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2 years ago

The author uses the example of the flameless candle to illustrate that

Engineering

2 answers:

Vesnalui [34]2 years ago

6 0

Answer:

Engineering makes products safer

Explanation:

krek1111 [17]2 years ago

4 0

Answer:

I don't necessarily know the context of the text you are talking about, but a flameless candle could represent how there no more light in the author's life, which could represent a lack of happiness or a loss. Hope this helps!

Explanation:

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A turboprop engine consists of a diffuser, compressor, combustor, turbine, and nozzle. The turbine drives a propeller as well as

Yuri [45]

Answer:

Power =10004.57kw

Velocity = 296.4m/s

Explanation:

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7 0

3 years ago

The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

miss Akunina [59]

Answer:

a) Thermal conductivity of skin: $k_{skin}=1.5W/mK$

b) Temperature of interface: $T_{interface}=35.6\°C$

Heat flux through skin: $\frac{Q}{A}=2100W/m^2$

Explanation:

$k=\frac{QL}{A(T_{2}-T_{1})}$

Where: $k$ is thermal conductivity of a material, $\frac{Q}{A}$ is heat flux through a material, $L$ is the thickness of the material, $T_{1}$ is the temperature on the first side and $T_{2}$ is the temperature on the second side

$k_{skin}=\frac{QL}{A(T_{2}-T_{1})}$

$k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}$

$k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}$

$k_{skin}=1.5W/mK$

$k_{insulation}=\frac{k_{skin}}{2}$

$k_{insulation}=\frac{1.5}{2}$

$k_{insulation}=0.75W/mK$

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

$\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}$

$\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}$

$\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}$

$55500-1500T_{interface}=375T_{interface}-11250$

$1875T_{interface}=66750$

$T_{interface}=35.6\°C$

$\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}$

$\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}$

$\frac{Q}{A}=2100W/m^2$

3 0

3 years ago

In the ______ phase of the organizational life cycle, the organization is usually very small and agile, focusing on new products

Vlad [161]

Answer:

Entrepreneurship

Explanation:

Different phases of organizational life cycles can be observed depending on the stage of the company. These cycles are,

Entrepreneurship
Survival and Early Success
Sustained Success
Renewal (or Decline)

In the initial stage which is Entrepreneurship Phase, all of the founders take part in various activities, no formality is formed between founders and employees. Generally lots of ideas are present and company is actively searching to define correct market and products to focus on.

5 0

2 years ago

1. A cylindrical casting is 0.3 m in diameter and 0.5 m in length. Another casting has the same metal is rectangular in cross-se

Lorico [155]

Based on the Chvorinov's rule, the diference in the solidification times of the two castings is 14.092 times the solidification time of the prism casting.

<h3>How to apply the Chvorinov's rule for casting processes</h3>

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a reference time. This rule states that cooling time (t) is directly proportional to the square of the volume (V), in cubic meters, divided to the surface area (A), in square meters. Now we proceed to model each casting:

<h3>Cylindrical casting</h3>

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]² (1)

<h3>Prism casting</h3>

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]² (2)

<h3>Relationship between the cross sections of both castings</h3>

3 · T² = 0.25π · D² (3)

Where:

t - Cooling time of the cylindrical casting, in time unit.
t' - Cooling time of the prism casting, in time unit.
C - Cooling factor, in time unit per square meter.
D - Diameter of the cylinder, in meters.
L - Length of the casting, in meters.
T - Width of the cross section of the prism casting, in meters.

If we know that D = 0.3 m, then the thickness of the prism casting is:

$T = \sqrt{\frac{\pi}{12} }\cdot D$

T ≈ 0.153 m

And (1) and (2) simplified into these forms:

<h3>Cylindrical casting</h3>

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C (1b)

<h3>Prism casting</h3>

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C (2b)

Lastly we find the percentual difference in the solidification times of the two castings by using the following expression:

r = (1 - t'/t) × 100 %

r = (1 - 0.00218/0.0329) × 100 %

r = 93.374 %

The cooling time of the prism casting is 6.626 % of the solidification time of the cylindrical casting. The diference in the solidification times of the two castings is 14.092 times the solidification time of the prism casting. $\blacksquare$

To learn more on solidification times, we kindly invite to check this verified question: brainly.com/question/13536247

3 0

2 years ago

Passband to baseband conversion: Consider the following passband signal,

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Answer:fr

Explanation:

5 0

3 years ago

The author uses the example of the flameless candle to illustrate that​

The author uses the example of the flameless candle to illustrate that