Answer:
0.16 micron per day
Explanation:
Given:
The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m
Initial tensile stress, σ₁ = 120 MPa
Final stress = 30 MPa
now from Griffith's equation, we have
![\sigma=[\frac{G_cE}{\pi\ a}]^\frac{1}{2}](https://tex.z-dn.net/?f=%5Csigma%3D%5B%5Cfrac%7BG_cE%7D%7B%5Cpi%5C%20a%7D%5D%5E%5Cfrac%7B1%7D%7B2%7D)
where,
Gc and E are the material constants
now,
for the initial stage
![120=[\frac{G_cE}{\pi\ (0.1\times10^{-6}}]^\frac{1}{2}](https://tex.z-dn.net/?f=120%3D%5B%5Cfrac%7BG_cE%7D%7B%5Cpi%5C%20%280.1%5Ctimes10%5E%7B-6%7D%7D%5D%5E%5Cfrac%7B1%7D%7B2%7D)
........{1}
and for the final case
![30=[\frac{G_cE}{\pi\ a_2}]^\frac{1}{2}](https://tex.z-dn.net/?f=30%3D%5B%5Cfrac%7BG_cE%7D%7B%5Cpi%5C%20a_2%7D%5D%5E%5Cfrac%7B1%7D%7B2%7D)
............{2}
on dividing 1 by 2, we get
![\frac{120}{30}=[\frac{a_2}{0.1\times10^{-6}}]^\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B120%7D%7B30%7D%3D%5B%5Cfrac%7Ba_2%7D%7B0.1%5Ctimes10%5E%7B-6%7D%7D%5D%5E%5Cfrac%7B1%7D%7B2%7D)
or
a₂ = 4² × 0.1 × 10⁻⁶ m
or
a₂ = 1.6 micron
Now,
the change from 0.1 micron to 1.6 micron took place in 10 days
therefore, the rate at which the crack is growing = 
or
average rate of change of crack = 0.16 micron per day
Answer:
Q = -68.859 kJ
Explanation:
given details
mass 
initial pressure P_1 = 104 kPa
Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K
final pressure P_2 = 1068 kPa
Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K
we know that
molecular mass of 
R = 8.314/44 = 0.189 kJ/kg K
c_v = 0.657 kJ/kgK
from ideal gas equation
PV =mRT
WORK DONE
w = 586*(0.1033 -0.514)
W =256.76 kJ
INTERNAL ENERGY IS
HEAT TRANSFER
= 187.902 +(-256.46)
Q = -68.859 kJ
Answer: Fiber Optic Network Fiber-optic networks have been used for decades to transmit large volumes of traffic across the country. The economics of fiber networks have only recently allowed for connecting the fiber directly to the home, creating a fiber-to-the-home (FTTH) network.
Explanation:
