Answer:
The architectural pattern i will use for the school management is the client-server pattern.
This pattern would consist of a server and many clients. wherein the server component would provide services to that of the clients and its components as specified and also there would be a client request service from that of the server.
Explanation:
Solution
A school management system would always involve the client server pattern as this pattern would have a server and many clients wherein the server component would give services to that of the clients and its components as specified and also there would be a client request service from that of the server. This server would share the appropriate services to such clients and also listen to the client's requests.
Such kind of pattern would mostly be used for for the online platforms or application like that of document.
Answer:
manda a senha do brainly bloquearam os amigos e você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo uma academia de lima e você vai ver o meu perfil completo a minha amiga Jeciane estamos completo a minha amiga Jeciane estamos fazendo uma academia de lima e você vai ver o meu perfil você vai ver o meu perfil você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo
Explanation:
Marcar como melhor porfavo
Okay I believe you I swear
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ )
/ 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
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.