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2 years ago

What input is required from ADC to allow the INS to calculate W/V?

Engineering

1 answer:

Kazeer [188]2 years ago

4 0

TAS is an important input which is required from ADC to allow the INS calculate W/V.

INS is abbreviation for Inertial Navigation System and it can be defined as a navigation device that makes use of motion sensors, a computer, and rotation sensors, so as to continuously calculate by dead reckoning the velocity, position, and orientation of a moving object.

In the Aviation and Engineering filed, True Airspeed (TAS) is an important input which is required from ADC to allow the Inertial Navigation System (INS) calculate W/V.

Read more on Inertial Navigation System here: brainly.com/question/26052911

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A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain. The h

sertanlavr [38]

Answer:

Explanation:

Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)

fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)

vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)

BFFS = 50+5, BFFS =55 (given) fLW= 6.6

TLC=6+3=9 fLC= 0.65

fM= 0.0

fA= 1.0

FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)

Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)

After: fA= 3.0

FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)

Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln

8 0

3 years ago

A thin-film gold conductor for an integrated circuit in a satellite application is deposited from a vapor, and the deposited gol

ahrayia [7]

Answer:

Explanation:

Considering the relation of the equilibrium vacancy concentration ;

nv/N = exp (-ΔHv/KT)

Where T is the temperature at which the vacancy sites are formed

K = Boltzmaan constant

ΔHv = enthalpy of vacancy formation

Rearranging the equation and expressing in term of the temperature and plugging the values given to get the temperature. The detailed steps is as shown in the attached file

5 0

3 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz

Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s and ρ= 850 kg/m³

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ = ρv

=850 x 0.00062

= 0.527 kg/m·s

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

$Q = \dfrac{\Delta P \pi D^4}{128 \mu L}$

$Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}$

Q = 5.06 x 10⁻⁸ m³/s

4 0

3 years ago

A 1.7 cm thick bar of soap is floating in water, with 1.1 cm of the bar underwater. Bath oil with a density of 890.0 kg/m{eq}^3

PIT_PIT [208]

Answer:

The height of the oil on the side of the bar when the soap is floating in only the oil is 1.236 cm

Explanation:

The water level on the bar soap = 1.1 m mark

Therefore, the proportion of the bar soap that is under the water is given by the relation;

Volume of bar soap = LW1.7

Volume under water = LW1.1

Volume floating = LW0.6

The relative density of the bar soap = Density of bar soap/(Density of water)

= m/LW1.7/(m/LW1.1) = 1.1/1.7

Given that the oil density = 890 kg/m³

Relative density of the oil to water = Density of the oil/(Density of water)

Relative density of the oil to water = 890/1000 = 0.89

Therefore, relative density of the bar soap to the relative density of the oil = (1.1/1.7)/0.89

Relative density of the bar soap to the oil = (1.1/0.89/1.7) = 1.236/1.7

Given that the relative density of the bar soap to the oil = Density of bar soap/(Density of oil) = m/LW1.7/(m/LWX) = X/1.7 = 1.236/1.7

Where:

X = The height of the oil on the side of the bar when the soap is floating in only the oil

Therefore;

X = 1.236 cm.

3 0

4 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago