Answer:
<h2>it's damage kindly be careful</h2>
A diagram showing a use case diagrams for these requirements is given in the image attached.
<h3>What is system
Case diagram?</h3>
A use case diagram is known to be a kind of graphical illustration of a users in terms of their various possible association or interactions within any given system.
A use case diagram in banking can be used to prepare, depict and also to know all the functional requirements of the banking system.
Therefore, Give the use case specification for the banking system services and paying a bill online is given in the image attached.
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Answer:
A. smallest wire is No. 12
Answer:
Explanation:
<u><em>General Considerations</em></u>
The design of the yard will affect the natural surface and subsurface drainage pattern of a watershed or individual hillslope. Yard drainage design has as its basic objective the reduction or elimination of energy generated by flowing water. The destructive power of flowing water increases exponentially as its velocity increases. Therefore, water must not be allowed to develop sufficient volume or velocity so as to cause excessive wear along ditches, below culverts, or along exposed running surfaces, cuts, or fills.
A yard drainage system must satisfy two main criteria if it is to be effective throughout its design life:
1. It must allow for a minimum of disturbance of the natural drainage pattern.
2.It must drain surface and subsurface water away from the roadway and dissipate it in a way that prevents excessive collection of water in unstable areas and subsequent downstream erosion
The diagram below ilustrate diffrent sturcture of yard to be consider before planing to utiliza rainwater
Answer:
Q=67.95 W
T=119.83°C
Explanation:
Given that
For air
Cp = 1.005 kJ/kg·°C
T= 20°C
V=0.6 m³/s
P= 95 KPa
We know that for air
P V = m' R T
95 x 0.6 = m x 0.287 x 293
m=0.677 kg/s
For gas
Cp = 1.10 kJ/kg·°C
m'=0.95 kg/s
Ti=160°C ,To= 95°C
Heat loose by gas = Heat gain by air
[m Cp ΔT] for air =[m Cp ΔT] for gas
by putting the values
0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )
T=119.83°C
T is the exit temperature of the air.
Heat transfer
Q=[m Cp ΔT] for gas
Q=0.95 x 1.1 x ( 160 -95 )
Q=67.95 W