Answer:
So, the evaporating pressure of the R410A = 118 psig
Explanation:
Solution:
For R410A system:
Data Given:
Evaporator Outlet Temperature = 50°F
Evaporator Superheat = 10°F
Required:
Evaporating Pressure in the system = ?
For this, first of all, we need to calculate inlet temperature on R410A system from the given value of outlet temperature.
Evaporator inlet temperature is the difference of outlet temperature and evaporator superheat.
Evaporator inlet temperature = Outlet Temperature - Evaporator Superheat
Evaporator inlet Temperature = 50°F - 10°F
Evaporator inlet Temperature = 40°F
Now, as we have the inlet temperature and the R410A system. We can consult the pressure temperature chart or PT chart, which I have attached and highlighted the value of evaporating pressure for 40°F inlet temperature.
So, the evaporating pressure of the R410A = 118 psig
Two hundred 432000 was broke down into aalll these parts
A theory can be represented as a model
Answer:
D). A sudden change in temperature increased the density of the atmosphere.
Explanation:
The claim that can be made and supported with evidence regarding the temporary decrease in light would be 'a sudden change in temperature will increase the density of the atmosphere.' <u>As the temperature increases, the molecules in the atmosphere begin to move fastly which causes a decrease in atmosphere density while it increases with the fall in temperature. As a result, the intensity of sunlight gets affected due to molecules being heavier and floating in the atmosphere</u>. Thus, <u>option D</u> is the correct answer.
Answer:
τ ≈ 0.90 N•m
F =
Explanation:
I = ½mR² = ½(10)0.5² = 1.25 kg•m²
α = ω²/2θ = 3.0² / 4π = 0.716... rad/s²
τ = Iα = 1.25(0.716) = 0.8952... ≈ 0.90 N•m
τ = FR
Now we have the unanswered question of reference frame.
80° from what?
If it's 80° from the radial
F = τ/Rsinθ = 0.90/0.5sin80 = 1.818... ≈ 1.8 N
If it's If it's 80° from the tangential
F = τ/Rcosθ = 0.90/0.5cos80 = 10.311... ≈ 10 N
There are an infinite number of other potential solutions
