At room temperature what is the strength of the electric field in a 12-gauge copper wire (diameter 2.05 mm that is needed to cau
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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 0.9
5500 = T₂ 0.974
T₂ = 5646 K
Answer:
The radius is
Explanation:
From the question we are told that
The magnetic field is
The electron kinetic energy is
Generally for the collision to occur the centripetal force of the electron in it orbit is equal to the magnetic force applied
This is mathematically represented as
=>
Where m is the mass of electron with values
v is the escape velocity which is mathematically represented as
So
apply indices
substituting values