Answer: precision
Explanation: Because accuracy is where you keep on getting it right but precision is where you get closer and closer
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.
Answer:
The condition does not hold for a compression test
Explanation:
For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.
<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test
Answer:
Multiplying impulse response by t ( option D )
Explanation:
We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .
When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .
Watts I believe is the answer
