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3 years ago

WHAT IS THE EFFECT OF ICE ACCRETION ON THE LONGITUDINAL STABILITY OF AN AIRCRAFT?

Engineering

1 answer:

soldier1979 [14.2K]3 years ago

5 0

Answer:

The major effects of ice accretion on the aircraft is that it disturbs the flow of air and effects the aircraft's performance.

Explanation:

The ice accretion effects the longitudinal stability of an aircraft as:

1. The accumulation of ice on the tail of an aircraft results in the reduction the longitudinal stability and the elevator's efficacy.

2. When the flap is deflected at $10^{\circ}$ with no power there is an increase in the longitudinal velocity.

3. When the angle of attack is higher close to the stall where separation occurs in the early stages of flow, the effect of ice accretion are of importance.

4. When the situation involves no flap at reduced power setting results in the decrease in aircraft's longitudinal stability an increase in change in coefficient of pitching moment with attack angle.

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What combustion after effects do actuators work to control

Rufina [12.5K]

Answer: I would help you but I don’t know the answer, so sorry

4 0

3 years ago

Read 2 more answers

What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF

exis [7]

Answer:

(d) 2 pF

Explanation: the charge on capacitor is given by the expression

Q=CV

where Q=charge

C=capacitance

V=voltage across the plate of the capacitor

here we have given Q=500 pF, V=250 volt

using this formula C= $\frac{Q}{V}$

=500× $10^{-12}$ × $\frac{1}{250}$

=2× $10^{-12}$

=2 pF

3 0

3 years ago

How high a building could fire hoses effectively spray from the ground? Fire hose pressures are around 1 MPa. (It is also said t

Mrac [35]

Answer:

$z_{2} = 91.640\,m$

Explanation:

The phenomenon can be modelled after the Bernoulli's Principle, in which the sum of heads related to pressure and kinetic energy on ground level is equal to the head related to gravity.

$\frac{P_{1}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}= z_{2}+\frac{P_{2}}{\rho\cdot g}$

The velocity of water delivered by the fire hose is:

$v_{1} = \frac{(300\,\frac{gal}{min} )\cdot(\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot (0.3\,m)^{2}}$

$v_{1} = 0.267\,\frac{m}{s}$

The maximum height is cleared in the Bernoulli's equation:

$z_{2}= \frac{P_{1}-P_{2}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}$

$z_{2}= \frac{1\times 10^{6}\,Pa-101.325\times 10^{3}\,Pa}{(1000\,\frac{kg}{m^{3}} )\cdot(9.807\,\frac{m}{s^{2}} )} + \frac{(0.267\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}$

$z_{2} = 91.640\,m$

7 0

3 years ago

(Gas Mileage) Drivers are concerned with the mileage their automobiles get. One driver has kept track of several trips by record

Effectus [21]

Answer:

import java.util.*;

public class Main {

public static void main(String[] args) {

double milesPerGallon = 0;

int totalMiles = 0;

int totalGallons = 0;

double totalMPG = 0;

Scanner input = new Scanner(System.in);

while(true){

System.out.print("Enter the miles driven: ");

int miles = input.nextInt();

if(miles <= 0)

break;

else{

System.out.print("Enter the gallons used: ");

int gallons = input.nextInt();

totalMiles += miles;

totalGallons += gallons;

milesPerGallon = (double) miles/gallons;

totalMPG = (double) totalMiles / totalGallons;

System.out.printf("Miles per gallon for this trip is: %.1f\n", milesPerGallon);

System.out.printf("Total miles per gallon is: %.1f\n", totalMPG);

}

Explanation:

Initialize the variables

Create a while loop that iterates until the specified condition is met inside the loop

Inside the loop, ask the user to enter the miles. If the miles is less than or equal to 0, stop the loop. Otherwise, for each trip do the following: Ask the user to enter the gallons. Add the miles and gallons to totalMiles and totalGallons respectively. Calculate the milesPerGallon (divide miles by gallons). Calculate the totalMPG (divide totalMiles by totalGallons). Print the miles per gallon and total miles per gallon.

6 0

2 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5

IgorC [24]

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

$\epsilon = \deltaL /L$ (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

$\sigma = E/ \epsilon$ (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

$L= \delta L/ \epsilon$

$\epsilon =\sigma /E$

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

$\sigma=F/A$