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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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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

We are given a CSP with only binary constraints. Assume we run backtracking search with arc consistency as follows. Initially, w

sweet-ann [11.9K]

We are given a CSP with only binary can concentrate assume we run backtrackingSearch with ARC

8 0

2 years ago

A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th

SCORPION-xisa [38]

Answer:

$F_{thrust}$ ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

$F_{drag}$ = $\frac{1}{2}$ × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, $F_{drag}-F_{thrust} = 0$

We can as well say:

$F_{drag}= F_{thrust}$

We can now replace $F_{thrust} with F_{drag}$ in the above equation.

Therefore, $F_{thrust}$ = $\frac{1}{2}$ × CρAv²

The A which stands as the area of the jet is given by the formula:

$A=\frac{\pi d^2}{4}$

We can now have a new equation after substituting our A into the previous equation as:

$F_{thrust}$ = $\frac{1}{2}$ × Cρ $(\frac{\pi d^2}{4})v^2$

Substituting our data from above; we have:

$F_{thrust}$ = $\frac{1}{2}$ × $(0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2$

$F_{thrust}$ = $\frac{1}{8} (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2$

$F_{thrust}$ = 110,990N

$F_{thrust}$ in N (newton) to KN (kilo-newton) will be:

$F_{thrust}$ = $(110,990N)*\frac{1KN}{1,000N}$

$F_{thrust}$ = 110.990 KN

$F_{thrust}$ ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0

3 years ago

Por favor. alguien me comunique con fatimalisethmateual

lions [1.4K]

Wait why do you want me to

7 0

2 years ago

Zander worked at a pet shop in high school and during college began taking classes in veterinary medicine. To pay the bills duri

anastassius [24]

True.
To understand it better
First job : Pet shop
Second job : pizza place
The first job supports his career path he has experience.
The second job support life in making sure he gets to his career path/ does help financially for him to get there.
And it’s called career pathway.

3 0

2 years ago