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3 years ago

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You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is movi

ng at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel

2 answers:

11111nata11111 [884]3 years ago

8 0

The solution is in the attachment

Vedmedyk [2.9K]3 years ago

6 0

Answer:

please find attached.

Explanation:

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Which is the most effective way to reduce discrimination at the workplace? A. Appoint employees of a single race or religion. B.

sveticcg [70]

Answer:

C is the correct answer coz the others don't make sense

6 0

2 years ago

Read 2 more answers

Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

Katyanochek1 [597]

Answer:

$K_{I}=5.21 MPa\sqrt{m}$

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

so

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

Y = $\frac{0.8792}{0.89}$

Y = 0.9878

so put here value in equation 1

$K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}$

$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

$K_{I}$ = 5210.45 × 10³ $Pa\sqrt{m}$

$K_{I}$ = 5.21 MPa $\sqrt{m}$

5 0

3 years ago

E = 50V, R= 1000 ohms, then I =

Margarita [4]

Explanation:

<em>Current</em><em> </em><em>(</em><em>I)</em><em>=</em><em> </em><em><u>Voltage</u></em><em><u> </u></em><em><u>(</u></em><em><u>V)</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>Resistance </em><em>(Ω)</em><em>. </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em>I=</em><em> </em><em><u>5</u></em><em><u>0</u></em><em><u>V</u></em>

<em> </em><em> </em><em> </em><em> </em><em>1000</em><em>(Ω)</em>

<em>I=</em><em> </em><em><u>5</u></em><em><u>V</u></em>

<em> </em><em> </em><em> </em><em> </em><em>1</em><em>0</em><em>0</em><em>(Ω)</em>

<em>I=</em><em> </em><em>0</em><em>.</em><em>0</em><em>5</em><em> </em><em>(</em><em>Amp)</em><em>. </em><em>Answer</em>

<em>Voltage=</em><em>V</em>

<em>Current</em><em>=</em><em>I</em>

<em>Resistance</em><em>=</em><em>(Ω)</em><em>. </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(Ω)</em><em>=</em><em>Ohm</em>

<em>And </em><em>SI </em><em>unit</em><em> </em><em>of </em><em>current</em><em> </em><em>is</em><em>. </em><em>Ampere(</em><em>Amp)</em><em> </em>

<em><u>hope</u></em><em><u> this</u></em><em><u> helps</u></em><em><u> you</u></em>

<em><u>have</u></em><em><u> a</u></em><em><u> good</u></em><em><u> day</u></em><em><u>.</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em>

8 0

3 years ago

Consider a 20-cm X 20-cm X 20-cm cubical body at 477°C suspended in the air. Assuming the body closely approximates a blackbody,

Olin [163]

Answer:

a) The rate at which the cube emits radiation energy is 704.48 W

b) The spectral blackbody emissive power is 194.27 W/m²μm

Explanation:

Given data:

a = side of the cube = 0.2 m

T = temperature = 477°C

Wavelength = 4 µm

a) The surface area is:

$A_{s} =6a^{2} =6(0.2)^{2} =0.24m^{2}$

According Stefan-Boltzman law, the rate of emission is:

$E=\sigma T^{4} A_{s} =5.67x10^{-8} *(477)^{4} *0.24=704.48W$

b) Using Plank´s distribution law to get the spectral blackbody emissive power.

$E=\frac{C_{1} }{\lambda ^{5}(exp(\frac{C_{2} }{\lambda T}) -1 )} =\frac{3.743x10^{8} }{4^{5}(exp(\frac{1.4387x10x^{4} }{4*477})-1) } =194.27W/m^{2} \mu m$

8 0

4 years ago

A wind turbine designer is considering installing a horizontal axis wind turbine at a location in Michigan. To increase the powe

bogdanovich [222]

Answer:

False ( B )

Explanation:

considering that the wind turbine is a horizontal axis turbine

Power generated/extracted by the turbine can be calculated as

P = n * 1/2 *<em> p</em> *Av^3

where: n = turbine efficiency

<em>p = air density </em>

<em> </em>A = πd^2 / 4

v = speed

From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine

3 0

3 years ago