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2 years ago

Drag each label to the correct location on the table. Match to identify permanent and temporary structures.

Engineering

1 answer:

Mama L [17]2 years ago

8 0

Answer:

Permanent Structures:

Motel

Mall

Refinery

Temporary Structures:

Kiosk

Marquee

Exhibition Stand

Explanation: Got a 100 percent on the test

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The efficiency of a steam power plant can beincreased by bleeding off some of the steam thatwould normally enter the turbine and

3241004551 [841]

Answer:

Explanation:

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2 years ago

A 50 (ohm) lossless transmission line is terminated in a load with impedance Z= (30-j50) ohm. the wavelength is 8cm. Determine:

Serjik [45]

Answer:

Reflection Coefficient = $0.57e^{-i79.8}$

SWR=3.65

Position of $V_{max} =3.11cm$

position of $i_{max} =1.11cm$

Explanation:

To determine the above answers, let outline the useful formulas

refection coefficient $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$ .

where terminal impednce = (30-i50)Ω

characteristics impedance= 50Ω

Secondly, the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

Now let us substitute values and solve,

a. $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$

$p=\frac{(30-i50)-50}{(30-i50)-50} \\$

$p=\frac{-20-i50}{80-i50} \\$

multiplying the numerator and denominator by the conjugate of the denominator. we have

$p=\frac{-20-i50}{80-i50}*\frac{80+i50}{80+i50}\\$

by carrying out careful operation, we arrived at

$p=\frac{900-i5000}{8900} \\p=0.1011-i0.56179\\$

To express in polar form i.e $re^{i alpha}$

$r=\sqrt{0.1011^{2}+0.56179^{2}} \\r=0.57\\$

to get the angle

alpha= $tan^{-1} \frac{0.56179}{0.1011} \\alpha=-79.8\\$

hence the Reflection Coefficient,<em>p</em> = $0.57e^{-i79.8}$

b. we now determine the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

$swr=\frac{1+0.57}{1-0.57} =3.65\\$

c. to determine the position of the maximum voltage nearest to the load,

we use the equation

$Position of V_{max}=\frac{\alpha λ}{4\pi}+\frac{λ}{2}\\$

were $λ$ is the wavelength of 8cm

lets convert α to rad by multiplying by π/180

$Position of V_{max}=\frac{-79.8 *8cm*\pi}{4\pi*180 } +\frac{8cm}{2} \\$

$Position of V_{max}=-0.89+4.0=3.11cm\\$ .

d. also were we have minimum voltage,there the maximum current will exist, to find this position nearest to the load

$Position of v_{min}=Position of V_{max}-\frac{λ}{4} \\\\Position of v_{min}=3.11cm-\frac{8cm}{4}=1.11cm$ .

since the voltage minimum occure at 1.11cm. we can conclude that the current maximum also occur at this point i.e 1.11cm

3 0

3 years ago

ABS system is necessary?

Monica [59]

Explanation:

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3 years ago

Carbon dioxide steadily flows into a constant pressure heater at 300 K and 100 kPa witha mass flow rate of 9.2 kg/s. Heat transf

docker41 [41]

Answer:

Carbon dioxide temperature at exit is 317.69 K

Carbon dioxide flow rate at heater exit is 20.25 m³/s

Explanation:

Detailed steps are attached below.

8 0

3 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz

Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s and ρ= 850 kg/m³

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ = ρv

=850 x 0.00062

= 0.527 kg/m·s

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

$Q = \dfrac{\Delta P \pi D^4}{128 \mu L}$

$Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}$

Q = 5.06 x 10⁻⁸ m³/s

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3 years ago