Answer:
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Rubber block is not shown. I have attached an image of it.
Answer:
A) ε_x = 0.0075
B) ε_y = 0.00375
C) γ_xy = 0.0122 rad
Explanation:
We are given;
δ = 0.03 in
L = 4 in
ν_r = 0.5
θ = 89.3° = 89.3π/180 rad
Let's calculate ε_x in the direction of axis x
Thus, ε_x = δ/L = 0.03/4 = 0.0075
Let's calculate ε_y in the direction of axis y;
ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375
Now, shear strain is angle between π/2 rad surfaces at that point.
Thus,
γ_xy = π/2 - θ = π/2 - 89.3π/180
γ_xy = π(0.003889) = 0.0122 rad
Answer:
Los aditivos que deben incorporarse a la masa de concreto para aumentar su resistencia a los ciclos alternos de congelación y descongelación son;
1. Agentes de arrastre de aire (AEA) o
2. Materiales poliméricos súper absorbentes
Explanation:
La resistencia alterna de los ciclos de congelación y descongelación en el concreto puede aumentarse mediante la adición de agentes de arrastre de aire.(AEA) que es un surfactante, crea burbujas de aire muy pequeñas en el concreto resultante para mejorar la durabilidad y resistencia del cemento al ciclo repetido de congelación y descongelación o materiales poliméricos súper absorbentes
Ejemplos de agentes de arrastre de aire son;
Sulfonatos alcalinos
Acidos de resinas sulfonadas
Sales de ácidos grasos
Ejemplos de materiales poliméricos superabsorbentes son;
SAP0.26CT
SAP0.39PT.
Answer:
The answer is "
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Explanation:
Please find the complete question in the attached file.
Its change in temperature in pipes depends on rate heads and loss in pipes owing to pipe flow, contractual loss, etc.
The temperature change thus relies on V1 v2 p d D L.
Answer:
The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.
Explanation:
The net force over the window is calculated by multiplying the difference in pressure by the area of the window:
F = Δp*A
The pressure inside the plane is around 1 atm, hence the difference in pressure is:
Δp = 1atm - 0.35 atm = 0.65 atm
Expressing in the EE unit system:
Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2
Replacing in the force:
F = 9.55 lbf/in^2 * 80 in^2 = 764 lbf
For the international unit system, we re-calculate the window's area and the difference in pressure:
A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2
Δp = 0.65 atm * 101325 Pa = 65861 Pa = 65861 N/m^2
Replacing in the force:
F = 65861 N/m^2 *0.0516 m^2 = 3398 N
