Answer:
E = NA/R^n - N alpha e^2/ (4 piEoR')
Explanation:
Where N = number of positive ion pairs
e= magnitude
R= interatomic distance
n = the Born exponent
Answer:
Time take to deposit Ni is 259.02 sec.
Explanation:
Given:
Current 
A
Faraday constant 
Molar mass of Ni 
Mass of Ni 
g
First find the no. moles in Ni solution,
Moles of Ni 
mol
From the below reaction,
⇆ 
Above reaction shows "1 mol of 
requires 2 mol of electron to form 1 mol of "
So for finding charge flow in this reaction we write,
Charge flow 
C
For finding time of reaction,
Where 
charge flow
sec
Therefore, time take to deposit Ni is 259.02 sec.
Enormous O unpredictability is in reference to the most exceedingly terrible conceivable development rate of the calculation. So O(N log N) implies that it will never keep running in some time more terrible than O(N log N). So in spite of the fact that Al's calculation scales superior to Bob's quadratic algo, it doesn't really mean it is better for ALL info sizes.
Maybe there is critical overhead in building up it, for example, making a lot of clusters or factors. Remember that even an O(N log N) calculation could have 1000 non settled circles that official at O(N) and still be viewed as O(N log N) the length of it is the most exceedingly awful part.
1. No two magnetic lines intersect
2.They always form closed loops
3. Outside they seems to travel from north to south and inside south to north
4. It is vector quantity.
