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3 years ago

Why is there an Engineering Process?

Engineering

2 answers:

Inga [223]3 years ago

8 0

There is a engineering process to plan your work out better!

Studentka2010 [4]3 years ago

7 0

Answer:

well if there wasnt a process tony starks suit wouldn’t work properly, and it would probably fall apart. it is to make sure everything works exactly as they planned, our civilization has gotten this far because we created rules on how to build stuff.

Explanation:

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Which of the following was a sustainable power source used during the Middle Ages?

spin [16.1K]

The answer is windmill it was used for electricity

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3 years ago

Read 2 more answers

How many grams of perchloric acid, HClO4, are contained in 37.6 g of 70.5 wt% aqueous perchloric acid? How many grams of water a

kap26 [50]

Answer :

The mass of perchloric acid is, 26.5 grams.

The mass of water in the same solution is, 11.1 grams

Explanation :

As we are given that 70.5 wt % aqueous perchloric acid that means 70.5 grams of perchloric acid present in 100 grams of solution.

Now we have to determine the mass of perchloric acid in 37.6 grams of aqueous perchloric acid.

As, 100 grams of aqueous perchloric acid (solution) contains 70.5 grams of perchloric acid.

So, 37.6 grams of aqueous perchloric acid (solution) contains $\frac{37.6}{100}\times 70.5=26.5$ grams of perchloric acid.

Thus, the mass of perchloric acid is, 26.5 grams.

Now we have to determine the mass of water are in the same solution.

Total mass of solution = 37.6 g

Mass of perchloric acid = 26.5 g

Mass of water = Total mass of solution - Mass of perchloric acid

Mass of water = 37.6 g - 26.5 g

Mass of water = 11.1 g

Thus, the mass of water in the same solution is, 11.1 grams

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3 years ago

What setting do i dry my jordan max arua 2s on in the dryer <br>will mark brainliest

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3 years ago

A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir

Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

= 500 - 2(20)

= 460 mm

So, internal radius, r = 230 mm

= 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

= $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

= 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

$=2 \pi (0.23)^2$

$=0.3324 \ m^2$

$F=\rho g A h$

$=7200 \times 9.81 \times 0.3324 \times 0.3$

= 7043.42 N

3 0

2 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

3 years ago