Answer:
V1 = 1.721 * V2
Explanation:
To start with, we assume that both lift forces are equal, such that
L2 = L1
1 is that of the level at 10000 m, and 2 is that of the level at sea level.
Next, we try and substitute the general formula for both forces such that
C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A
On further simplification, we have
ρ1.V1² = ρ2.V2², making V1 subject of formula, we have
V1 = √(ρ2/ρ1). V2²
Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have
V1 = √(1.225/0.4135) * V2
V1 = √2.9625 * V2
V1 = 1.721 * V2
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: For #1 I'm going to go with A because that has to do with biology
For #2 I'm going to go with B oceans because that has to do with plant life (and life in general).
For #3 I'll say marine/maritime engineer (you can just say marine)
Hope it helps!
Answer:
Explanation:
From the given information:
Water freezing temp. 
Heat rejected temp 
Recall that:
The coefficient of performance is:
Again:
The efficiency given by COP can be represented by:
Finally; the power input in an hour can be determined by using the formula:
In hp; since 1 kW = 1.34102 hp
6.86kW will be = (6.86 × 1.34102) hp
= 9.199 hp
