the answer is (c)
After the vehicle is involved in a car accident or fire
(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b) ρ = n X (AM) / v X Nₐ
<u>Explanation:</u>
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Given-
Lattice parameter of Li = 3.5089 X 10⁻⁸ cm
1 vacancy per 200 unit cells
Vacancy per cell = 1/200
(a)
Number of vacancies per cubic cm = ?
Vacancies/cm³ = vacancy per cell / (lattice parameter)³
Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³
Vacancies/cm³ = 1.157 X 10²⁰
Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b)
Density is represented by ρ
ρ = n X (AM) / v X Nₐ
where,
Nₐ = Avogadro number
AM = atomic mass
n = number of atoms
v = volume of unit cell
Answer:
See explaination
Explanation:
Let's define tuple as an immutable list of Python objects which means it can not be changed in any way once it has been created.
Take a look at the attached file for a further detailed and step by step solution of the given problem.
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:
Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.
Explanation:
