Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = 
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
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:
Tire rotation is the least likely cause of tire wear. So, the option D is correct.
Explanation:
Step1
Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.
Step2
On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.
Step3
Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.
Step4
Tire rotation has least amount of wear and tear due to no rubbing action. It has less amount surface contact with the surface in rotation.
Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.
Those that harden under strain, such as the aluminum-magnesium alloys used in beverage cans and the copper-zinc alloy, brass, used for cartridges, which show more strain hardening than pure copper or aluminum, respectively.
When a material is deformed under a substantial amount of strain, strain hardening is seen as a strengthening process. Lamellar crystals and chain molecule orientation on a vast scale are the culprits. When plastic materials are stretched past their yield point, this phenomena is frequently seen. When a metal is stretched past its yield point, strain hardening occurs. The metal appears to get stronger and harder to deform as more stress is needed to cause additional plastic deformation. Strain hardening is directly related to fatigue.
Learn more about strain hardening here-
brainly.com/question/15058191
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Answer: the answer to you’re problem is Pythagorean Theorem
a²+b²=c² Is the Pythagorean Theorem
Explanation:
