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3 years ago

Which of the following statements is true of a mature technology?

Engineering

1 answer:

hoa [83]3 years ago

8 0

Answer:

I think it is( More expensive than immature technologies) I took engineering class and this question is still quite tricky.

Explanation:

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Joinn my zo om lets play some blookets 98867 708157 9dPQPW

pav-90 [236]

Answer:k

Explanation:

6 0

3 years ago

An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator

MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit $i_m=385mA=385\times 10^{-3}A=0.385A$

Inductance of the inductor $L=400mH=400\times 10^{-3}h=0.4H$

Capacitance $C=4.43\mu F=4.43\times 10^{-3}F$

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be $x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm$

Capacitive reactance will be equal to $X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm$

Impedance of the circuit will be $Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm$

So maximum voltage will be $\Delta V_{max}=0.385\times 865.44=333.194volt$

(B) Phase difference will be given as $\Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70$

So current will be leading by an angle 54.70

5 0

3 years ago

A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i

Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature $T_{1}$ = 300 F = 422.04 K

Duct temperature $T_{2}$ = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant $\sigma$ = 5.67 × $10^{-8}$ $\frac{W}{m^{2} K^{4} }$

Heat transfer per foot length is given by

Q = ∈ $\sigma$ A ( $T_{1}^{4} - T_{2} ^{4}$ ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × $10^{-8}$ × 3.14 × 0.0508 × 1 × ( $422.04^{4} - 294.26^{4}$ )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

4 0

3 years ago

Help now please evaluate using the commutative property: 40 (32) (10) (25)

Ilia_Sergeevich [38]

Answer:

#See solution for details.

Explanation:

#The commutative property of multiplication tells us that it doesn't matter in what order you multiply numbers. The formula for this property is a * b = b * a:

$40\times32\times10\times 25=320,000\\\\25\times10\times32\times 40=320,000\\\\10\times32\times25\times 40=320,000\\\\32\times25\times40\times 10=320,000$

Hence, the product of the four numbers remains 320,000 irrespective of their order.

4 0

3 years ago

The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si

alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

A) Classifying the soil according to USCS system

( using 2nd image attached below )

description of sand :

The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0

3 years ago