Sheeeeeesh bro same name ayoooo??
Answer:
(a) attached below
(b)

(c) 
(d)
Ω
(e)
and 
Explanation:
Given data:





(a) Draw the power triangle for each load and for the combined load.
°
°
≅ 

≅ 
The negative sign means that the load 2 is providing reactive power rather than consuming
Then the combined load will be


(b) Determine the power factor of the combined load and state whether lagging or leading.

or in the polar form
°

The relationship between Apparent power S and Current I is

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.
(c) Determine the magnitude of the line current from the source.
Current of the combined load can be found by


(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω


Ω
(e) Compute the magnitude of the current in each capacitor and the line current from the source.
Current flowing in the capacitor is

Line current flowing from the source is

Answer:3.47 m
Explanation:
Given
Temperature(T)=300 K
velocity(v)=1.5 m/s
At 300 K


And reynold's number is given by



x=3.47 m
The heat transferred to and the work produced by the steam during this process is 13781.618 kJ/kg
<h3>
How to calcultae the heat?</h3>
The Net Change in Enthalpy will be:
= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg
Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)
= 1/2 x ( 75 + 225) x (5 - 2)
W = 450 KJ
From the First Law of Thermodynamics, Q = U + W
So, Heat Transfer = Change in Internal Energy + Work Done
= 13331.618 + 450
Q = 13781.618 kJ/kg
Learn more about heat on:
brainly.com/question/13439286
#SP1
Answer:
public static int average(int j, int k) {
return (int)(( (long)(i) + (long)(j) ) /2 );
}
Explanation:
The above code returns the average of two integer variables
Line 1 of the code declares a method along with 2 variables
Method declared: average of integer data type
Variables: j and k of type integer, respectively
Line 2 calculates the average of the two variables and returns the value of the average.
The first of two integers to average is j
The second of two integers to average is k
The last parameter ensures average using (j+k)/2