Motor oil is responsible for

The following C program asks the user for two input null-terminated strings, each stored in uninitialized 100-byte buffer, and c

marissa [1.9K]

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

3 0

3 years ago

g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

0 - 1" -20 -15 -10 5 0 1 2 3 0

faust18 [17]

Answer:

#WeirdestQuestionOfAllTime

Explanation:

8 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".