Answer:
See the attached picture for answer.
Explanation:
See the attached picture for explanation.
Answer:
The Current will decrease by a factor of 2
Explanation:
Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.
Now, as the question says, the load is reduced to half its original value, we can write:


Since, P2 = P1/2,

Dividing equations (1) and (2), we get,
P1 / (P1/2) = I1/ I2

Hence, it is proved that the current in the transmission line will decrease by a factor of 2 when load is reduced to half.
Answer:
2)
3) 
Explanation:
1) Expressing the Division as the summation of the quotient and the remainder
for
118, knowing it is originally a decimal form:
118:2=59 +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

2) 
Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.
|-49|=49
49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1
100011

3) 
The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

Answer:
Δ enthalpy = -23 Btu/Ibm
Explanation:
Given data:
Pressure ( P1 ) = 250 psi
Initial Temperature ( T1 ) = 175°F
Final temperature ( T2 ) = 20°F
<u>Calculate the change in the enthalpy of R-134a </u>
From R-134 table
h1 = 129.85 Btu/Ibm
s1 = 0.23281 Btu/Ibm.R
note : entropy is constant
hence ; s1 = s2
by interpolation ; h2 = 106.95
Δ enthalpy = h2 - h1
= ( 106.95 - 129.85 ) = -23 Btu/Ibm