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:
what do u mean.
Answer:
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Explanation:
yww <33
The expression of V(m³)=e^(t(s)) to make V in in³ and t in minutes is;
V(in³) = (¹/₆₁₀₂₄)a
We are given that;
Volume of microbial culture is observed to increase according to the formula;
V = e^(t)
where;
t is in seconds
V is in m³
We want to now express V in in³ and t in minutes.
Now, from conversions;
1 m³ = 61024 in³
Also; 1 second = 1/60 minutes
according to formula for exponential decay, we know that;
V = ae^(bt)
Thus, we have;
61024V = ae^(¹/₆₀b(t(h))
V(in³) = (¹/₆₁₀₂₄)a
Read more about subject of formula at; brainly.com/question/790938
Answer:
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