Answer:
What is the difference Plastic vs elastic deformation
Explanation:
The elastic deformation occurs when a low stress is apply over a metal or metal structure, in this process, the stress' deformation is temporary and it's recover after the stress is removed. In other words, this DOES NOT affects the atoms separation.
The plastic deformation occurs when the stress apply over the metal or metal structure is sufficient to deform the atomic structure making the atoms split, this is a crystal separation on a limited amount of atoms' bonds.
Complete Question
The complete question is shown on the first uploaded image.
Answer:
The answer is shown on the second uploaded image
Explanation:
The explanation is also shown on the second uploaded image
Answer
For isotropic material plastic yielding depends upon magnitude of the principle stress not on the direction.
Tresca and Von Mises yield criteria are the yield model which is widely used.
The Tresca yield criterion stated that yielding will occur in a material only when the greatest maximum shear stress reaches a critical value.
max{|σ₁ - σ₂|,|σ₂ - σ₃|,|σ₃ - σ₁|} = σ_f
under plane stress condition
|σ₁ - σ₂| = σ_f
The Von mises yielding criteria stated that the yielding will occur when elastic energy of distortion reaches critical value.
σ₁² - σ₁ σ₂ + σ₂² = σ²_f
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus 
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
![w_{b, out} =m([tex]u_{1} -u_{2](https://tex.z-dn.net/?f=w_%7Bb%2C%20out%7D%20%3Dm%28%5Btex%5Du_%7B1%7D%20-u_%7B2)
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
