Where are your answer options?
Answer:
ΔU = 53 BTU
Explanation:
In thermodynamics we consider heat entering a system as positive and heat leaving as negative.
Also, work performed by the system is positive and work applied to the system is negative.
The total heat applied to the system is:
Q = 72 - 7 = 65 BTU
The work is:
L = 12 BTU
The first law of thermodynamics states that:
Q = L + ΔU
Where ΔU is the change in internal energy.
Rearranging:
ΔU = Q - L
ΔU = 65 - 12 = 53 BTU
Answer:
This question assesses how well an engineer can explain complicated designs to people who work in other industries. It's important that your mechanical engineer be able to articulate their idea to marketing and executive teams in your company before production begins.
What to look for in an answer:
Evidence they can communicate complicated designs to other teams in a way that makes sense
Willingness to break down complex issues without becoming frustrated
Understanding of the importance of communicating engineering points to others
Example: "Axles serve two main purposes. They help bear some of the weight of the car, and they help the steering system turn your wheels. So, when you turn your steering wheel to the right, the axle helps turn the tires and absorbs any weight shift."
The unique model production line is responsible for producing identical pieces. For this purpose the balancing of the assembly line is only responsible for assembling a model throughout the line.
This is a considerable difference compared to the mixed model assembly line where many models are assembled during the same production line, that is, it produces parts or products that have slight changes accommodated in them, with slight variations in their model or products of soft variety
The choice of the type of production depends on the type of company and its own demand, always prioritizing the efficiency in the operation. Generally, the mixed model tends to be chosen when demand is very large and customer demand is required to be met. In others it is considered a plant model in which half of the line is mixed and the other one is the only model in order to keep the efficiency balanced.
To resolve this problem we have,
![R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm](https://tex.z-dn.net/?f=R%3D3.5mm%5C%5CF_f1%3D950N%5C%5CL_1%3D50mm%5C%5Cb%3D12mm%5C%5CL_2%3D40mm)
is unknown.
With these dates we can calculate the Flexural strenght of the specimen,
![\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa](https://tex.z-dn.net/?f=%5Csigma%7Bfs%7D%3D%5Cfrac%7BF_%7Bf1%7DL%7D%7B%5Cpi%20R%5E3%7D%5C%5C%5Csigma%7Bfs%7D%3D%5Cfrac%7B%28950%29%2850%2A10%5E%7B-3%7D%29%7D%7B%5Cpi%203.5%2A10%5E%7B-3%7D%7D%5C%5C%5Csigma%7Bfs%7D%3D352.65Mpa)
After that, we can calculate the flexural strenght for the square cross section using the previously value.
![\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN](https://tex.z-dn.net/?f=%5Csigma%7Bfs%7D%3D%5Cfrac%7BF_%7Bf2%7DL%7D%7B%5Cpi%20R%5E3%7D%5C%5C%28352.65%2A10%5E6%29%3D%5Cfrac%7B3Ff%2840%2A10%5E%7B-3%7D%29%7D%7B2%2812%2A10%5E%7B-10%7D%29%7D%5C%5CF_%7Bf2%7D%3D%5Cfrac%7B352.65%2A10%5E6%7D%7B34722.22%7D%5C%5CF_%7Bf2%7D%3D10156.32N%5C%5CF_%7Bf2%7D%3D10.2kN)