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:
Explanation:
Answer: With crumple zones at the front and back of most cars, they absorb much of the energy (and force) in a crash by folding in on itself much like an accordion. ... As Newton's second law explains force = Mass x Acceleration this delay reduces the force that drivers and passengers feel in a crash.Sep 30, 2020
Answer:
a) Q = 251.758 kJ/mol
b) creep rate is 
Explanation:
we know Arrhenius expression is given as
where
Q is activation energy
C is pre- exponential constant
At 700 degree C creep rate is
% per hr
At 800 degree C creep rate is
% per hr
activation energy for creep is 
= 
![\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%5C%25%7D%7B5.5%20%5Ctimes%2010%5E%7B-2%7D%5C%25%7D%20%3D%20e%5E%7B%5B%5Cfrac%7B-Q%7D%7BR%28800%2B273%29%7D%5D%20-%5B%5Cfrac%7B-Q%7D%7BR%28800%2B273%29%7D%5D%7D)
![\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B0.01%7D%7B5.5%5Ctimes%2010%5E%7B-4%7D%7D%20%3D%20ln%20%5Be%5E%7B%5Cfrac%7BQ%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B1073%7D%20-%20%5Cfrac%7B1%7D%7B973%7D%5D%7D%5D)
solving for Q we get
Q = 251.758 kJ/mol
b) creep rate at 500 degree C
we know
Answer:
a)5.28 Å , b)3.73 Å , c)3.048 Å
Explanation:
the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.
Therefore, a particular unit cell consist only 1/8th part of an atom.
The lattice constant of a simple cubic primitive cell is 5.28 Å
We know formula of distance,
d = 
a)(100)
a=5.28 Å
Distance = 
=5.28 Å
b)(110)
Distance = 
= 3.73 Å
c)(111)
Distance= 
= 3.048 Å
