If you have a 100W electrical load and you drive 100W plus efficiency losses, say 110W, into the generator, things will be in a state of equilibrium, with 100W being converted from mechanical input power into electricity, and the other 10W of mechanical input power being eaten up by losses.
Now suddenly put 1kW of mechanical power into the machine; at that instant, before the rotational speed can change, the 100W electrical load will continue to present the same mechanical load to the prime mover. Things will not be in equilibrium, and the machine's rotational speed will accelerate. Depending on circumstances, this may or may not increase the electrical load. Certainly the generated voltage will go up, and any simple resistive load will therefor absorb more power, but maybe you have some regulation such that the load continues to draw exactly 100W.
So assume the load continues to draw exactly 100W. Where does the extra 900W of mechanical power go then? The machine's speed must increase until the losses equal the mechanical driving power; so it ends up turning extremely fast, the increased power going into increases in friction in the bearings, windage loss due to the rotating parts, eddy currents in the magnetics (and doubtless a couple of other things I forget at the moment), none of which are desirable.
You would find that, without exceeding the machine's electrical rating, you would quickly exceed its mechanical ratings, i.e., probably long before you got to 1000W, the rotation speed would be several times the suggested speed, and catastrophic failure would likely result. Note you can do this with no electrical load on the generator at all.
Reading the question and the comments, there may be a conceptual misunderstanding : the attenuator WILL attenuate any noise presented on its input (even from just a 50 ohm source impedance), to the same extent it attenuates the signal.
However it also generates noise of its own, which may be represented as the noise from a perfect resistor equal to its own output impedance, and this is added at the output to the (attenuated) input signal and noise. So if input and output Z are both 50 ohms, the net result is attenuated signal + marginally increased noise (i.e. NF = attenuation).
But if its output impedance is lower, the added noise is also lower, thus improving the noise voltage as Andy states.
So represent the attenuator as a perfect attenuator (attenuating noise) in series with a Johnson noise voltage source equal to the output impedance. The rest is just applying the formulae.
EDIT: re: updated question.
(1) There is nothing special about 290K except that it's a realistic temperature for the operation of a passive circuit. The reason they chose it is that the article quotes a noise floor ( -174dBm/Hz) which is correct for a specific temperature : yes, 290k.
(2) While any resistance in the attenuator will contribute noise, I realise that it is not a satisfactory explanation as to why you get the same noise out of an attenuator, because (as Andy says) you could make a capacitive attenuator which is not a Johnson noise generator. So we have to look a little deeper, and remember these noise sources are the statistics of the individual electrons that make up the current.
So, let's say we build a (50 ohm in, 50 ohm out) attenuator, and attempt to cheat Johnson by using a capacitive divider. That implies a node within the attenuator which conducts some of the input current to ground. At this node, we have two current paths; a fraction of the current flows to output, the rest to ground. What determines which path an individual electron will take? Essentially, chance. Collectively? Statistics. So this is a noise source.
Or let's just add series capacitance to provide enough attenuation : we thereby avoid dividing the current flow and eliminate the noise source, right? At the cost of reducing the signal current; our statistics now operate with a smaller sample size and consequently greater variance : more noise.
These results are the best you can do, there is no way round them.
Best Answer
This circuit amplifies the noise caused by the transistor Emitter-Base junction avalanching. The reverse breakdown voltage is usually 6-7V so it definitely won't work at 5V.
If you want something to work from a 5V supply you can step up the 5V to ~9V, using something like an ICL7660 (charge pump) or a boost regulator, or you could use a totally different principle and build a PRNG (pseudo-random number generator) using a microcontroller.