Technician B is correct because the way aluminum collapses can be predicted.
Hardened steel and aluminum are two metals used for different purposes including:
- Construction.
- Appliances.
- Small utensils.
- Airplanes.
- Vehicles.
These two materials have slightly different features in terms of resistance, flexibility, etc.
In the case of hardened steel, this is considered to be malleable but strong. This means it is possible to change its shape under some conditions but it can resist great forces and pressure. Moreover, if the hardening process is carried out properly all the areas should be equally strong.
On the other hand, aluminum is recognized due to its durability and for being lighter than other materials. Despite this, aluminum is more flexible than steel and collapses under weaker forces. This has been widely studied because aluminum collapse shows a predictable pattern.
Based on this, only technician B is correct.
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Answer:
the time, in hours = 4.07hrs
Explanation:
The detailed step by step derivation and appropriate integration is as shown in the attached files.
Technician b says that hydrogen has better specific energy than fossil fuels. so technician b is correct.
<h3>
Does hydrogen have the highest specific energy?</h3>
- Although hydrogen has the highest energy density per mass of any fuel, it has the lowest energy density per unit volume due to its low ambient temperature density, necessitating the development of new storage technologies with the potential for higher energy density.
- Hydrogen is utilized as a rocket fuel and in fuel cells to generate power on some spacecraft because it is beneficial as an energy source or fuel due to its high energy content per unit of weight.
- Since hydrogen has a significantly higher energy content than fossil fuels, less of it is required to carry out any given task.
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Answer:
a. 30°
b. 0.9MPa
Explanation:
The slip will occur along that direction for which the Schmid factor is maximum. The three possible slip directions are mentioned as 30°, 48°, 78°
The cosines for the possible λ values are given as
For 30°, cos 30 = 0.867
For 48°, cos 48 = 0.67
For 78°, cos 78 = 0.21
Among the three-calculated cosine values, the largest cos(λ) gives the favored slip direction
The maximum value of Schmid factor is 0.87. Thus, the most favored slip direction is 30° with the tensile axis.
The plastic deformation begins at a tensile stress of 2.5MPa. Also, the value of the angle between the slip plane normal and the tensile axis is mentioned as 65°
Thus, calculate the value of critical resolved shear stress for zinc:
From the expression for Schmid’s law:
τ = σ*cos(Φ)*cos(λ)
Substituting 2.5MPa for σ, 30° for λ and 65° for Φ
We obtain The critical resolved shear stress for zinc, τ = 0.9 MPa
The answer is the B because that is the correct answer
