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
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A good source of insight to the technician about an engine's condition is the noises coming from the engine
The correct option that gives the exception to the noises heard during a cranking sound diagnosis is the option;
- b. <u>The engine revving and starting up</u>
Reason:
During a cranking sound diagnosis, the engine need to be disabled to
prevent it from starting, which is done by removing the relay for the fuel
pump, which will prevent fuel from potentially hydro-locking the engine
The key is then used to crank the engine, during which the cranking sound
is listened to
The sources of the noises produced includes;
- The starter is misaligned
- The clunk of a spun crankshaft bearing noise
- The sound given by a low compression cylinder, which is an uneven sound
- The no compression condition fast cranking sound, due to broken time belt that results in bent valves
Therefore;
The exception to the sounds heard during a cranking sound diagnosis is; <u>The engine revving and starting up</u>
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Answer:
v=82 m/s
s=116m
Explanation:
a=20t
using condition given at t=0
c=-8
now equation becomes
v=10t²-8
v at t= 3s v=82 m/s
again
now using condition given s=50 at t=0
b=50
now equation becomes
calculating s at t=3s
s=116m
Answer:
Condition 2 is true.
we can not calculate the load with given information.
According to above values, Condition 2 is satisfied so we can not find the load with given information because material deformation lies in plastic region.
Explanation:
In order to check whether we can find the load or not we have to check the following conditions:
Condition 1:
If this condition is true then we can calculate the load.
Condition 2:
If this condition is true then we can not calculate the load with given information.
Calculating :
Calculating :
Hence:
Condition 2 is true.
According to above values, Condition 2 is satisfied so we can not find the load with given information because material deformation lies in plastic region.
Answer:
(a) 20 MHz
(b) 1.025 KW
(c) 3.33 ns
(d) 33 pF
Explanation:
(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>
(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>
(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>
(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>