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ehidna [41]
3 years ago
11

Cost and benefit, environmental effects, ethical or legal implications, health and safety considerations, and sustainability are

the five main categories of which of the following options?
the main constraints for engineers in modern times

concerns flagged by the U.S. Green Building Council

course requirements for engineering degrees

the major items for engineers to consider during the design process
Engineering
1 answer:
kykrilka [37]3 years ago
7 0

Answer:

the major items for engineers to consider during the design process.

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A mass of 7 kg undergoes a process during which there is heat transler frorn the mass at a rate of 2 kJ per kg, an elevation dec
irinina [24]

Answer:44.61 KJ

Explanation:

Let h_1,V_1,Z_1 be the initial specific enthalpy,velocity&elevation of the system

and h_2,V_2,Z_2 be the Final specific enthalpy,velocity&elevation of the system

mass(m)=7kg

Applying Steady Flow Energy Equation

m\left [ h_1+\frac{v_1^2}{2g}+gZ_1\right ]+Q=\left [ h_2+\frac{v_2^2}{2g}+gZ_2\right ]+W

h_1-h_2=4 KJ/kg

V_1=13m/s

V_2=23m/s

Z_1-Z_2=40m

substituting values

7\left [ h_1+\frac{13^{2}}{2g}+gZ_1\right ]+7\times2 = \left [ h_2+\frac{23^2}{2g}+gZ_2\right ]+W

W=7\left [h_1-h_2+\frac{V_1^2-V_2^2}{2000g}+g\frac{\left (Z_1-Z_2 \right )}{1000}\right ]+Q

W=7\left [4+\frac{13^2-23^2}{2000g}+g\frac{\left (40 \right )}{1000}[\right ]+2\times 7

W=44.61KJ

4 0
4 years ago
Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these d
Tems11 [23]

Complete Question

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these devises are available for weekend warriors who want to play with their chain saws. Let us model the illustrated tongs as a planar mechanism that carries a log of weight 210 N. Given the following dimensions: 35 mm 10 mm 40 mm 230 mm 85 mm 45 mm 10 mm 35 mm 345 mm determine the force in N and moment in Nm that our worker exerts on the tongs. Also determine the pinching force magnitude in N that the tongs exert on the log; i.e. determine the horizontal force that the tong's teeth exert on the log. Assume the  point E is centered between the tong's teeth.

The diagram for this question is shown on the first uploaded image

Answer:

The force P is P= 210 N

and the moment M is M = -48.3N \cdot m

The horizontal force that the tong teeth exerts is F_T =89.67N

Explanation:

First let denote the dimension to corresponding to the diagram

      a=35mm , b= 10mm, c= 40mm, d= 230mm, e= 85mm,f= 45mm,\\g= 10mm,h=35mm,i=345mm.

Next looking at the diagram let us consider the vertical direction

At equilibrium

               \sum F_{vertical} =0

This mean that

               P+ W = 0

Since they are acting in opposite direction the equation becomes

               P - W = 0

=>           P= W

=>            P= 210 N

At Equilibrium  Moment about F gives

               \sum M_f  = 0

=> F_T * (e +f + g+ h+ i) - F_T * (e+ f+g+ h+i) - W *d -M =0

=> M = -W *d

=> M = -210 * 0.230

=> M = -48.3N \cdot m

Here F_T is the horizontal force that the tong teeth exerts

Now let consider the part BAF of the system as shown on the second uploaded image

  Now the angle \theta is mathematically given as

             tan \theta = \frac{g+h}{a}

=>        \theta = arctan \frac{g+h}{a}

                = arctan (\frac{10+35}{35} )

               =52.125^o

Now at equilibrium the moment about A is

                \sum M_A = 0

          =>  P * (c+d) +M + F_{BC} cos \theta *f+F_{BC}sin\theta *(a+b) =0

                210 * (0.040 + 0.230)-48.3+F_{BC} cos (52.125^o)*0.045+ F_{BC}sin(52.125^o)* (0.035 +0.010) =0

    =>   10.29 +F_{BC} (0.02763+0.03552) =0

    =>     F_{BC} =\frac{10.29}{0.06315}

    =>      F_{BC} = - 162.925 N

Looking at the forces acting on the teeth as shown on the third uploaded image

 At Equilibrium the moment about D is

      \sum M_D = 0

=>  \frac{W}{2} *d - F_T *(i+h) -F_{BC} cos \theta *h -F_{BC} sin \theta * (b+c) =0

=>   \frac{210}{2} * 0.230 -F_T (0.345 +0.035) - (-162.925)cos(52.125^o) *0.035\\-(-162.925)sin(52.125^o)(0.010 +0.040) =0

=>    34.081  = F_T(0.345 +0.035)

=>   F_T =89.67N

         

   

         

7 0
4 years ago
Engineers please help im not good when it comes to drawing​
kakasveta [241]

Answer:

I'm good at drawing and computer-animated design

Explanation:

3 0
3 years ago
When buttons or switches are pressed by humans for arbitrary periods of time, we need to convert a signal level to a pulse. In t
ddd [48]

Answer:

The FSM uses the states along with the generation at the P output on each of the positive edges of the CLK. The memory stores the previous state in the machine and the decoder generates a P output based on the previous state.

Explanation:

The code is in the image.

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Why do historians continue to gather historical evidence for even well-known events?
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Answer:

lol

Explanation:

7 0
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