Answer:
The spring is compressed by 0.275 meters.
Explanation:
For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston
Mathematically we can write

we know that


Now the force exerted by an spring compressed by a distance 'x' is given by 
Using the above quatities in the above relation we get

Answer:
Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.
Answer:
$5.184
Explanation:
The cost can be calculated using the formula: 
Before using this, we require the following conversions:
<em>320 W → kW:</em>
<em>30 Days → Hours:</em>

Using the above stated formula:

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 diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm