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3 years ago

Based on the following passage on construction technology during the Middle Ages, why might a worker not be allowed

Engineering

2 answers:

Alex777 [14]3 years ago

6 0

Answer:

He was not born into a family of skilled laborers

Explanation:

KonstantinChe [14]3 years ago

4 0

Answer:

A guild are an association of skilled merchants, craftsmen and artisans that provide support and protection for their members and push their professional interest forward and further their craft

Because the guild guild members mainly gained their skill through knowledge passed to them by their family bloodlines, other unskilled workers were by not being schooled on the craft and by family tradition not allowed to join a guild based on the nature knowledge transfer the guild members have which is mainly several generations of family trade trainings

Explanation:

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D. projectile

Colt1911 [192]

Answer: parabola

Explanation:

•Parabolic Trajectory:

In conclusion, projectiles travel with a parabolic trajectory due to the fact that the downward force of gravity accelerates them downward from their otherwise straight-line, gravity-free trajectory.

5 0

3 years ago

A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para

mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c) 54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

$V_p=\frac{3810.5}{\sqrt{3} }=2200 V$

The complex power S is given as:

$S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA$

$where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA$

The line current I is given as:

$I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A$

The phase voltage at the sending end is:

$V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV$

The magnitude of the line voltage at the source end of the line ( $V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V$

b) The Total real and reactive power loss in the line is:

$S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000$

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

$S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000$

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0

4 years ago

An experiment is performed to determine the boiling characteristics of a special coating applied to the exposed surface. Under s

Dmitry [639]

Answer:

hello your question incomplete attached below is the complete question and detailed solution

Answer : Csf = 0.0131

Explanation:

Attached below is the detailed solution

Given data :

ΔTe = 17.1⁰c calculated as ;Ts - Tsat = ( 117.1 - 100 )

Pe = 957.9 kg/m^3

Cp1e = 4217 j/kgk

<em>U</em>e = 279 * 10^-6 n. s / m^2

Pre = 1.76

hfg = 2.257 * 10^6 J/kg

Pv = 0.5955 kg/m^3

б = 0.0589 N/m

q" = 664 * 10^3 w/m^2 ( calculated )

Input these values into equation 1 as contained in the detailed solution

Csf = 0.0131

3 0

3 years ago

The following figures were obtained in a standard tensile test on a specimen of low carbon steel with a circular sectional area:

liq [111]

Answer:

See Explaination

Explanation:

1)here for given stress strain curve graph is given as follows

where for getting stress,S=F/A=4F/(pi*(50*10^-3)^2)

for strain=e=dl/l=dl*10^-3/100 mm/mm or m/m

2)so graph is as follows

3)for getting youngs modulus of elasticity we must know slope of graph stress verses strain and for straight line in elastic region upto 12 point we have elastic region and from that we get E as

E=slope of graph for first 12 points=S/e=14.5665*10^9/.812=17.9390*10^9 N/m2

4)for getting ultimate tensile stress at which specimen bears maximum load without failure so we get UTS as

UTS=maximum load/area=40*10^6/1.9634=20.3728*10^6 N/m2

5)percentage reduction in area is given by

percentage reduction in area=[original area-final area/original area]*100

Percent reduction=[5062-10^2]*100/50^2=96%

6)percentage elongation is given by

percent elongation=[final length-original length/original length]*100

final length at fractureis=14.56+100=114.56 mm

so we get percent elongation as=[114.56-100/100]*100=14.56%

7)true fracture stress is given by load at fracture devided by true area at fracture

Sf=load/(true area)=4*28*10^3/(pi*(10*10^-3)^2)=356.5070*10^6 N/m2

8 0

4 years ago

A gas turbine has a thermal efficiency of 20% and develops a power output of 8 MW. Determine the fuel consumption rate if heatin

AlexFokin [52]

Answer:

0.8 kilograms of fuel are consumed each second.

Explanation:

As turbines are steady-state devices, the thermal efficiency of a turbine is equal to the percentage of the ratio of the output power to fluid power, that is:

$\eta_{th} = \frac{\dot W}{\dot E} \times 100\,\%$

The fluid power is:

$\dot E = \frac{\dot W}{\eta} \times 100\,\%$

$\dot E = \frac{8\,MW}{20\,\%}\times 100\,\%$

$\dot E = 40\,MW$

Which means that gas turbine consumes 40 megajoules of fluid energy each second, which is heated and pressurized with help of the fuel, whose amount of consumption per second is:

$50\,\frac{MJ}{kg} = \frac{40\,MJ}{m_{fuel}}$

$m_{fuel} = 0.8\,kg$

0.8 kilograms of fuel are consumed each second.

3 0

3 years ago