Answer:
False I'm pretty sure sorry If its wrong
The technician that is correct about either testing lights for simple tests or to check SRS Circuits is; Technician A.
<h3>Which Technician is Correct?</h3>
First of all it is pertinent to note that test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits.
Now, the two values of voltage and current are high and sufficient to light up the bulb. However, in digital circuits, the current is very small in the order of milliamps, and as a result there is not enough power to turn on the lights.
Thus, we can conclude that Technician A is correct.
Read more about Correct Technician at; brainly.com/question/14449935
Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.
<u>Given the following data:</u>
- Jam density = 122 veh/km.
<h3>How to calculate the
maximum flow.</h3>
According to Greenshield Model, maximum flow is given by this formula:
<u>Where:</u>
is the free flow speed.
is the Jam density.
In order to calculate the free flow speed, we would use this formula:
Substituting the parameters into the model, we have:
Max flow = 3,599 veh/hr.
Read more on traffic flow here: brainly.com/question/15236911
Answer:
B) 5.05
Explanation:
The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:
Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2
Given that:
Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05
Maximum outer diameter = 35 + 0.05 = 35.05
Minimum inner diameter = 25 - 0.05 = 24.95
Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05
or
Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05
Therefore the LMC wall thickness is 5.05
Answer:
Check the explanation
Explanation:
Code
.ORIG x4000
;load index
LD R1, IND
;increment R1
ADD R1, R1, #1
;store it in ind
ST R1, IND
;Loop to fill the remaining array
TEST LD R1, IND
;load 10
LD R2, NUM
;find tw0\'s complement
NOT R2, R2
ADD R2, R2, #1
;(IND-NUM)
ADD R1, R1, R2
;check (IND-NUM)>=0
BRzp GETELEM
;Get array base
LEA R0, ARRAY
;load index
LD R1, IND
;increment index
ADD R0, R0, R1
;store value in array
STR R1, R0,#0
;increment part
INCR
;Increment index
ADD R1, R1, #1
;store it in index
ST R1, IND
;go to test
BR TEST
;get the 6 in R2
;load base address
GETELEM LEA R0, ARRAY
;Set R1=0
AND R1, R1,#0
;Add R1 with 6
ADD R1, R1, #6
;Get the address
ADD R0, R0, R1
;Load the 6th element into R2
LDR R2, R0,#0
;Display array contents
PRINT
;set R1 = 0
AND R1, R1, #0
;Loop
;Get index
TOP ST R1, IND
;Load num
LD R3,NUM
;Find 2\'s complement
NOT R3, R3
ADD R3, R3,#1
;Find (IND-NUM)
ADD R1, R1,R3
;repeat until (IND-NUM)>=0
BRzp DONE
;load array address
LEA R0, ARRAY
;load index
LD R1, IND
;find address
ADD R3, R0, R1
;load value
LDR R1, R3,#0
;load 0x0030
LD R3, HEX
;convert value to hexadecimal
ADD R0, R1, R3
;display number
OUT
;GEt index
LD R1, IND
;increment index
ADD R1, R1, #1
;go to top
BR TOP
;stop
DONE HALT
;declaring variables
;set limit
NUM .FILL 10
;create array
ARRAY .BLKW 10 #0
;variable for index
IND .FILL 0
;hexadecimal value
HEX .FILL x0030
;stop
.END
