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2 years ago

In a morphological matrix, which of the following contains the parameters that are essential to a design?

Engineering

1 answer:

choli [55]2 years ago

8 0

In a morphological matrix, the parameters that are essential for a design are in the left column.

<h3>What is a morphological matrix?</h3>

The morphological matrix is a matrix where columns and rows represent the various parameters for solving a problem. The first column is used for the characteristics relevant to the problem; the horizontal lines are filled with possibilities for each of these parameters.

With this information, we can conclude that in a morphological matrix, the parameters that are essential for a project are in the left column.

Learn more about morphological matrix in brainly.com/question/21120930

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Would be much appreciated if someone could help with this will give brainiest.

Mashcka [7]

Answer: both mm and inches on each dimension in a sketch (with the main dimension in one format and the other in brackets below it), in the way you can have dual dimensions shown when detailing an idw view.

personally think it would look a mess/cluttered with even more text all over the sketch environment, but everyone's differenent.

If it's any help - you know you can enter dimensions in either format? If you're working in mm you can still dimension a line and type "2in" and vice-versa. Probably know this already, but no harm saying it, just in case.

You can enter the units directly in or mm and Inventor will convert to current document settings (which you can change - maybe someone can come up with a simple toggle icon to toggle the document settings). Tools>Document Settings>Units

Unlike SolidWorks when you edit the dimension the original entry shows in the dialog box so it makes it easy to keep track of different units even if they aren't always displayed. (SWx does the conversion or equation and then that is what you get.)

I work quite a bit in inch and metric and combination (ex metric frame motor on inch machine) and it doesn't seem to be a real difficulty to me.

4 0

2 years ago

There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.

sergejj [24]

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

$C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})$

$C_{1}=1.2606\times10^{-5}$ $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

$C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})$

$C_{2}=3.334\times10^{-5}$ $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

$C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})$

$C_{3}=1.66\times10^{-5}$ $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

5 0

2 years ago

A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

3 0

2 years ago

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:

Yakvenalex [24]

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to $\large\mathsf{\red{\underline{Pascal(pa)}}}$

7 0

2 years ago

Read 2 more answers

A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum

antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

Density = 61 veh/km.

Speed = 59 km/hr.

Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

$q_{max}=\frac{V_f \times K_i}{4}$

<u>Where:</u>

$V_f$ is the free flow speed.

$K_i$ is the Jam density.

In order to calculate the free flow speed, we would use this formula:

$V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr$

Substituting the parameters into the model, we have:

$q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}$

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

6 0

1 year ago