Laser beam machining and water jet cutting are the following processes would be appropriate for cutting a narrow slot, less than 0.015 inch wide, in a 3/8- inch thick sheet of fiber-reinforced plastic. Hence option d and f is correct.
<h3>
What is fiber reinforced plastic?</h3>
Fiber reinforced plastic is defined as a polymer-based composite material that is strengthened by fiber support. Fiber-reinforced plastics (FRP) are composite materials that use glass or carbon fibers as reinforcement and polymer resins as a matrix.
Because laser cutting is a far less aggressive and abrasive process than water jet cutting, it is much more accurate. A laser can carefully and safely cut materials as thin as 0.006 inches, whereas water jet cutters can't handle cutting through surfaces smaller than 0.02 inches.
Thus, laser beam machining and water jet cutting are the following processes would be appropriate for cutting a narrow slot, less than 0.015 inch wide, in a 3/8- inch thick sheet of fiber-reinforced plastic. Hence option d and f is correct.
To learn more about fiber-reinforced plastic, refer to the link below:
brainly.com/question/11941367
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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
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Answer:
4 times around
Explanation:
The total number of teeth involved will be the same for each gear. If the front gear is connected to the pedal and it goes around twice, then 2·24 = 48 teeth will have passed the reference point.
If the rear gear is attached to the wheel, and 48 teeth pass the reference point, then it will have made ...
(48 teeth)/(12 teeth/turn) = 4 turns
Answer:
(a) BP = 11.99 KPa
(b) h = 2 m
Explanation:
(a)
Since, the fluid pressure and blood pressure balance each other. Therefore:
BP = ρgh
where,
BP = Blood Pressure
ρ = density of fluid = 1020 kg/m³
g = acceleration due to gravity = 9.8 m/s²
h = height of fluid = 1.2 m
Therefore,
BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)
<u>BP = 11995.2 Pa = 11.99 KPa</u>
(b)
Again using the equation:
P = ρgh
with data:
P = Gauge Pressure = 20 KPa = 20000 Pa
ρ = density of fluid = 1020 kg/m³
g = acceleration due to gravity = 9.8 m/s²
h = height of fluid = ?
Therefore,
20000 Pa = (1020 kg/m³)(9.8 m/s²)h
<u>h = 2 m</u>