Answer:
yes, it is
Explanation:
The sequence: (+4)
23,27,31,35,39,43,47,51,55,59,63,67,71,75,79,83
Hope this helps! :)
Answer:
B. rotor
Explanation:
The correct answer Is rotor because the others are part of a cars drivetrain
Answer:
The correct option is;
c. the exergy of the tank can be anything between zero to P₀·V
Explanation:
The given parameters are;
The volume of the tank = V
The pressure in the tank = 0 Pascal
The pressure of the surrounding = P₀
The temperature of the surrounding = T₀
Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change
The exergy balance equation is given as follows;
![X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}](https://tex.z-dn.net/?f=X_2%20-%20X_1%20%3D%20%5Cint%5Climits%5E2_1%20%7B%7D%20%5C%2C%20%5Cdelta%20Q%20%5Cleft%20%281%20-%20%5Cdfrac%7BT_0%7D%7BT%7D%20%5Cright%20%29%20-%20%5BW%20-%20P_0%20%5Ccdot%20%28V_2%20-%20V_1%29%5D-%20X_%7Bdestroyed%7D)
Where;
X₂ - X₁ is the difference between the two exergies
Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system
Therefore, the exergy of the tank can be anything between zero to P₀·V.
Answer:

Just draw a line from point D join to point E
The triangle formed DME will be congruent to AMC
Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃

taking natural log on both sides

Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)

Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0