Answer:
The average velocity is 0.203 m/s
Explanation:
Given;
initial displacement, x₁ = 20 yards = 18.288 m
final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m
change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s
The average velocity is given by;
V = change in displacement / change in time
V = (x₂ - x₁) / Δt
V = (18.288 - 6.096) / 60
V = 0.203 m/s
Therefore, the average velocity is 0.203 m/s
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
Answer:
Taking as a basis of calculation 100 mol of gas leaving the conversion reactor, draw andcompletely label a flowchart of this process. Then calculate the moles of fresh methanol feed,formaldehyde product solution, recycled methanol, and absorber off-gas, the kg of steamgenerated in the waste-heat boiler, and the kg of cooling water fed to the heat exchangerbetween the waste-heat boiler and the absorber. Finally, calculate the heat (kJ) that must beremoved in the distillation column overhead condenser, assuming that methanol enters as asaturated vapor at 1 atm and leaves as a saturated liquid at the same pressure.
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Explanation:
Unless cylinders are firmly secured on a special carrier intended for this purpose, regulators shall be removed and valve protection caps put in place before cylinders are moved. A suitable cylinder truck, chain, or other steadying device shall be used to keep cylinders from being knocked over while in use.
Answer:
R=1923Ω
Explanation:
Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.
Coil length(L) of the wire=37.0m
Cross-sectional area of the conductor or wire (A) = πr^2
A= π * (2.053/1000)/2=3.31*10^-6
To calculate for the resistance (R):
R=ρ*L/A
R=(1.72*10^8)*(37.0)/(3.31*10^-6)
R=1922.65Ω
Approximately, R=1923Ω