Answer:
≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
= 
× CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, 
We can as well say:
We can now replace 
in the above equation.
Therefore, = × CρAv²
The A which stands as the area of the jet is given by the formula:
We can now have a new equation after substituting our A into the previous equation as:
= × Cρ 
Substituting our data from above; we have:
= × 
= 
= 110,990N
in N (newton) to KN (kilo-newton) will be:
= 
= 110.990 KN
≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.
Answer:
identify function of the system unit and its components
Explanation:
Styrene is a vinyl monomer in which there is a carbon carbon double bond.
The polymerization of the styrene, which is initiated by using a free radical which reacts with the styrene and the compound thus forms react again and again to form polystyrene (PS).
The equation is shown below as:
⇒ ![\begin{matrix}&C_6H_5 \\&|\\ -[-H_2C & -CH-]-_n\end{matrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bmatrix%7D%26C_6H_5%20%5C%5C%26%7C%5C%5C%20-%5B-H_2C%20%26%20-CH-%5D-_n%5Cend%7Bmatrix%7D)
Correcto no se muy bien de que se trata el tema porque está en inglés.
Sorry
Answer:
B. The thickness of the heated region near the plate is increasing.
Explanation:
First we know that, a boundary layer is the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant. The fluid is often slower due to the effects of viscosity. Advection i.e the transfer of heat by the flow of liquid becomes less since the flow is slower, thereby the local heat transfer coefficient decreases.
From law of conduction, we observe that heat transfer rate will decrease based on a smaller rate of temperature, the thickness therefore increases while the local heat transfer coefficient decreases with distance.
