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2 years ago

Select four geometric shapes that are commonly used by engineers in their designs.

Engineering

1 answer:

stealth61 [152]2 years ago

3 0

Answer:

Arc ,sphere ,line ,prism

Explanation:

Arc is used for semicircular or short term projects
Sphere is used for design of spherical surfaces or projects
Prism is used for creating buliding designs
Lines are especially used for creating roads, flyover design

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The minimum fresh air requirement of a residential building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1

Natalka [10]

We know that

$A=200m^2\\h=2.7m\\\upsilon= 5.5m/s\\\%_{air} = 35%$

So, the volume of the entire building is

$V=2.7*200 = 540m^3$

The flow capacity of the fan

$\dot{V} = \frac{0.35*540}{60}$

$\dot{V} = 3.15m^3/min$

As $1L=10^{-3}m^3,$

$\dot{V}=3150L/min$

For the other part we know

$\dot{V}=\frac{\pi d^2}{4}V$

The diameter is,

$d=\sqrt{\frac{4\dot{V}}{\pi V}}$

$d=\sqrt{\frac{4*3.15}{\pi* 5.5* 60}}$

<em>**Note 60 is for the minutes</em>

$d= 0.1101m$

<em />

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3 years ago

Blank is an art form in which the focus is to make life and society better

Ganezh [65]

Engineering is an art form in which the focus is to make life and society better

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2 years ago

Find the requested quantities for the circuit. We used the mesh-current method to identify the meshes. We then identified the me

Irina18 [472]

Answer:

Explanation:

The image that is supposed to be attached to the question is displayed in the diagram below.

Applying Nodal Analysis at node 1;

$\dfrac{V_o -50}{12.5*10^{-3}} + \dfrac{V_o}{50*10^3}+\dfrac{V_o-7500 \ in}{10*10^3}=0$

where;

$in = \dfrac{V_o}{50*10^3}$ (from the circuit)

= $\dfrac{V_o-50}{12.5}+\dfrac{V_o}{50} + V_o -\dfrac{7500 *V_o }{\frac{50*10^3}{10}}=0$

= $V_o [ \dfrac{1}{12.5}+\dfrac{1}{50}+\dfrac{1}{10}-\dfrac{75}{500}] = \dfrac{50}{12.5}$

= $V_o[ \dfrac{500*500+12.5*5000+12.5*5000*5-75*12.5*500}{12.5*50*10*500}]= \dfrac{50}{12.5}$

= $V_o = 80 \ volts$

$in = \frac{80}{50*10^3}= 1.6 mA \\ \\ 7500*in = 120 volts \\ \\ I = \frac{120-80}{10(10^3} =4*10^{-3} Amps \\ \\ \\ \\ P_{generated} = 75000*in*I \\ \\ P_{generated} = 120*4*10^{-3} \\ \\ P_{generated} = 480 \ MW$

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4 years ago

IDENTIFY THE POINTER READING FOR EACH SCALE. COMPUTE THE READING USING THE ASSIGNED RANGE MULTIPLIER FOR OHMMETER SCALE.

daser333 [38]

Answer:

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Lesson 2 - T.L.E Learning Module

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3 years ago

Read 2 more answers

From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displaye

tia_tia [17]

Answer:

(a) Precipitation hardening - 1, 2, 4

(b) Dispersion strengthening - 1, 3, 5

Explanation:

The correct options for each are shown as follows:

Precipitation hardening

From the first statement; Dislocation movement is limited by precipitated particles. This resulted in an expansion in hardness and rigidity. Precipitates particles are separated out from the framework after heat treatment.

The aging process occurs in the second statement; because it speaks volumes on how heated solutions are treated with alloys above raised elevated temperature. As such when aging increases, there exists a decrease in the hardness of the alloy.

Also, for the third option for precipitation hardening; This cycle includes the application of heat the alloy (amalgam) to a raised temperature, maintaining such temperature for an extended period of time. This temperature relies upon alloying components. e.g. Heating of steel underneath eutectic temperature. Subsequent to heating, the alloy is extinguished and immersed in water.

Dispersion strengthening

Here: The effect of hearting is not significant to the hardness of alloys hardening by the method in statement 3.

In statement 5: The process only involves the dispersion of particles and not the application of heat.

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3 years ago