Answer:
The condition does not hold for a compression test
Explanation:
For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.
<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test
Click and drag it down to the bottom bro
Answer:
The specific weight of unknown liquid is found to be 15 KN/m³
Explanation:
The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.
Total Pressure = Pressure of oil + Pressure of unknown liquid
65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)
65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)
(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²
(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m
<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>
Answer:
<em>The temperature will be greater than 25°C</em>
Explanation:
In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.
mathematically
Change in the internal energy of a system ΔU = ΔQ + ΔW
in an adiabatic process, ΔQ = 0
therefore
ΔU = ΔW
where ΔQ is the change in heat into the system
ΔW is the work done by or done on the system
when work is done on the system, it is conventionally negative, and vice versa.
also W = pΔv
where p is the pressure, and
Δv = change in volume of the system.
In this case,<em> work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C </em>
Answer:
1123.6 pounds/ square inch.
Explanation:
Fluid pressure is the ratio of force or weight applied by the fluid per unit area.
i.e Fluid pressure = 
The maximum load of the jack is obtained at its maximum capacity = 5000 pounds
Area of the large piston on the jack = 4.45 square inches
Thus,
Fluid pressure = 
= 1123.5955
Fluid pressure = 1123.6 pounds/ square inch
Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.
