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:
It helps ensure the security of client data.
Explanation:
Cloud computing is one of the security features in networking which is being adopted by big corporations and organizations inorder to ensure smooth running of the organization.
Also, due to the sensitivity of data which some organizations deals in, they tried everything possible to protect themselves and their clients' information through use of cloud computing.<em> Data are stored in the clouds, and requires special administrative rights for it to be accessed by some of the staffs working in any given organization.</em>
Answer:
All 3 principal stress
1. 56.301mpa
2. 28.07mpa
3. 0mpa
Maximum shear stress = 14.116mpa
Explanation:
di = 75 = 0.075
wall thickness = 0.1 = 0.0001
internal pressure pi = 150 kpa = 150 x 10³
torque t = 100 Nm
finding all values
∂1 = 150x10³x0.075/2x0,0001
= 0.5625 = 56.25mpa
∂2 = 150x10³x75/4x0.1
= 28.12mpa
T = 16x100/(πx75x10³)²
∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]
= 1/2[84.37±√791.2969+5.827396]
= 1/2[84.37±28.33]
∂1 = 1/2[84.37+28.33]
= 56.301mpa
∂2 = 1/2[84.37-28.33]
= 28.07mpa
This is a 2 d diagram donut is analyzed in 2 direction.
So ∂3 = 0mpa
∂max = 56.301-28.07/2
= 14.116mpa
Answer:
combining scientific knowledge, careful reasoning, and artistic invention in a flexible approach to problem-solving
Explanation:
