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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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Once the magma found at location "E" cools and crystalizes, it will
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Turn into stone. this will cause erosion for magma to turn to stone
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3 years ago
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron
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Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62   distance traveled after collision .. crumpling of truck

Part a

V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s

Part b

vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s

7 0
3 years ago
Read 2 more answers
Heat transfer in liquids or gases that happens due to currents of hot and cold is called
yKpoI14uk [10]

Answer:

Convection

Explanation:

three types of heat transfer

Heat is transfered via solid material (conduction), liquids and gases (convection), and electromagnetical waves (radiation).

3 0
3 years ago
Read 2 more answers
Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test c
amid [387]

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F

Then, it means that the net force acting on the test charge has a magnitude of √2F.

7 0
3 years ago
What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af
sladkih [1.3K]

Answer:

L=0.0045\ kg-m^2/s

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, \omega=4300\ rpm=450.29\ rad/s

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

L=I\omega

Where I is moment of inertia

For sphere, I=\dfrac{2}{5}mr^2

L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s

So, the magnitude of the angular momentum of the sphere is 0.0045\ kg-m^2/s.

4 0
2 years ago
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