Answer:

boiling point and freezing point

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2 years ago
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

5 0
3 years ago
A very long conducting cylinder (length L) of radius R(R<R and (b)r
mina [271]

Answer:

Explanation:

We have to find electric potential V at a distance r.

a) For r>R,

The electric field in the cylinder is given by

E.A equating it to the other electric field given by

б.A/ε₀

Here the area of cylinder is given by= 2*3.14*r*L

While for the outside, the area= 2*3.14*R*L

Equating both, we get

E= бR/rε₀

Now,

The potential difference is given as:

ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.

Where ΔV= V₀-V

So solving we get

V₀=V-бR/ε₀ln (r/R)

b) For r<R i.e. inside the cylinder

There will be no electric field produced as E=0

So ultimately Vin= V

c) V=0 at r= infinity.

4 0
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