The covers and guardrails are required in all of the following areas except ditches.
<h3>Do open pits require covers and guardrails?</h3>
Yes, they do need a covers and/or guardrails and it is one that is known or shall be provided to help in the protection of personnel from the hazards that pertains to open pits, tanks, vats and others.
Note that they are often needed that is covers and/or guardrails need to be provided to help to protect personnel.
Therefore, The covers and guardrails are required in all of the following areas except ditches.
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Answer:
The answer is "583.042533 MPa".
Explanation:
Solve the following for the real state strain 1:
Solve the following for the real stress and pressure for the stable.
![K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5Csigma_%7Br1%7D%7D%7B%5B%5CIn%20%5Cfrac%7BI_%7Bil%7D%7D%7BI_%7B01%7D%7D%5D%5En%7D)
Solve the following for the true state stress and stress2.
![=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Csigma_%7Br1%7D%7D%7B%5B%5CIn%20%5Cfrac%7BI_%7Bil%7D%7D%7BI_%7B01%7D%7D%5D%5En%7D%20%5Ctimes%20%5B%5CIn%20%5Cfrac%7BI_%7Bi2%7D%7D%7BI_%7B02%7D%7D%5D%5En%5C%5C%5C%5C%3D%5Cfrac%7B399%20%5C%20MPa%7D%7B%5BIn%20%5Cfrac%7B54.4%7D%7B47.7%7D%5D%5E%7B0.2%7D%7D%20%5Ctimes%20%5BIn%20%5Cfrac%7B57.8%7D%7B47.7%7D%5D%5E%7B0.2%7D%5C%5C%5C%5C%20%3D%5Cfrac%7B399%20%5C%20MPa%7D%7B%5B%20In%20%281.14046122%29%5D%5E%7B0.2%7D%7D%20%5Ctimes%20%5BIn%20%281.21174004%29%5D%5E%7B0.2%7D%5C%5C%5C%5C%20%3D%5Cfrac%7B399%20%5C%20MPa%7D%7B%5B%20In%20%281.02663509%29%5D%7D%20%5Ctimes%20%5BIn%201.03915873%5D%5C%5C%5C%5C%3D%5Cfrac%7B399%20%5C%20MPa%7D%7B0.0114161042%7D%20%5Ctimes%200.0166818905%5C%5C%5C%5C%3D%20399%20%5C%20MPa%20%5Ctimes%201.46125948%5C%5C%5C%5C%3D583.042533%5C%20%5C%20MPa)
My dad said possibly memory
Answer:
u/v = S (y²w) / m sinwt + y/h
Explanation:
see attached image
Answer:
The rate of entropy change of the air is -0.10067kW/K
Explanation:
We'll assume the following
1. It is a steady-flow process;
2. The changes in the kinetic energy and the potential energy are negligible;
3. Lastly, the air is an ideal gas
Energy balance will be required to calculate heat loss;
mh1 + W = mh2 + Q where W = Q.
Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;
Rate of Entropy Change = -Q/T
Where Q = 30Kw
T = Temperature of air = 25°C = 298K
Rate = -30/298
Rate = -0.100671140939597 KW/K
Rate = -0.10067kW/K
Hence, the rate of entropy change of the air is -0.10067kW/K
