Answer:
β = 
= 0.7071 ≈ 1 ( damping condition )
closed-form expression for the response is attached below
Explanation:
Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s
computing a solution :
M = 1,
c = 2,
k = 2,
Wn = 
= 
next we determine the damping condition using the damping formula
β = = 0.7071 ≈ 1
from the condition above it can be said that the damping condition indicates underdamping
attached below is the closed form expression for the response
Answer:
D
Explanation: She hopes to be able to make this, however she hasn't yet...therefore she is thinking of a concept and it's development
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:
3270 N/m^2
Explanation:
we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure
Py = - Ph ( g ± a )
for a downward movement
Py = - Ph ( g - a ) ------ ( 1 )
From the above data given will be
p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81
input values into equation 1 becomes
Py = -Ph ( g - 2g ) = Phg ------ ( 3 )
Py = 1000 * 0.33 * 9.81
= 3270 N/m^2
