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Lesechka [4]
3 years ago
6

A hard real-time system has been developed to for a fly-by-wire aviation system. It has sensors for the following pilot interfac

e systems: System Sampling Frequency (Hz) CPU time required (ms) Yoke 20 5 Rudder pedals 15 2 Throttle 10 1 Overhead of the hard real-time system on the architecture selected for the plane is .21. Can these tasks be scheduled? Show your work. Regardless of whether or not the tasks are schedulable, what should a designer do if they are not.
Engineering
1 answer:
Jlenok [28]3 years ago
4 0

Answer:

Answer for the question:

A hard real-time system has been developed to for a fly-by-wire aviation system. It has sensors for the following pilot interface systems: System Sampling Frequency (Hz) CPU time required (ms) Yoke 20 5 Rudder pedals 15 2 Throttle 10 1 Overhead of the hard real-time system on the architecture selected for the plane is .21. Can these tasks be scheduled? Show your work. Regardless of whether or not the tasks are schedulable ,  what should a designer do if they are not.

is given below.

Explanation:

Yea. Scheduling of these tasks can be done  in below order:

  1. Throttle
  2. Rudder pedals.
  3. Yoke.

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A lower coefficient of thermal conductivity along with a higher coefficient of thermal expansion and higher electrical resistanc
HACTEHA [7]

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stainless steel

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5 0
3 years ago
A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6
amm1812

1) a_x=4.287+2.772x\\a_y=-5.579+2.772y

2) 8.418

Explanation:

1)

The two components of the velocity field in x and y for the field in this problem are:

u=1.85+2.05x+0.656y

v=0.754-2.18x-2.05y

The x-component and y-component of the acceleration field can be found using the following equations:

a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}

a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}

The derivatives in this problem are:

\frac{du}{dt}=0

\frac{dv}{dt}=0

\frac{du}{dx}=2.05

\frac{du}{dy}=0.656

\frac{dv}{dx}=-2.18

\frac{dv}{dy}=-2.05

Substituting, we find:

a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x

And

a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y

2)

In this part of the problem, we want to find the acceleration at the point

(x,y) = (-1,5)

So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

a_x=4.287+2.772x\\a_y=-5.579+2.772y

And so we find:

a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281

And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:

a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418

4 0
4 years ago
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