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3 years ago

In the engineering design and prototyping process, what is the advantage of drawings and symbols over written descriptions?

Engineering

1 answer:

MrMuchimi3 years ago

4 0

The advantages that can be associated to

drawings and symbols over written descriptions in engineering design and prototyping process are;

Communicate design ideas as well as technical information to engineers.

Symbols and drawings can be universal which means it is easy to interpret any where by professionals.

An engineering drawing serves as complex dimensional object and symbol use by engineer to communicate.

Drawings and symbols makes it easier to communicate design ideas and technical information to engineers and and how the process will go.

Therefore, drawings and symbols is universal to all engineer unlike written one.

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Sorry here is my second one

lesya692 [45]

Answer:

brainlyest if it right and sorry if it was wrong

Explanation:

1:a

2:b

3:a

4:a

5:c

6:a

4 0

3 years ago

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin

Leni [432]

Answer:

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^{-1}0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^{-1}0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

6 0

3 years ago

How does a carburetor work?

Pachacha [2.7K]

Answer:

Air flows into the top of the carburetor from the car's air intake, passing through a filter that cleans it of debris. When the throttle is open, more air and fuel flows to the cylinders so the engine produces more power and the car goes faster. The mixture of air and fuel flows down into the cylinders.

Explanation:

4 0

4 years ago

The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable

Fittoniya [83]

Answer:

a) 35%

b) yes it can be improved by moving the tray near the top

Tray should be located ( 1 to 2 meters below surface )

max removal efficiency ≈ 70%

c) The maximum removal will drop as the particle settling velocity = 0.5 m/h

Explanation:

Given data:

flow rate = 8000 m^3/d

Detention time = 1h

depth = 3m

Full length movable horizontal tray : 1m below surface

a) Determine percent removal of particles having a settling velocity of 1m/h

velocity of critical sized particle to be removed = Depth / Detention time

= 3 / 1 = 3m/h

The percent removal of particles having a settling velocity of 1m/h ≈ 35%

b) Determine if the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray and the maximum removal efficiency

The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the

Total Maximum removal efficiency

= percent removal $_{above}$ + percent removal $_{below}$