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
........{1}
and for the final case
............{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 = ![\frac{1.6-0.1}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B1.6-0.1%7D%7B10%7D)
or
average rate of change of crack = 0.16 micron per day
Answer:
the length of the bridge B is L = m - 5555 ft
Explanation:
Denoting the length of the bridge B as L we get
length of bridge B = length of bridge A - 5555 ft
since length of bridge A=m
L = m - 5555 ft
Answer:
i)ω=3600 rad/s
ii)V=7059.44 m/s
iii)F=1245.8 N
Explanation:
i)
We know that angular speed given as
![\omega =\dfrac{d\theta}{dt}](https://tex.z-dn.net/?f=%5Comega%20%3D%5Cdfrac%7Bd%5Ctheta%7D%7Bdt%7D)
We know that for one revolution
θ=2π
Given that time t= 2 hr
So
ω=θ/t
ω=2π/2 = π rad/hr
ω=3600 rad/s
ii)
Average speed V
![V=\sqrt{\dfrac{GM}{R}}](https://tex.z-dn.net/?f=V%3D%5Csqrt%7B%5Cdfrac%7BGM%7D%7BR%7D%7D)
Where M is the mass of earth.
R is the distance
G is the constant.
Now by putting the values
![V=\sqrt{\dfrac{GM}{R}}](https://tex.z-dn.net/?f=V%3D%5Csqrt%7B%5Cdfrac%7BGM%7D%7BR%7D%7D)
![V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}](https://tex.z-dn.net/?f=V%3D%5Csqrt%7B%5Cdfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%7D%7B8000%5Ctimes%2010%5E3%7D%7D)
V=7059.44 m/s
iii)
We know that centripetal fore given as
![F=\dfrac{mV^2}{R}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7BmV%5E2%7D%7BR%7D)
Here given that m= 200 kg
R= 8000 km
so now by putting the values
![F=\dfrac{mV^2}{R}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7BmV%5E2%7D%7BR%7D)
![F=\dfrac{200\times 7059.44^2}{8000\times 10^3}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B200%5Ctimes%207059.44%5E2%7D%7B8000%5Ctimes%2010%5E3%7D)
F=1245.8 N
Answer:
C. assembly line workers.
Explanation: