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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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What is the resolution limit for a projection type of photolithographic system if the incident wavelength is 365 nm (the i-line
Vadim26 [7]

Answer:

Depth of field  = 347.619 nm

Explanation:

wavelenght =  365nm

N.A =0.63

k1= 0.6

so we have that the resolution limit is:

R=k1*A

R=0.6*365

R=219 nm

and the Depth of field needed for the best resolution is:

DoF = Resolution / N.A.

DoF= R/N.A

DoF= 219/0.63

DoF= 347.619 nm

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(25 points).The heat is turned on in a building in a cold morning, but the heater does not burnthe natural gas properly and emit
nydimaria [60]

Answer: Yes, it will be around 40ppm(at steady state)

Explanation:

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A direct-coupled amplifier has a low-frequency gain of 40 dB, poles at 2 MHz and 20 MHz, a zero on the negative real axis at 200
rewona [7]

Answer:

Explanation:

Low frequency gain is= 40db= 20logK=>100 poles at 2MHz,20MHz

Zero at -200MHz, zero at infinity.

A) A(s) = 100FH(s)

B) Poles (1): 2 pi × 2 × 10^6= 4pi × 10^6MHz

              (2): 2pi × 20 × 10^6= 4pi × 10^6 MHz

 Zeroed: 2pi × 10^6 × 200= 400pi × 10^6, at infinity.

T/(S) = (1 + S/400π × 10^6)/S(1 + S/4π × 10^6)(1 + S/4π × 10^6)

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3 years ago
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yulyashka [42]

Answer: OHMMETER & MEGOHMMETER:

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