There have actually been cases of (accidental!) exposure to near-vacuum conditions. Real life does not conform to what you see in the movies. (Well, it depends on the movie; Dave Bowman's exposure to vacuum in 2001 was pretty accurate.)
Long-term exposure, of course, is deadly, but you could recover from an exposure of, say, 15-30 seconds. You don't explode, and your blood doesn't immediately boil, because the pressure is held in by your skin.
In one case involving a leaking space suit in a vacuum chamber in 1965:
He remained conscious for about 14 seconds, which is about the time it
takes for O2 deprived blood to go from the lungs to the brain. The
suit probably did not reach a hard vacuum, and we began repressurizing
the chamber within 15 seconds. The subject regained consciousness at
around 15,000 feet equivalent altitude. The subject later reported
that he could feel and hear the air leaking out, and his last
conscious memory was of the water on his tongue beginning to boil
(emphasis added)
UPDATE: Here's a YouTube video regarding the incident. It includes video of the actual event, and the test subject's own description of bubbling saliva.
Another incident:
The experiment of exposing an unpressurized hand to near vacuum for a
significant time while the pilot went about his business occurred in
real life on Aug. 16, 1960. Joe Kittinger, during his ascent to
102,800 ft (19.5 miles) in an open gondola, lost pressurization of his
right hand. He decided to continue the mission, and the hand became
painful and useless as you would expect. However, once back to lower
altitudes following his record-breaking parachute jump, the hand
returned to normal.
If you attempt to hold your breath, you could damage your lungs. If you're exposed to sunlight you could get a nasty sunburn, because the solar UV isn't blocked by the atmosphere (assuming the exposure happens in space near a star). You could probably remain conscious for about 15 seconds, and survive for perhaps a minute or two.
The considerations are about the same in interstellar or interplanetary space, or even in low Earth orbit (or a NASA vacuum chamber). The major difference is the effect of sunlight. As far as temperature is concerned -- well, a vacuum has no temperature. There would be thermal effects as your body cools by radiating heat, but over the short time span that you'd be able to survive, even intergalactic space isn't much different from being in shadow in low Earth orbit.
Reference: http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970603.html
There seem to be a lot of human body mechanical models, such as this one: 
As for applications, I have heard that sub-audio frequency vibrations have been considered as nonlethal weapons for riot control.
Addendum:
Guys, stop upvoting this. The image was not composed by me. I found it so long ago there's no chance to find the original source. Google reverse image search says it might be newbedev.com. In the "related images" section there are other similar interesting sketches on human resonant frequency.
Best Answer
I am not 100% sure if I understand your question but this might clarify the concept of center of mass (COM) of an object. It is defined by the following equation $$\int_{\mathbb R^3} \rho(\vec r) \cdot (\vec r- \vec r_{com}) \, \mathrm d V =\vec 0 $$
where $\rho(\vec r)$ is the density of the object at a position $\vec r$ from the origin. If you modify this integral you get:
$$ \int_{\mathbb R^3} \rho(\vec r) \cdot \vec r \, \mathrm d V = \int_{\mathbb R^3} \rho(\vec r) \cdot \vec r_{com} \, \mathrm d V = \vec r_{com} \int_{\mathbb R^3} \rho(\vec r) \cdot \, \mathrm d V = \vec r_{com }\cdot M $$ $$ \implies \vec r_{com} = \frac 1 M \int_{\mathbb R^3} \rho(\vec r) \cdot \vec r \, \mathrm d V $$
where $M=\int_{\mathbb R^3} \rho(\vec r) \, \mathrm d V$ is the total mass of the object This integral tells you to do one thing: Take density-weighed average of the position vectors.
You see however there is one (technical) problem with this definition. The equation tells you if you give me $\rho(\vec r)$ at every point in space. I'll give you $\vec r_{com}$. For a complex object like a human being $\rho(\vec r)$ is a very very complicated function. Your muscles have different density from your bones and your fat tissue etc. There is one more problem and that is you don't always have the same fat tissue or whatever at the same position for every person. Thus calculating that integral is for all practical purposes impossible for a human being. However the nature doesn't care about the complexity of the object. There is still a unique COM at each fraction of time for each person.**
Let's take a very rough model of a human being. I.e. a box with height $h$ and a square base with side length $a$. Let the density constant $\rho(\vec r) = \rho$ and zero outside of the box. Let's choose the origin to be on the bottom corner of the box.
$$ M \cdot \vec r_{com} = \int_{\mathbb R^3} \rho(\vec r) \cdot \vec r \, \mathrm d V = \iiint\limits_{\text{Box}} \rho \cdot \vec r \, \mathrm d V = \int_{z=0}^h \int_{y=0}^{a} \int_{x=0}^a \rho \cdot \left( \begin{smallmatrix} x \\ y \\ z\end{smallmatrix} \right) \, \mathrm dx \mathrm dy \mathrm dz$$
$$ \implies M \cdot \vec r_{com,\ standing} = \frac \rho 2 \left( \begin{smallmatrix} a^3h \\ a^3h \\ a^2h^2\end{smallmatrix} \right) \implies \vec r_{com} = \frac 1 2 \left( \begin{smallmatrix} a \\ a \\ h\end{smallmatrix} \right)$$
Suppose you have raised your hands above. We can very crudely model this a box wiht a base of a rectangle with sides $a/2$ and $a$ and height $2h$. Note that the volume and thus the mass of the box is the same. If you calculate the integral you'll get
$$\vec r_{com,\ hands\ raised} = \left( \begin{smallmatrix} a/2 \\ a/4 \\ h\end{smallmatrix} \right) \neq \vec r_{com,\ standing}$$
Note that in both cases you have a unique center of mass. However the COM of a person raising hands is clearly different from a person just standing.
What you can do however is to measure the COM experimentally and get a rough idea of where the COM of a human being might be in certain positions.
*In order for this to make physical sense $\rho(\vec r) =0 $ for all $\lVert \vec r \rVert > R$ i.e. your object has to end somewhere.
** Your COM is not the same when your stomach is full and empty. That is why you have to take a small fraction of time to make sure that there is nothing funny going on.