Las hojas de cálculo en Excel facilitan los cálculos numéricos a través del uso de fórmulas; de manera fácil y rápida se pueden hacer operaciones aritméticas sobre cientos de miles de datos numéricos; por lo que se puede actualizar o corregir cualquiera de los datos numéricos y las operaciones se recalculan
Answer:
This question is comprising many parts (a to r). That is impossible to answer in one sheet. Following are attached images having answers to most of the parts.
I hope it will help you a lot.
Explanation:
Answer:
The essence including its problem is listed throughout the clarification section following.
Explanation:
Projects build deliverable that seem to be the products of the venture or indeed the implementation of the project. This ensures that perhaps the agile methodology may be as broad as either the goal of the study itself as well as the coverage that would be part of a much larger venture.
For every other production to have been marked as "deliverable" within the same project, this should satisfy a few eligibility requirements:
- It should be within the development of the work.
- The interested parties-external or internal-must consent to the above. This is perhaps the product of hard effort.
So that the above seems to be the right answer.
This question is incomplete, the complete question is;
For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.
Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.
Answer:
the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Explanation:
Given the data in the question;
treatment time t₁ = 11.3 hours
Carbon concentration = 0.444 wt%
thickness at surface x₁ = 1.8 mm = 0.0018 m
thickness at identical steel x₂ = 4.9 mm = 0.0049 m
Now, Using Fick's second law inform of diffusion
/ Dt = constant
where D is constant
then
/ t = constant
/ t₁ = / t₂
t₂ = t₁
t₂ = t₁ /
t₂ = ( / )t₁
t₂ = / × t₁
so we substitute
t₂ = 0.0049 / 0.0018 × 11.3 hrs
t₂ = 7.41 × 11.3 hrs
t₂ = 83.733 hrs
Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs