Answer:
i think it is A, i could not really understand your question.
Explanation:
just a question
(wheres the article?)
Answer:
Explanation:
The situation being described completely fails in regard to the importance of metrology. This is because the main importance of metrology is making sure that all of the measurements in a process are as accurate as possible. This accuracy allows an entire process to function efficiently and without errors. In a food production plant, each individual department of the plant relies on the previous function to have completed their job with the correct and accurate instructions so that they can fulfill their functions correctly and end up with a perfect product. If the oven (like in this scenario) is a couple of degrees off it can cause the product to come out burned or undercooked, which will then get transferred to the next part of production which will also fail due to the failed input (burned or undercooked product). This will ultimately lead to an unusable product at the end of the process and money wasted. Which in a large production plant means thousands of products in a single batch are thrown away.
Answer:
2.77mpa
Explanation:
compressive strength = 20 MPa. We are to find the estimated flexure strength
We calculate the estimated flexural strength R as
R = 0.62√fc
Where fc is the compressive strength and it is in Mpa
When we substitute 20 for gc
Flexure strength is
0.62x√20
= 0.62x4.472
= 2.77Mpa
The estimated flexure strength is therefore 2.77Mpa
Answer: what is the formula
Explanation:
I can’t figure nothing out with out the formula
Answer:

Explanation:
Given that
Shear modulus= G
Sectional area = A
Torsional load,

For the maximum value of internal torque

Therefore

Thus the maximum internal torque will be at x= 0.25 L
![t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}](https://tex.z-dn.net/?f=t%28x%29_%7Bmax%7D%20%3D%20%5Cint_%7B0%7D%5E%7B0.25L%7Dp%20sin%28%20%5Cfrac%7B2%5Cpi%7D%7B%20L%7D%20x%29dx%5C%5Ct%28x%29_%7Bmax%7D%20%3D%5Cleft%20%5Bp%5Ctimes%20%5Cdfrac%7B-cos%28%20%5Cfrac%7B2%5Cpi%7D%7B%20L%7D%20x%29%7D%7B%5Cfrac%7B2%5Cpi%7D%7B%20L%7D%7D%20%20%5Cright%20%5D_0%5E%7B0.25L%7D%5C%5Ct%28x%29_%7Bmax%7D%20%3D%5Cdfrac%7Bp%5Ctimes%20L%7D%7B2%5Ctimes%20%5Cpi%7D)