Can I use angular diameter to determine the apparent size of an object at a given distance? I believe this question uses Angular Diameter to determine the answer but I'm not sure how to use it.  
Question: If a widget is 10 meters wide and it is positioned 1,000 meters from the observer how wide will it appear?  How wide will it appear at 2,000 meters?  Is Angular Diameter used for this?
Using this online calculator I inputted 10 meters for Linear Size and 1000 meters for Distance to Object.  It returned an Angular Diameter of 0.572953020554149.  But how does this help me determine the apparent size of an object at a given distance?  Or alternately am I using the wrong formula?
Backstory:
I can find many references to Angular Diameter on both Google and this forum.  But I can't find one that explains how to use this simple concept to determine on object's apparent size.
 A: From a physics perspective, angular diameter and "apparent size" are the same thing. They both essentially tell you how much of your vision the object is taking up.

Even simpler at what distance will a widget that is 10 meters wide appear to be 1 meter wide? 

There's no simple relationship between angular diameter and the apparent physical size of an object in meters. The apparent size of an object is psychological, not physical.
For instance, an adult human will appear to be in the ballpark of $1.6~\rm m$ tall from any distance, because you know from experience that's about how tall people are. Your brain uses many cues to estimate the physical size of objects in the distance.
A: You need to add Ratio to your equations. If you know how big something is than add the ratio of that to the equation and do the math there... P.s. I'm horrible at math. It's just a thought.
A: If you want an equation, here it is:
$$\tan{\theta} = \frac{D}{L}, $$
where $\theta$ is the angular diameter, $D$ is the true diameter, and $L$ is the distance. For things with small angular diameters, one can approximate:
$$ \tan{\theta} \approx \theta,$$
with the caveat that $\theta$ is measured in radians. (Note, your post had a unitless angle, which was in fact in degrees). For your hypothetical numbers:
$$ \theta \approx \frac{10\,m}{1000\,m} = 0.01\,rad = \frac{180}{100\pi}\, deg = 0.5729^{\circ},$$
where the numbers of significant figures reflects the accuracy of the small angle approximation.
