Answer:
ΔU =195 KJ
W=0 KJ
Explanation:
Given that
Mass of air m=3 kg
heat absorb by air =195 KJ
Cv=0.741 KJ/kgK
If we assume that air is as ideal gas so
Given that container is rigid it means that volume of system is constant so W=0
From first law of thermodynamics
Q=ΔU + W
⇒Q= ΔU (W=0)
So ΔU =195 KJ
And work done will be zero because because it is a constant volume process.
The answer is false. Square means a corner is a 90 degrees. Plumb means it vertically perpendicular to the surface.
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
Answer: 1.38g
Explanation:
Width of planting area = 100m
Field size = 6-hectares(14.425 acres)
Weight of 1000 black eye seed = 230g
1 lb = 453.4g
1 black eye seed = 230g/1000 = 0.23g = 0.00023kg
1 hectare = 10,000sq metre
6 hectare = 60,000sq metre
(Weight/Area) kg/m2
0.00023kg / 10,000 = 2.3×10^-8kg/m^2
And field size = 6hectares = 60,000m^2
(2.3×10^-8) × 60,000 = 0.00138kg of black eye seed
0.00138kg × 1000 = 1.38g
Answer:
Test the working prototype
Explanation:
After the prototype is built, it is put through basic operational tests and purpose of applicability or performance test
The prototype is also laboratory tested to determine how reliable it is with tests performed to determine the modes of failure. Performance of tests are carried out by subjecting the prototype to stresses beyond its designed stress such as to discover failure modes
The system is however tested in isolation at this stage and the performance and risks of the prototype are quantified after the test, with improvement requirement outlined for modification of the design.