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Ludmilka [50]
3 years ago
7

Leading edge flaps can be used to decrease (or eliminate) the leading edge suction peak at a desired lift coefficient. When airf

oils are designed for cruise performance, however, a better strategy is to design an airfoil that produces the correct lift with no suction peak using a cambered airfoil (i.e. without including leading edge flaps). To see that this is possible, we will consider the NACA 44XX airfoils. Also, p is the location of the maximum camber and is second digit/10. Apply thin airfoil theory to answer the following questions: (a) Determine the angle of zero lift for the 44XX airfoils (b) Determine the angle at which the suction peak is eliminated. We will call this the design angle of attack for the 44XX airfoils (c) What is the design lift coefficient for the 44XX airfoils (i.e. the lift coefficient at the design angle of attack)?
Physics
1 answer:
Alik [6]3 years ago
5 0

Answer:

An investigation is made to determine the performance of simple thin airfoils in the slightly supersonic flow region with the aid of the nonlinear transonic theory first developed by von Kármán[1]. Expressions for the pressure coefficient across an oblique shock and a Prandtl-Meyer expansion are developed in terms of a transonic similarity parameter. Aerodynamic coefficients are calculated in similarity form for the flat plate and asymmetric wedge airfoils, and curves are plotted. Sample curves for a flat plate and a specific asymmetric wedge are plotted on the usual coordinate grid of Cl, Cd,andCmc/4versus angle of attack and Cl versus Mach Number to illustrate the apparent features of nonlinear flow.

Explanation:

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You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica
Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A} , where:

l is the length

m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3

We divide the first equation by the second equation to get:

\frac{m}{R} = \frac{d A^2}{\rho}

A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8}   \ m^2

Using this Area, we find the diameter of the wire:

D = \sqrt{\frac{4A}{\pi}}

D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}

D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm

To find the length, we multiply the two equations stated initially:

mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}

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8 0
3 years ago
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So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)
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16. Calculate the mass in kg of an object that is being accelerated 12 m/s with a force of 654 N.
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slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a
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Answer:

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Explanation:

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v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}

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