Answer:
ºC
Explanation:
First, let's write the energy balance over the duct:

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

So, let's isolate
:

The Cp of the air at 27ºC is 1007
(Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are
and Q.
Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.
The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:
Perimeter:

Surface area:

Then, the heat Q is:

Finally, find the exit temperature:

=27.0000077 ºC
The temperature change so little because:
- The mass flow is so big compared to the heat flux.
- The transfer area is so little, a bigger length would be required.
Answer:
the order of importance must be b e a f c
Explanation:
Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.
Based on the aforementioned, let's analyze the statements in order of importance
b) True. Since the moon is material evaporated from Earth, its compassion is similar
e) True. If the moon is material volatilized from the earth it must train a finite receding speed
a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies
f) False. The moon's rotation and translation are equal has no relation to its formation phase
c) false. The amount of vaporized material on the moon is large
Therefore, the order of importance must be
b e a f c
Answer:
Explanation:
For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.
X axis
-Aₓ - f_e +T = 0 (1)
Y axis
N₁ - W_y = 0 ( 2)
let's use trigonometry for the weight components
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
We write the diagram for the second body.
Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards
W₂ -T = 0 (3)
we add the equations is 1 and 3
- W₁ sin θ - μ N₁ + W₂ = 0
from equation 2
N₁ = W₁ cos θ
we substitute
-W₁ sin θ - μ (W₁ cos θ) + W₂ = 0
W₂ = m₁ g (without ea - very expensive)
This is the smallest value that supports the equilibrium system
Answer:
uKkskdjod 7q and the rays are the best in all the ways ❤ ♥