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2 years ago

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

Engineering

1 answer:

oksano4ka [1.4K]2 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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tigry1 [53]

Is the control group

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3 years ago

Two soils are fully saturated with liquid (no gas present) and the soils have the same void ratio. One soil is saturated with wa

mariarad [96]

Answer:

water sample have more water content

Explanation:

given data

soil 1 is saturated with water

unit weight of water = 1 g/cm³

soil 2 is saturated with alcohol

unit weight of alcohol = 0.8 g/cm³

solution

we get here water content that is express as

water content = $S_s \times \gamma _w$ ....................1

here soil is full saturated so $S_s$ is 100% in both case

so put here value for water

water content = 100 % × 1

water content = 1 g

and

now we get for alcohol that is

water content = 100 % × 0.8

water content = 0.8 g

so here water sample have more water content

5 0

3 years ago

What is the number model for three families live in the same apartment

Gnesinka [82]

Three families live in the same apartment building. They decided to share a giant 220-ounce container.

3 0

2 years ago

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Disk A has a mass of 8 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3.5 kg and is initially at

garri49 [273]

Answer:

A. αa= 9.375 rad/s^2, αb= 28.57 rad/s^2

B. ωa=251rpm, ωb=333rpm

Explanation:

Mass of disk A, m1= 8kg

Mass of disk B, m2= 3.5kg

The initial angular velocity of disk A= 360rpm

The horizontal force applied = 20N

The coefficient of friction μk = 0.15

While slipping occurs, a frictional force is applied to disk A and disk B

T= 1/2Mar^2a

T=1/2 (8kg) (0.08)^2

T= 0.0256kg-m^2

N=P=20N

F= μN

F= 0.15 × 20

F=3N

We have

Summation Ma= Summation(Ma) eff

Fra= Iaαa

(3N)(0.08m)= (0.0256kg-m^2)α

αa= 9.375 rad/s^2

The angular acceleration at disk A is αa= 9.375 rad/s^2 is acting in the anti-clockwise direction.

For Disk B,

T= 1/2Mar^2a

T= 1/2(3.5) (0.06)^2

=0.0063kg-m^2

We have,

Summation Mb= Summation(Mb) eff

Frb= Ibαb

(3N)(0.06m)= (0.0063kg-m^2) α

αb= 28.57 rad/s^2

B) ( ωa)o= 360rpm(2 pi/60)

= 1 pi rad/s

The disk will stop sliding where

ωara=ωbrb

(ωao-at)ra=αtr

(12pi-9.375t) (0.08)=28.57t(0.06)

(37.704-9.375t)(0.08)= 1.7142t

3.01632-0.75t= 1.7142t

t=1.22s

Now,

ωa=(ωa)o- t

12pi - 9.375(1.22)

37.704-11.4375

=26.267 rad/s

26.267× (60/2pi)

= 250.80

251rpm

The angular velocity at a, ω= 251rpm

Now,

ωb= αbt

=28.57(1.22)

=34.856rad/s

34.856rad/s × (60/2pi)

=332.807

= 333rpm

Therefore the angular velocity at b ω=333rpm

4 0

3 years ago

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The author uses the example of the flameless candle to illustrate that

krek1111 [17]

Answer:

I don't necessarily know the context of the text you are talking about, but a flameless candle could represent how there no more light in the author's life, which could represent a lack of happiness or a loss. Hope this helps!

Explanation:

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2 years ago

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