Answer:
The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.
Explanation:
Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
__
<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
__
<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
__
<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
Examples of quality assurance activities include process checklists, process standards, process documentation and project audit. Examples of quality control activities include inspection, deliverable peer reviews and the software testing process. You may like to read more about the quality assurance vs quality control.
The person that is correct based on the 2 statements from Tech A and Tech B is; Tech B
A mass flow sensor is defined as a sensor that is used to measure the mass flow rate of air entering a fuel-injected internal combustion engine and then sends a voltage that represents the airflow to the electronic control circuit.
However, for Tech A is incorrect and so the correct answer is that Tech B is right because his statement corresponds with the definition of mass flow sensor.
Read more about fuel injection engines at; brainly.com/question/4561445
Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1

For pipe 2

Q=2.8 l/s
![Q=2.8\times 10^{-3]](https://tex.z-dn.net/?f=Q%3D2.8%5Ctimes%2010%5E%7B-3%5D)
We know that Q=AV




head loss (h)

Now putting the all values

So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.