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zepelin [54]
2 years ago
7

How does a rudder help maneuver an airplane?

Engineering
1 answer:
Nonamiya [84]2 years ago
4 0
It’s just helps it like
You might be interested in
In a four bar mechanism, L, is a fixed link; L2 is driver crank; L3 is coupler and L4 is follower crank. L=27 cm, L3 =5 cm and L
Lera25 [3.4K]

Answer:

L_1>32 cm

Explanation:

Given that

L_2=27 cm

L_3=5 cm

L_4=10 cm

Here shortest link is link3

To make the linkage as a double rocker mechanism we have to fix longest link that link is link1

So for double rocker mechanism

L_3+L_1>L_2+L_4

Now put the values

5+L_1>27+10

L_1>32 cm

It means that length of link1 should be greater than 32 cm.

6 0
3 years ago
Define volume flow rate Q of air flowing in a duct of area A with average velocity V
Shalnov [3]

Answer:

The volume flow rate of air is Q=A\times V

Explanation:

A random duct is shown in the below attached figure

The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time

Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters

From the attached figure we can see that

The volume of the prism that the flow occupies in 1 second equals

Volume=Area\times V=A\times V

Hence the volume flow rate is Q=V\times A

3 0
2 years ago
A punch must cut a hole 30mm diameter in a sheet of steel 2mm thick. The ultimate shear
Anettt [7]

2+2=3

4+5=7

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7 0
2 years ago
An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20 degree C to 1000 kPa and 300 degree C.
Oksana_A [137]

Answer:

work=281.4KJ/kg

Power=4Kw

Explanation:

Hi!

To solve follow the steps below!

1. Find the density of the air at the entrance using the equation for ideal gases

density=\frac{P}{RT}

where

P=pressure=120kPa

T=20C=293k

R= 0.287 kJ/(kg*K)= gas constant ideal for air

density=\frac{120}{(0.287)(293)}=1.43kg/m^3

2.find the mass flow by finding the product between the flow rate and the density

m=(density)(flow rate)

flow rate=10L/s=0.01m^3/s

m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s

3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow

Work

w=Cp(T1-T2)

Where

Cp= specific heat for air=1.005KJ/kgK

w=work

T1=inlet temperature=20C

T2=outlet temperature=300C

w=1.005(300-20)=281.4KJ/kg

Power

W=mw

W=(0.0143)(281.4KJ/kg)=4Kw

5 0
3 years ago
Question #4
inn [45]

Answer:

Deconstruction

Explanation:

8 0
3 years ago
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