Answer:
true
Explanation:
Creep is known as the time dependent deformation of structure due to constant load acting on the body.
Creep is generally seen at high temperature.
Due to creep the length of the structure increases which is not fit for serviceability purpose.
When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.
When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.
Answer:
The Poisson's Ratio of the bar is 0.247
Explanation:
The Poisson's ratio is got by using the formula
Lateral strain / longitudinal strain
Lateral strain = elongation / original width (since we are given the change in width as a result of compession)
Lateral strain = 0.15mm / 40 mm =0.00375
Please note that strain is a dimensionless quantity, hence it has no unit.
The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.
Longitudinal strain = 4.1 mm / 270 mm = 0.015185
Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247
The Poisson's Ratio of the bar is 0.247
Please note also that this quantity also does not have a dimension
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:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu
