Well, because the axis of the rotation of the Earth is not the same as the axis of rotation of the disk of the Milky Way (and also because we're transforming a 2-dimensional spherical map into a 2-dimensional cartesian map), the path of the disk of the Milky Way galaxy looks something like this:
So, there is actually a wide range in declination that the Milky Way can be seen at. The range of declination you can see depends on your latitude (for a review of RA and declination, coordinates used in the celestial coordinate system, see this post). For example, here in Philadelphia (just about $+40^{circ}$ declination), I'd be able to see from from $-50^{circ}$ to $+90^{circ}$ in declination. For Nebraska, find the latitude of your location. To calculate the lower limit, add your latitude (which will be positive since you're in the northern hemisphere) to $-90^{circ}$ (mine was $+40^{circ}$, so: $-90^{circ} + 40^{circ} = -50^{circ}$). To find the upper limit in declination, it's even easier. Since you're in the northern hemisphere, you can actually see the north celestial pole. This means that the upper limit is simply the maximum it can possibly be, which is $+90^{circ}$. The larger the latitude the more circumpolar your night sky gets.
The good news is that you should definitely be able to see it in Nebraska. The only question during what season will you see it at its best? Take a look at the constellations which contain parts of the Milky Way and find the season you can see it in - now you know where to look.
The last thing I want to add is that you need nice dark conditions. Local weather conditions or sources of light pollution may very easily hide the Milky Way. From Philadelphia, even on perfectly clear nights we have no chance of seeing it. If you've got a city on one of your horizons, try to plan around that - either go to a darker location or try looking opposite in the sky from any sources of light pollution.
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