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3 years ago

Introduce yourself to the class and

Engineering

1 answer:

lubasha [3.4K]3 years ago

6 0

Answer:

What If There Were No U.S. Government?

Whenever Congress and a president of either party lock horns over something of vital national interest, it is easy to declare "a pox on both their houses" and reflect fondly on those nonexistent days when the U.S. government worked brilliantly and without friction.

In 2021, though, the public mood is a little different. The United States Capitol complex was attacked on January 6th—an action that polling suggests around 20% of Americans shockingly approve.1 Self-defined libertarians, pseudo-anarchists, right-wing groups like QAnon that are openly insurrectionist, and other formerly fringe groups have come into national prominence in part by openly questioning the federal government's scope and role in America.

Explanation:

You might be interested in

A beam spans 40 feet and carries a uniformly distributed dead load equal to 2.2 klf (not including beam self-weight) and a live

salantis [7]

Answer:

A beam is a structural member that is subjected primarily to transverse loads ... stress equal to σy, then the section Moment - Curvature (M-φ) response for .... Example 2.2 Design a simply supported beam subjected to uniformly distributed dead ... (live load). • Step I. Calculate the factored design loads (without self-weight).

Explanation:

6 0

3 years ago

Unwanted resistance is being discussed.

krok68 [10]

I don’t know if I’m right but I’m guessing B

6 0

4 years ago

The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsq

Citrus2011 [14]

Answer:

Given that

$V=\dfrac{\Delta Pd^2}{32\mu L}$

LHS of above given equation have dimension $[M^oL^{1}T^{-1}]$ .

Now find the dimension of RHS

Dimension of P = $[ML^{-1}T^{-2}]$ .

Dimension of d= $[M^{0}L^{1}T^{0}]$ .

Dimension of μ = $[ML^{-1}T^{-1}]$ .

Dimension of L= $[M^{0}L^{1}T^{0}]$ .

$\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}$

$\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]$

It means that both sides have same dimensions.

5 0

4 years ago

The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correct

Zanzabum

The complete stress distribution obtained by superposing the stresses produced by an axial force and a bending moment is correctly represented by F/A - (My)/(Iz).

<h3>What is the distribution of pressure at some stage in bending?</h3>

Compressive and tensile forces expand withinside the path of the beam axis beneath neath bending loads. These forces set off stresses at the beam. The most compressive pressure is observed on the uppermost fringe of the beam whilst the most tensile pressure is positioned on the decrease fringe of the beam.

The bending pressure is computed for the rail through the equation Sb = Mc/I, wherein Sb is the bending pressure in kilos in keeping with rectangular inch, M is the most bending second in pound-inches, I is the instant of inertia of the rail in (inches)4, and c is the space in inches from the bottom of rail to its impartial axis.

Read more about beam;

brainly.com/question/25329636

#SPJ1

7 0

2 years ago

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$