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3 years ago

How is air pressure affected by the shape of an aircraft wing

Engineering

1 answer:

oksano4ka [1.4K]3 years ago

3 0

Answer:

Airplanes' wings are curved on top and flatter on the bottom. That shape makes air flow over the top faster than under the bottom. As a result, less air pressure is on top of the wing. This lower pressure makes the wing, and the airplane it's attached to, move up.

Explanation:

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Explain how use of EGR is effective in reducing NOx emissions 4. In most locations throughout the U.S., the octane number of reg

TiliK225 [7]

Answer:please see attached file

Explanation:

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4 years ago

Heather is troubleshooting a computer at her worksite. She has interviewed the computer’s user and is currently trying to reprod

Luba_88 [7]

Answer:

The correct option is A

Explanation:

Heather is trying to establish a theory of probable cause. In this step of the troubleshooting process, the person troubleshooting questions the obvious and then test the theory or response given by the user to really determine the cause. Once confirmation of this theory has been achieved, the troubleshooter then tries to establish a resolution to the problem. However in the event whereby the theory is not confirmed, the troubleshooter then tries to establish a new theory.

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4 years ago

2.1 What is the minimum number of pins required for a so-called dual-op-amp IC package, one containing two op amps? What is the

cupoosta [38]

Answer:

8 for dual-op-amp package, and 14 for quad-op-amp

Explanation;

This is because every op-amp has 2 input terminal 4 pns

So one output terminal that is 2 pins which are required for power

and the same for a minumum number of pins required by quad op amp which is 14

5 0

3 years ago

Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

Sholpan [36]

Answer:

Frequency = $3.5294\times 10^{14}s^{-1}$

Wavenumber = $1.1765\times 10^6m^{-1}$

Energy = $2.3365\times 10^{-19}J$

Energy = 1.4579 eV

Energy = $2.3365\times 10^{-12}erg$

Explanation:

As we are given the wavelength = 850 nm

conversion used : $(1nm=10^{-9}m)$

So, wavelength is $850\times 10^{-9}m$

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where, c is the speed of light having value = $3\times 10^8m/s$

So, Frequency is:

$Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}$

$Frequency=3.5294\times 10^{14}s^{-1}$

Wavenumber is the reciprocal of wavelength.

So,

$Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}$

$Wavenumber=1.1765\times 10^6m^{-1}$

Also,

$Energy=h\times frequency$

where, h is Plank's constant having value as $6.62\times 10^{-34}J.s$

So,

$Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})$

$Energy=2.3365\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=1.4579eV$

Also,

$1J=10^7erg$

So,

$Energy=(2.3365\times 10^{-19})\times 10^7erg$

$Energy=2.3365\times 10^{-12}erg$

5 0

3 years ago

A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the g

olga55 [171]

Answer:

$P_2_{abs}=160\ psia$ (absolute).

Explanation:

Given that

Pressure ratio r

r=8

$r=\dfrac{P_2_{abs}}{P_1_{abs}}$

$8=\dfrac{P_2_{abs}}{P_1_{abs}}$ -----1

P₁(gauge) = 5.5 psig

We know that

Absolute pressure = Atmospheric pressure + Gauge pressure

Given that

Atmospheric pressure = 14.5 lbf/in²

P₁(abs) = 14.5 + 5.5 psia

P₁(abs) =20 psia

Now by putting the values in the above equation 1

$8=\dfrac{P_2_{abs}}{20}$

$P_2_{abs}=8\times 20\ psia$

$P_2_{abs}=160\ psia$

Therefore the exit gas pressure will be 160 psia (absolute).

7 0

3 years ago