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Lina20 [59]
3 years ago
6

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

Engineering
1 answer:
Mashcka [7]3 years ago
4 0

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.

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Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h
yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we  will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A  = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ;  B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

8 0
3 years ago
Which of the following best reflects a shield system?
Dafna11 [192]
C, because a narrow structure evacuation below surface ground isn’t the best and a structure holding forces and isn’t to do with the question at all and d doesn’t matter if there include away or not
8 0
3 years ago
An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are
Vika [28.1K]

This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle W_{net = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ^{Y-1 = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = ( T₂ / T₁ )^{\frac{1}{Y-1}

so we substitute

⇒ V₁ / V₂ = (  973 K / 303 K  )^{\frac{1}{1.4-1}

= (  3.21122  )^{\frac{1}{0.4}

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c)  the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂  = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

8 0
3 years ago
Create a variable pounds to store weight in pounds. Convert this to kilograms and assign the result to a variable kilos. The con
vodka [1.7K]

Answer:

>>pounds=13.2

>>kilos=pounds/2.2

Explanation:

Using Matlab to write the program, consider at any time when the weight in pounds is 13.2 lb, this variable of weight is created in MATLAB by typing >>pounds=13.2. To convert it from lb to Kg, we simply divide it by 2.2 hence the second command to created is kilos. For this, the output of the program will be 6 Kg.

5 0
3 years ago
An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f
Gelneren [198K]

Answer:

(a) Q=332 kvar and C=5.66 uF

(b) pf=0.90 lagging

Explanation:

Given Data:

P=600kW

V=12.47kV

f=60Hz

pf_{old} =0.75

pf_{new} =0.95

(a) Find the required kVAR rating of a capacitor

\alpha _{old}=cos^{-1}(0.75) =41.41°

\alpha _{new}=cos^{-1}(0.95) =18.19°

The required compensation reactive power can be found by

Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))

Q=600(tan(41.41) - tan(18.19))

Q=332 kvar

The corresponding capacitor value can be found by

C=Q/2\pi fV^{2}

C=332/2*\pi *60*12.47^{2}

C=5.66 uF

(b) calculate the resultant supply power factor

First convert the hp into kW

P_{mech} =250*746=186.5 kW

Find the electrical power (real power) of the motor

P_{elec} =P_{mech}/n

where n is the efficiency of the motor

P_{elec} =186.5/0.80=233.125 kW

The current in the motor is

I_{m} =(P/\*V*pf)

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

I_{m} =(233.125/12.47*0.85)

I_{m}=18.694+11.586j A

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

I_{load} =(600/12.47*0.75)

I_{load} =48.115-42.433j A

Now the supply current is the current flowing in the load plus the current flowing in the motor

I_{supply} =I_{m} + I_{load}

I_{supply}= (18.694+11.586)+(48.115-42.433)

I_{supply} =66.809-30.847j A

or in polar form

I_{supply} =73.58°

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

pf=cos(24.78)=0.90 lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0
3 years ago
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