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:
Read the passage. Then, answer the questions about the metaphor in boldface in the text.
Lately, I've been so overwhelmed with school and sports. There was a time when I enjoyed going to classes and going to practice every afternoon. Now, everything is piling up and wearing me down. Thankfully, I get to see you every day. You are truly the sunshine of my life. Thank you for making me laugh when I'm feeling down.
What is the context of the passage?
What is being compared in the metaphor?
What is the meaning of the metaphor?
Explanation:
Answer:
Either D or C
Both of these masks are used for dust, but since half masks are generally cheaper and easier to use, I'd go with C.
If this is correct, I'd appreciate a brainliest.
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)