The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.
Answer:
They find problems and solutions by working together
Explanation:
Answer:
7.7 kN
Explanation:
The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.
It can be expressed by using the formula:
![K = \sigma Y \sqrt{\pi a}](https://tex.z-dn.net/?f=K%20%3D%20%5Csigma%20Y%20%5Csqrt%7B%5Cpi%20a%7D)
where;
fracture toughness K = 137 MPa![m^{1/2}](https://tex.z-dn.net/?f=m%5E%7B1%2F2%7D)
geometry factor Y = 1
applied stress
= ???
crack length a = 2mm = 0.002
∴
![137 =\sigma \times 1 \sqrt{ \pi \times 0.002 }](https://tex.z-dn.net/?f=137%20%3D%5Csigma%20%5Ctimes%201%20%20%5Csqrt%7B%20%5Cpi%20%5Ctimes%200.002%20%7D)
![137 =\sigma \times 0.07926](https://tex.z-dn.net/?f=137%20%3D%5Csigma%20%5Ctimes%200.07926)
![\dfrac{137}{0.07926} =\sigma](https://tex.z-dn.net/?f=%5Cdfrac%7B137%7D%7B0.07926%7D%20%3D%5Csigma)
![\sigma = 1728.489 MPa](https://tex.z-dn.net/?f=%5Csigma%20%3D%201728.489%20MPa)
Now, the tensile impact obtained is:
![\sigma = \dfrac{P}{A}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7BP%7D%7BA%7D)
P = A × σ
P = 1728.289 × 4.5
P = 7777.30 N
P = 7.7 kN
Explanation:
He would work on the thing like in the method you work on your question.
Architects must have a professional bachelor's or master's degree in architecture from a program that has been accredited by the National Architectural Accrediting Board, and a state license.