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3 years ago

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How should the flight controls be held while taxiing a tricycle-gear equipped airplane with a left quartering tailwind?

1 answer:

AlexFokin [52]3 years ago

5 0

<span>The flight controls must be held with left aileron up and elevator neutral while taxiing a tricycle-gear equipped airplane with a left quartering tailwind. In aircraft, ailerons are placed on the trailing edge of each wing near the wingtips and can be moved up and down. So when the left aileron is up, the movement of the airplane moves to the left and turns the wheel in a counterclockwise direction while at the same time, the right aileron is down.
</span>

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Which term is defined as the ratio of the speed of light in a vacuum to the speed of light in the material it is passing through

KATRIN_1 [288]

Answer:

Index of refraction

Explanation:

or refractive index

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3 years ago

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sertanlavr [38]

Answer: sliding

Explanation:

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3 years ago

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What is the frequency of a radio wave with an energy of 3.686 × 10−24 j/photon?answer in units of hz?

Makovka662 [10]

To solve this problem, we have to use the formula:

E = h f

where E is total energy, h is Plancks constant 6.626x10^-34 J s, f is frequency

f = E / h

f = 3.686 × 10−24 J / (6.626x10^-34 J s)

<span>f = 5.56 x 10^9 Hz</span>

4 0

3 years ago

Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10

strojnjashka [21]

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

$n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42$

So, the refractive index of the medium is 1.42.

5 0

3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago