Answer:
Increases
Explanation:
By inhibiting the motion of dislocations by impurities in a solid solutions, is a strengthening mechanism. In solid solutions it is atomic level strengthening resulting from resistance to dislocation motion. Hence, the strength of the alloys can differ with respect to the precipitate's property. Example, the precipitate is stronger (ability to an obstacle to the dislocation motion) than the matrix and it shows an improvement of strength.
Answer:
The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.
Explanation:
From Fluid Mechanics, we remember that absolute pressure (
), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:
(1)
Where:
- Atmospheric pressure, measured in pounds per square inch.
- Working pressured of the boiler (gauge pressure), measured in pounds per square inch.
If we suppose that 
and 
, then the absolute pressure is:
The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.
All of the dimensions on an aircraft drawing are _________ to the bottom of the drawing
Answer: parallel
Answer:
S = 0.5 km
velocity of motorist = 42.857 km/h
Explanation:
given data
speed = 70 km/h
accelerates uniformly = 90 km/h
time = 8 s
overtakes motorist = 42 s
solution
we know initial velocity u1 of police = 0
final velocity u2 = 90 km/h = 25 mps
we apply here equation of motion
u2 = u1 + at
so acceleration a will be
a = 
a = 3.125 m/s²
so
distance will be
S1 = 0.5 × a × t²
S1 = 100 m = 0.1 km
and
S2 = u2 × t
S2 = 25 × 16
S2 = 400 m = 0.4 km
so total distance travel by police
S = S1 + S2
S = 0.1 + 0.4
S = 0.5 km
and
when motorist travel with uniform velocity
than total time = 42 s
so velocity of motorist will be
velocity of motorist = 
velocity of motorist = 
velocity of motorist = 42.857 km/h
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
