R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
Read more about resistance here:
brainly.com/question/17563681
#SPJ1
Answer:
Kawasaki Ninja H2R – top speed: 222 mph. This one is another beast in the form of a bike. ...
MTT Turbine Superbike Y2K – top speed: 227 mph. This bike is one of the most powerful production motorcycles. ...
Suzuki Hayabusa – top speed: 248 mph. 1340cc
Answer:
Explained
Explanation:
This situation can occur because of various factors such as:
- Gradual deterioration of lubrication and coolant.
- change of environmental condition such as temperature, humidity, moisture, etc.
- Change in the properties of incoming raw material
- An increase or decrease in the temperature of the heat treating operation
- Debris interfering with the manufacturing process.
I think it’s rationalization.
Hope this helps
Answer:
The work of the cycle.
Explanation:
The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.
The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.
