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2 years ago

13

The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu

ral frequency is _____.

1 answer:

BartSMP [9]2 years ago

8 0

Answer:

hello your question is incomplete attached below is the missing part

answer : short period oscillations frequency = 0.063 rad / sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= $( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns} ) = 0$

where :

$w_{np} = Natural frequency of plugiod oscillation$

$\alpha _{p} = damping ratio of plugiod oscilations$

comparing the general form with the given equation

$w^{2} _{np}$ = 18.2329

$w^{2} _{ns} = 0.003969$

hence the short period oscillation frequency ( $w_{ns}$ ) = 0.063 rad/sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

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3 years ago

You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.

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Answer:

Explanation:

a)Magnitude = $\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }$

84= $\sqrt{(0- (-67))^{2} + (x-0)^{2} }$

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b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.

To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.

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c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - $Tan^{-1}(67/130.67)degrees$ i.e 152.85 degrees from the +ve x-axis.

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3 years ago

a whistle you use to call your hunting dog has a frequency of 21 khz, but your dog is ignoring it. you suspect the whistle may n

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$f = f_0 (\frac{v-v_0}{v})$

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$v$ = Velocity of the speed

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$v = 340m/s \rightarrow \text{Speed of sound}$

$f = 20kHz \rightarrow \text{Apparent frequency}$

$f_0 = 21kHz \rightarrow \text{Original frequency}$

Using the previous equation,

$f = f_0 (\frac{v-v_0}{v})$

Rearrange to find the velocity of the observer

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3 years ago

A small plane flies 37.0 km in a direction 45° north of east and then flies 28.0 km in a direction 25° north of east.

Karo-lina-s [1.5K]

Answer:

d= 64.1 km θ = 36.4º

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If we choose to put our origin at the initial point of trajectory, we have that (x₀, y₀) = (0, 0)

In order to find the position of the plane after finishing the flight, we need to find its final coordinates (x₁, y₁).

In order to get x₁, we need to add the x-coordinate after flying 45º north of east, and the Δx after completing the flight in a direction 25º of east, that we can find applying trigonometry, as follows:

x₁ = 37.0 km * cos 45º + 28.0 km* cos 25º = 51.6 Km

Appying the same considerations for the y-coordinate, we have:

y₁ = 37.0 km * sin 45º + 28.0 km* sin 25º = 38.0 km

Now, as the initial position coincides with the origin, the distance in a straight line from this point to the origin, is just the hypotenuse of the triangle determined by the coordinates (x₁, y₁) and (0,0), as follows:

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The geographic direction of the displacement vector (which coincides in magnitude with the distance we have just found), is just the angle that this distance forms with the east axis, that we can find getting the tangent of this angle as follows:

tg θ = $\frac{y1}{x1} = \frac{38km}{56.1km} =0.736$

⇒ θ = 36.4º North of East (counterclockwise from the east axis).

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2 years ago