Answer:
c. V2 equals V1
Explanation:
We can answer this question by using the continuity equation, which states that:
(1)
where
A1 is the cross-sectional area in the first section of the pipe
A2 is the cross-sectional area in the second section of the pipe
v1 is the velocity of the fluid in the first section of the pipe
v2 is the velocity of the fluid in the second section of the pipe
In this problem, we are told that the pipe has a uniform cross sectional area, so:
A1 = A2
As a consequence, according to eq.(1), this means that
v1 = v2
so, the velocity of the fluid in the pipe does not change.
Uh, I’d assume so because Brainly has a whole section of questions for them.
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
Hopefully that helps you out and is this for history or science?
Technical Drawings give a better understanding of what is needed and required in the project.
Explanation:
