Answer:
Code is given below:
Explanation:
.data
str1: .space 20
str2: .space 20
msg1:.asciiz "Please enter string (max 20 characters): "
msg2: .asciiz "\n Please enter string (max 20 chars): "
msg3:.asciiz "\nSAME"
msg4:.asciiz "\nNOT SAME"
.text
.globl main
main:
li $v0,4 #loads msg1
la $a0,msg1
syscall
li $v0,8
la $a0,str1
addi $a1,$zero,20
syscall #got string to manipulate
li $v0,4 #loads msg2
la $a0,msg2
syscall
li $v0,8
la $a0,str2
addi $a1,$zero,20
syscall #got string
la $a0,str1 #pass address of str1
la $a1,str2 #pass address of str2
jal methodComp #call methodComp
beq $v0,$zero,ok #check result
li $v0,4
la $a0,msg4
syscall
j exit
ok:
li $v0,4
la $a0,msg3
syscall
exit:
li $v0,10
syscall
methodComp:
add $t0,$zero,$zero
add $t1,$zero,$a0
add $t2,$zero,$a1
loop:
lb $t3($t1) #load a byte from each string
lb $t4($t2)
beqz $t3,checkt2 #str1 end
beqz $t4,missmatch
slt $t5,$t3,$t4 #compare two bytes
bnez $t5,missmatch
addi $t1,$t1,1 #t1 points to the next byte of str1
addi $t2,$t2,1
j loop
missmatch:
addi $v0,$zero,1
j endfunction
checkt2:
bnez $t4,missmatch
add $v0,$zero,$zero
endfunction:
jr $ra
Answer:
A) 282.34 - j 12.08 Ω
B) 0.0266 + j 0.621 / unit
C)
A = 0.812 < 1.09° per unit
B = 164.6 < 85.42°Ω
C = 2.061 * 10^-3 < 90.32° s
D = 0.812 < 1.09° per unit
Explanation:
Given data :
Z ( impedance ) = 0.03 i + j 0.35 Ω/km
positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km
A) calculate Zc
Zc = 
= 
= 
= 282.6 < -2.45°
hence Zc = 282.34 - j 12.08 Ω
B) Calculate gl
gl = 
d = 500
z = 0.03 i + j 0.35
y = j4.4*10^-6 S/km
gl = 
= 
= 0.622 < 87.55 °
gl = 0.0266 + j 0.621 / unit
C) exact ABCD parameters for this line
A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )
C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )
D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
where : cos h (gl) = 
sin h (gl) = 
Answer:
401.3 kg/s
Explanation:
The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).
qw = 0.85 * q2
q2 = 0.64 * q1
p = 0.36 * q1
q1 = p /0.36
q2 = 0.64/0.36 * p
qw = 0.85 *0.64/0.36 * p
qw = 0.85 *0.64/0.36 * 600 = 907 MW
In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)
The latent heat for the vaporization of water is:
SLH = 2.26 MJ/kg
So, to dissipate 907 MW
G = qw * SLH = 907 / 2.26 = 401.3 kg/s
Mass affects the weight of an airplane because everything has mass
