To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.
Where,
Q = Heat
U = Internal Energy
By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore
Work is done ON the system
Heat is extracted FROM the system
Therefore the value of the Work done on the system is -158.0J
Answer
given,
flow rate = p = 660 kg/m³
outer radius = 2.8 cm
P₁ - P₂ = 1.20 k Pa
inlet radius = 1.40 cm
using continuity equation
A₁ v₁ = A₂ v₂
π r₁² v₁ = π r₁² v₂
Applying Bernoulli's equation
v₂ = 1.97 m/s
b) fluid flow rate
Q = A₂ V₂
Q = π (0.014)² x 1.97
Q = 1.21 x 10⁻³ m³/s
<span>Reactants
Hope this helps.</span>
The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925
To answer the question, we need to know what polarization of light is.
<h3>What is polarization of light?</h3>
This is when the electric field vector of light is oscillating in one plane.
- Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
- Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.
<h3>Intensity of light through each polarized filter</h3>
Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.
So, since the light is initially unpolarized,
- The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.
- The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
- The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
- The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
- The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
- The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
- The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.
<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>
Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.
So, It/I₀ = 1/2cos¹²17°
= 1/2(0.9563)¹²
= 1/2 × 0.5850
= 0.2925
So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925
Learn more about intensity of polarized light here:
brainly.com/question/25402491
