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Dominik [7]
3 years ago
12

Differentiate between "Threshold and Resolution" with suitable examples.

Engineering
1 answer:
9966 [12]3 years ago
6 0

Answer:

to make the bace of a building more sturdy

Explanation:

example: the bace of the empire state building is stone very sturdy

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thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat
eduard

Answer:

q_{in} = 528.6\,\frac{kJ}{kg}

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

q_{in} = h_{g} - h_{mix}

The specific enthalpies are:

Liquid-Vapor Mixture:

h_{mix} = 2217.2\,\frac{kJ}{kg}

Saturated Vapor:

h_{g} = 2745.8\,\frac{kJ}{kg}

The thermal energy per unit mass required to heat the steam is:

q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}

q_{in} = 528.6\,\frac{kJ}{kg}

7 0
3 years ago
What is christmas really about
Ray Of Light [21]

Answer:

Christmas is celebrated to remember the birth of Jesus Christ, who Christians believe is the Son of God.

Explanation:

Christmas is celebrated to remember the birth of Jesus Christ, who Christians believe is the Son of God. The name 'Christmas' comes from the Mass of Christ (or Jesus). A Mass service (which is sometimes called Communion or Eucharist) is where Christians remember that Jesus died for us and then came back to life.

5 0
3 years ago
Read 2 more answers
A bearing is to be used as shaft support to carry both radial and thrust forces. Calculations show that a 02 series ball bearing
shepuryov [24]

Answer:

The correct answer is "20.8 kN" and "31 kN". A further explanation is given below.

Explanation:

  • The angular touch bearing seems to be a fine replacement while accommodating radial and even some displacement pressures.
  • You may receive static as well as dynamic scores from either the manufacturer's collections.  

The load ratings should be for the SKF bearing including its predetermined distance:

Static load

= 20.8 kN

Dynamic load

= 31 kN

4 0
3 years ago
Select the correct answer.
andreev551 [17]

Answer:

c

Explanation:

8 0
3 years ago
Read 2 more answers
An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u=umaxðAy2 + By+
nexus9112 [7]

Answer:

the volume flow rate per unit depth is:

\frac{Q}{b} = \frac{2}{3} u_{max} h

the ratio is : \frac{V}{u_{max}}=\frac{2}{3}

Explanation:

From the question; the  equations of the velocities profile in the system are:

u = u_{max}(Ay^2+By+C)   ----- equation (1)

The above boundary condition can now be written as :

At y= 0; u =0           ----- (a)

At y = h; u =0            -----(b)

At y = \frac{h}{2} ; u = u_{max}     ------(c)

where ;

A,B and C are constant

h = distance between two plates

u = velocity

u_{max} = maximum velocity

y = measured distance upward from the lower plate

Replacing the boundary condition in (a) into equation (1) ; we have:

u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0

Replacing the boundary condition (b) in equation (1); we have:

u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*h^2+B*h+C) \\ \\ 0 = Ah^2 +Bh + C \\ \\ 0 = Ah^2 +Bh + 0 \\ \\ Bh = - Ah^2 \\ \\ B = - Ah   \ \ \ \ \   --- (d)

Replacing the boundary condition (c) in equation (1); we have:

u = u_{max}(Ay^2+By+C) \\ \\ u_{max}= u_{max}(A*(\frac{h^2}{2})+B*\frac{h}{2}+C) \\ \\ 1 = \frac{Ah^2}{4} +B \frac{h}{2} + 0 \\ \\ 1 =  \frac{Ah^2}{4} + \frac{h}{2}(-Ah)  \\ \\ 1=  \frac{Ah^2}{4}  - \frac{Ah^2}{2}  \\ \\ 1 = \frac{Ah^2 - Ah^2}{4}  \\ \\ A = -\frac{4}{h^2}

replacing A = -\frac{4}{h^2} for A in (d); we get:

B = - ( -\frac{4}{h^2})hB = \frac{4}{h}

replacing the values of A, B and C into the velocity profile expression; we have:

u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)

To determine the volume flow rate; we have:

Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)

Replacing u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y) \ for \ u

\frac{Q}{b} = \int\limits^h_0 u_{max}(-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\  \frac{Q}{b} = u_{max}  \int\limits^h_0 (-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} (-\frac{-4}{h^2}\frac{y^3}{3} +\frac{4}{h}\frac{y^2}{y})^ ^ h}}__0  }} \\ \\ \frac{Q}{b} =u_{max} (-\frac{-4}{h^2}\frac{h^3}{3} +\frac{4}{h}\frac{h^2}{y})^ ^ h}}__0  }} \\ \\ \frac{Q}{b} = u_{max}(\frac{-4h}{3}+\frac{4h}2} ) \\ \\ \frac{Q}{b} = u_{max}(\frac{-8h+12h}{6}) \\ \\ \frac{Q}{b} =u_{max}(\frac{4h}{6})

\frac{Q}{b} = u_{max}(\frac{2h}{3}) \\ \\ \frac{Q}{b} = \frac{2}{3} u_{max} h

Thus; the volume flow rate per unit depth is:

\frac{Q}{b} = \frac{2}{3} u_{max} h

Consider the discharge ;

Q = VA

where :

A = bh

Q = Vbh

\frac{Q}{b}= Vh

Also;  \frac{Q}{b} = \frac{2}{3} u_{max} h

Then;

\frac{2}{3} u_{max} h = Vh \\ \\ \frac{V}{u_{max}}=\frac{2}{3}

Thus; the ratio is : \frac{V}{u_{max}}=\frac{2}{3}

5 0
3 years ago
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