Answer:
the range of K can be said to be : -3.59 < K< 0.35
Explanation:
The transfer function of a typical tape-drive system is given by;
![KG(s) = \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]}](https://tex.z-dn.net/?f=KG%28s%29%20%3D%20%5Cdfrac%7BK%28s%2B4%29%7D%7Bs%5Bs%2B0.5%29%28s%2B1%29%28s%5E2%2B0.4s%2B4%29%5D%7D)
calculating the characteristics equation; we have:
1 + KG(s) = 0
![1+ \dfrac{K(s+4)}{s[s+0.5)(s+1)(s^2+0.4s+4)]} = 0](https://tex.z-dn.net/?f=1%2B%20%20%20%5Cdfrac%7BK%28s%2B4%29%7D%7Bs%5Bs%2B0.5%29%28s%2B1%29%28s%5E2%2B0.4s%2B4%29%5D%7D%20%3D%200)
![{s[s+0.5)(s+1)(s^2+0.4s+4)]} +{K(s+4)}= 0](https://tex.z-dn.net/?f=%7Bs%5Bs%2B0.5%29%28s%2B1%29%28s%5E2%2B0.4s%2B4%29%5D%7D%20%2B%7BK%28s%2B4%29%7D%3D%200)


We can compute a Simulation Table for the Routh–Hurwitz stability criterion Table as follows:
1 5.1 2+ K
1.9 6.2 4K
1.83
0
4K 0
S
0 0


We need to understand that in a given stable system; all the elements in the first column is usually greater than zero
So;
11.34 - 1.9(X) > 0


7.54 +2.1 K > 0
2.1 K > - 7.54
K > - 7.54/2.1
K > - 3.59
Also
4K >0
K > 0/4
K > 0
Similarly;
XY - 7.32 K > 0
![(\dfrac{3.8+1.9K-4K}{1.9})[11.34 - 1.9(\dfrac{3.8+1.9K-4K}{1.83}) > 7.32 \ K]](https://tex.z-dn.net/?f=%28%5Cdfrac%7B3.8%2B1.9K-4K%7D%7B1.9%7D%29%5B11.34%20%20-%201.9%28%5Cdfrac%7B3.8%2B1.9K-4K%7D%7B1.83%7D%29%20%3E%207.32%20%5C%20K%5D)
0.54(2.1K+7.54)>7.32 K
11.45 K < 4.07
K < 4.07/11.45
K < 0.35
Thus the range of K can be said to be : -3.59 < K< 0.35
Answer:
Explanation:
Considering the relation of the equilibrium vacancy concentration ;
nv/N = exp (-ΔHv/KT)
Where T is the temperature at which the vacancy sites are formed
K = Boltzmaan constant
ΔHv = enthalpy of vacancy formation
Rearranging the equation and expressing in term of the temperature and plugging the values given to get the temperature. The detailed steps is as shown in the attached file
Answer:
C. assembly line workers.
Explanation:
Oop
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