Answer:
Professional engineering license
Bachelor's degree
Computer science classes
job recommendations
Explanation:
The correct answers to the fill in the blanks would be;
1. Viscoelastic stress relaxation refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).
2. Viscoelastic creep refers to scenarios for which a polymer will permanently flow over time in response a constant applied stress.
The polymer whose properties have been mentioned above is commonly known as Kevlar.
It is mostly used in high-strength fabrics and its properties are because of several hydrogen bonds between polymer molecules.
Answer:
![0.14\ lb/in^{3}](https://tex.z-dn.net/?f=0.14%5C%20lb%2Fin%5E%7B3%7D)
Explanation:
Density is defined as mass ler unit volume, expressed as
![\rho=\frac {m}{v}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%20%7Bm%7D%7Bv%7D)
Where m is mass,
is density and v is the volume. For a sphere, volume is given as
![v=\frac {4\pi r^{3}}{3}](https://tex.z-dn.net/?f=v%3D%5Cfrac%20%7B4%5Cpi%20r%5E%7B3%7D%7D%7B3%7D)
Replacing this into the formula of density then
![\rho=\frac {m}{\frac {4\pi r^{3}}{3}}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%20%7Bm%7D%7B%5Cfrac%20%7B4%5Cpi%20r%5E%7B3%7D%7D%7B3%7D%7D)
Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then
![\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%20%7B0.82%7D%7B%5Cfrac%20%7B4%5Cpi%201.12%5E%7B3%7D%7D%7B3%7D%7D%3D0.13932044952427%5Capprox%200.14%20lb%2Fin%5E%7B3%7D)
Answer:
![T_2=315.69k](https://tex.z-dn.net/?f=T_2%3D315.69k)
Explanation:
Initial Temperature ![T_1=500K](https://tex.z-dn.net/?f=T_1%3D500K)
Initial Pressure ![P_1=1000kPa](https://tex.z-dn.net/?f=P_1%3D1000kPa)
Final Pressure ![P_2=200kPa](https://tex.z-dn.net/?f=P_2%3D200kPa)
Generally the gas equation is mathematically given by
![\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7BT_1%7D%3D%5Cfrac%7BP_2%7D%7BP_1%7D%5E%7B%5Cfrac%7Bn-1%7D%7Bn%7D%7D)
Where
n for ![CO=1.4](https://tex.z-dn.net/?f=CO%3D1.4)
Therefore
![\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7B500%7D%3D%5Cfrac%7B200%7D%7B1000%7D%5E%7B%5Cfrac%7B1.4-1%7D%7B1.4%7D%7D)
![T_2=315.69k](https://tex.z-dn.net/?f=T_2%3D315.69k)
Answer:
Current Liabilities
Explanation:
Accounts of Current Liabilities
- Account Payable
- Expense Payable
- Tax Outcome
- Tax Income
- Income tax payable