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3 years ago

Three possible career opportunities in embedded systems engineering

Engineering

1 answer:

timama [110]3 years ago

3 0

Answer:Embedded Software Engineer (firmware)

- System Software Engineer (kernal & RTOS)

- Application Software Engineer (device drivers)

- Software Test Engineer.

- Embedded Hardware Engineer.

- Embedded System Trainer.

- Marketing & Sales Executive.

Explanation:

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What are the two main what are the two main concerns in the research of fluid power efficiency?

Galina-37 [17]

Answer:

The correct option is;

Materials and Components

Explanation:

The efficiency of fluid power is influenced by the components and the materials used to deliver the power of the fluid as such fluid power control are focused on

1) Advances in fluid power

2) Making use of the advantages

3) Making use of the other externally available technological advantages

4) Giving allowance for disadvantages

Areas of interest in advances in fluid power are;

a. Computer optimized flow

b. The use of new and improved materials/coatings

c. The use of components that save energy, such as intelligent supply pressure adapting systems

3 0

2 years ago

Read 2 more answers

Argue the importance to society of incorporating green building into an engineer's designs, with at least two examples.

stellarik [79]

Answer:

Green Design is your answer

Explanation:

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2 years ago

Read 2 more answers

The pads are 200mm long, 150 mm wide and thickness equal to 12mm. 1- Determine the average shear strain in the rubber if the for

lord [1]

Answer:

a) 0.3

b) 3.6 mm

Explanation:

Given

Length of the pads, l = 200 mm = 0.2 m

Width of the pads, b = 150 mm = 0.15 m

Thickness of the pads, t = 12 mm = 0.012 m

Force on the rubber, P = 15 kN

Shear modulus on the rubber, G = 830 GPa

The average shear strain can be gotten by

τ(average) = (P/2) / bl

τ(average) = (15/2) / (0.15 * 0.2)

τ(average) = 7.5 / 0.03

τ(average) = 250 kPa

γ(average) = τ(average) / G

γ(average) = 250 kPa / 830 kPa

γ(average) = 0.3

horizontal displacement,

δ = γ(average) * t

δ = 0.3 * 12

δ = 3.6 mm

5 0

3 years ago

A 30 mm thick AISI 1020 steel plate is sandwiched between two 10 mm thick 2024-T3 aluminum plates and compressed with a bolt and

denis-greek [22]

Answer:

275 MPa

Explanation:

Regardless of what it is holding, the stiffness of a bolt depends on its own material properties and geometry.

The stiffness is:

$k = E * \frac{A}{l}$

I assume this one is made of steel, because regular bolts are steel.

The Young's modulus for steel is E = 210 GPa

The longitude is given. (But note that in a real application you have to consider the length up to the nut.)

The section is (using the nominal diameter of 10 mm)

$A = \frac{\pi * d^2}{4} = \frac{\pi * 0.01^2}{4} = 7.85e-5 m^2$

Then:

$k = 2.1e11 * \frac{7.85e-5}{0.06} = 275e6 Pa = 275 MPa$

5 0

3 years ago

A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

$F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}$

Substituting figures in the equation we have:

$= \frac{1}{1+0.1(2.5-1)+0(2-1)}$

= 0.75

Let's now calculate equivalent flow rate of the car using:

$Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}$

$= \frac{7500}{0.9*4*0.75*1}$

= 2777.7 pc/h/en

Calculating for traffic density, we have:

$D = \frac{Vp}{FFS}$

$D = \frac{2777.7}{70}$

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0

3 years ago