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inysia [295]
3 years ago
5

Please choose a specific type of stability or control surface (e.g., a canard) and explain how it is used, what it is used for,

and the pros and cons of the device or system.
Engineering
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

small forewing

<u>pro</u> : Can be used in place of tail plane configuration

<u>con</u> : Can be very complex and difficult to use

Explanation:

A canard is generally used to provide some form of stability to an unstable or semi stable system.

An example of a Canard is a small forewing placed in an aircraft that will help stabilize the aircraft when in motion( in air ). because an airplane is generally an unstable system on its own

<u>pro </u>: Can be used in place of tail plane configuration

<u>con</u> : Can be very complex and difficult to use

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Archy [21]

Answer:

The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.

Explanation:

Since we know that the force is the rate at which the momentum of an object changes.

Mathematically \overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}

The momentum of any body is defines as \overrightarrow{p}=mass\times \overrightarrow{v}

In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as

\overrightarrow{F}=\frac{m(\overrightarrow{v_{f}}-\overrightarrow{v_{i}})}{\Delta t}

\overrightarrow{F}=\frac{m(0-(-73.33)}{0.23}=\frac{1.8\times 73.33}{0.23}=573.88Pounds

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were the cord plugs in

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
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Answer:please see attached file

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4 years ago
What happens to the duty cycle for a GMAW Gun when 75Ar/25COzgas
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g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
4 years ago
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