Answer:
Yes. YES yes yes. Unless you are in Australia or something.
Answer:
D. a triangle and a T-Square
Explanation:
A T-Square is the best drawing tool to create squares. You would need a squares to create cubes.
Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=![\frac{s}{(s+3)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bs%7D%7B%28s%2B3%29%5E2%7D)
U(s)=
for unit step function
output =H(s)×U(s)
=
×![\frac{1}{s}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bs%7D)
=![\frac{1}{(s+3)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28s%2B3%29%5E2%7D)
taking inverse laplace of output
output=t×![e^{-3t}](https://tex.z-dn.net/?f=e%5E%7B-3t%7D)
at t=0 putting the value of t=0 in output
output =0
Answer:
Mechanical Efficiency = 83.51%
Explanation:
Given Data:
Pressure difference = ΔP=1.2 Psi
Flow rate = ![V=8ft^3/s\\](https://tex.z-dn.net/?f=V%3D8ft%5E3%2Fs%5C%5C)
Power of Pump = 3 hp
Required:
Mechanical Efficiency
Solution:
We will first bring the change the units of given data into SI units.
![P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW](https://tex.z-dn.net/?f=P%3D1.2%2A6.895%20%3D%208.274KPa%5C%5CV%3D8%2A0.00283%3D0.226%20m%5E3%2Fs%5C%5CP%3D3%2A0.746%3D2.238KW)
Now we will find the change in energy.
Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.
Thus change in energy is
![=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\](https://tex.z-dn.net/?f=%3D%28Mass%20%2A%20change%20in%20P%29%2Fdensity%5C%5C%3D%20%5Cfrac%7BM%2AP%7D%7Bp%7D%5C%5C%5C%5C)
As we know that Mass = Volume x density
substituting the value
Energy = Volume * density x ΔP / density
Change in energy = Volumetric flow x ΔP
Change in energy = 0.226 x 8.274 = 1.869 KW
Now mechanical efficiency = change in energy / work done by shaft
Efficiency = 1.869 / 2.238
Efficiency = 0.8351 = 83.51%
Answer:
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Explanation:
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