As an IT professional, you will have access to data that regular employees don't. You keep your mouth shut and you don't snoop. Period. You only look at as much as you have to diagnose and fix problems; the details are irrelevant.
It's called "being professional."
Think of it as the equivalent of lawyer-client or doctor-patient relationships.
Actually, something else is causing the seals to fail on the bearings and master bearing. The sampling pipe was the original theory but it could not account for the damage being done.
Even so, this shows what happens when you plan a one-shot operation with a single point of failure.
In this case, two: the drill itself and the seals. Either one means failure. When it's a one-shot operation with no provision for pause or repair, you're SOL. Fixable in this case? Yeah. For a fortune.
I have a feeling the only person here that is confused is you.
Again. Yes/No: do you claim the heated plate will remain at 150F after the second plate is added?
Repeat Latour's comment here:
However, the absorption rate of real bodies depends on whether the absorber T (radiating or not), is less than the intercepted radiation T, or not. If the receiver T [is less than] intercepted T, no [net] absorption occurs; if the receiver T [is less than] intercepted T the absorption rate may be as great as proportional to (T intercepted - T absorber), depending on the amounts reflected, transmitted or scattered.
(I have added [net] to indicate his argument is net heat transfer there, as Latour has explained many times elsewhere. I have also replaced "less than symbol" with [is less than] due to Slashdot's character restrictions.)
I repeat: you are completely (and wrongly) ignoring real-world conditions that apply to the experiment.
The experiment requires the passive plate to be some unspecified distance from the heat source. (This is a condition of the experiment; no contact is allowed since this is only about radiative transfer.) And no matter its configuration, it will radiate some of its absorbed energy outward to the chamber walls. This much we know.
The chamber walls C are actively cooled, although at a fixed power input. So it absorbs radiation from the internal space. We know at equilibrium T(c) will always be less than T at the source S: T(c) [is less than] T(s) because the wall is actively cooled. (We know for another reason too, but this is sufficient.)
We also know that the passive plate P will always be at a temperature less than that of the source, for the simple reason that no matter what its position, it does not absorb all the radiation from S. Or even if it did, as in the case of completely enclosing S, it would still re-radiate some of its absorbed energy out to the chamber walls C. Therefore as long as the conditions of the experiment are met, no matter what else varies such as relative mass, in this experiment T(p) will be lower than T(s). The amount is of no consequence, as long as it is non-zero (and it is).
Therefore, from S-B law, we can directly infer the following things:
T(c) [is less than] T(s) [net heat transfer is from source to walls, never the other way around]
T(p) [is less than] T(s) [net heat transfer is from source to plate, never the other way around]
We can also infer from the experimental conditions that T(c) [is less than] T(p), but that is irrelevant to the argument.
When equilibrium is achieved, these conditions still hold.
An elementary, obvious, and perfectly sound conclusion from this is that the source is not made hotter under any of these conditions. Even if plate P completely encloses source S, we know for at least two different reasons (greater radiative area, and the simple fact that it does radiate outward to the wall, which cools it) that T(p) will always be less than T(s), even at equilibrium.
Since the temperature of every other object is less than that of the heat source, there is no net heat flow TO the heat source, therefore the heat source does not become hotter. This is, and has been, the whole of Latour's argument, and it is valid. It is not crazy speculation by some nitwit, it is straightforward application of Stefan-Boltzmann law.
Q.E.D., indeed. If the above inequalities hold (and they do), Latour's conclusion is the only one that is mathematically valid.
More importantly, can we agree that in equilibrium, power in = power out?
No. I am not aware of any "conservation of power" law.
And if one service offers an obstensibly cheaper price but has deficiencies that could actually cost you more money, result in tragedy, etc., how do you know that?
The same way you know that for traditional taxi services: they don't stay in business.
Sheesh. Is this really a question? How do you know when you buy a plum in the supermarket that it isn't poisoned? Is it just a wild-assed guess? Or is it more likely that purveyors of poisoned plums didn't get any repeat business?
Were it "Netflix for games" you'd pay a flat monthly fee and be able to play whatever game(s) you want.
A look at how other online rating systems have been rigged suggests you're being hopelessly naive.
Because all those bloggers critical of President Obama are being rounded up as we speak...
Russia has a reputation for jailing or even killing critics of Putin or his allies. The last president of the United States accused of that sort of activity against opponents ended up resigning before his inevitable impeachment and conviction. Even in the latest IRS scandal, which may or may not represent targeting of critics by someone in the executive branch, the end result has been quite the opposite to what one would find in Russia.
The US has no lack of problems, and people in positions of power will always tend towards abusing it. But all in all, it's probably the safest place in the world to speak one's opinion without fear of state persecution.
Socialism isn't a system, it's a class of systems. It encompasses everything from social democratic state on Scandinavia to Marxist-Leninist states.
Don't be irreplaceable, if you can't be replaced, you can't be promoted.
