Answer:
(C) passive state.
Explanation:
The earth pressure is the pressure exerted by the soil on the shoring system. They are three types of earth pressure which are:
a) Rest state: In this state, the retaining wall is stationary, this makes the lateral stress to be zero.
b) Active state: In this state, the wall moves away from the back fill, this leads to an internal resistance. Hence the active earth pressure is less than earth pressure at rest
c) Passive state: In this state the wall is pushed towards the back fill, this leads to shearing resistance. Hence, the passive earth pressure is greater than earth pressure at rest
Answer:
a)
, b) Yes.
Explanation:
a) The maximum thermal efficiency is given by the Carnot's Cycle, whose formula is:


b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.
Maybe it’s a vending machine, I’m confused too
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)
Answer:
The volume of the gas is 11.2 L.
Explanation:
Initially, we have:
V₁ = 700.0 L
P₁ = 760.0 mmHg = 1 atm
T₁ = 100.0 °C
When the gas is in the thank we have:
V₂ =?
P₂ = 20.0 atm
T₂ = 32.0 °C
Now, we can find the volume of the gas in the thank by using the Ideal Gas Law:

(1)
Where R is the gas constant
With the initials conditions we can find the number of moles:
(2)
By entering equation (2) into (1) we have:

Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.
I hope it helps you!