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anyanavicka [17]
4 years ago
8

What is the terms for this definition? The unique layout and focus of each type of promotion

Engineering
1 answer:
saveliy_v [14]4 years ago
7 0

Answer:

Explanation:

The term for this definition is format.

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The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Ilya [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

8 0
3 years ago
Which size of impurity atom, smaller impurity atom or larger impurity atom, when located near a dislocation, will nullify some o
svp [43]

Answer:

Smaller impurity atom will nullify some of the compressive strain of a dislocation in a crystal. Because, smaller impurity atoms located near a dislocation creates tensile strain on atoms around it thereby partially nullifying compressive strain at the dislocation.

4 0
3 years ago
a hallow steel tube 3.5m long has external diameter of 120mm. in order to determine the internal diameter the tube was subjected
Dafna1 [17]

Answer:

Please check the photo (solve)

4 0
3 years ago
A steel cylindrical sample was subjected to a tensile test. The yield load was 2100N. The maximum load was 3400N and the failure
Misha Larkins [42]

Answer:

initial diameter of the sample is 2.95 mm

Explanation:

given data

yield load = 2100 N

maximum load = 3400 N

failure load = 2350 N

ultimate engineering stress = 497.4 MPa = 497 × 10^{6} N/m²

to find out

What was the initial diameter of the sample in mm

solution

we will apply here ultimate engineering stress formula that is express as

ultimate engineering stress = \frac{Pmax}{A}    ...............1

here A is area and P max is maximum load applied

so area = \frac{\pi }{4} d^2

here d is initial diameter

so put all value in equation 1

497 × 10^{6}  = \frac{3400}{\frac{\pi }{4} d^2}

solve it we get d

d = 2.95 × 10^{-3} m

so initial diameter of the sample is 2.95 mm

7 0
4 years ago
Which claim does president Kennedy make in speech university rice ?
mafiozo [28]

Answer:  The United States must lead the space race to prevent future wars.

Explanation: Hope this helps

4 0
3 years ago
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