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You need some flowmeters. The type depends on the quantity of the gas that you use, and the type. You would then need to come up with a full description of a program to use those values. For example, you might just want to store in a file the quantity of gas supplied to furnaces as a function of time, updated every second (or minute). Or you might want to ...


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Let me tell you the most known phenomenon. You may try this even at home. Take a metallic tumbler and put your mobile phone in it then close the lid so that no part is exposed out of the metallic container. Now with another mobile try calling your first one. You should not be able to connect to the mobile inside the tumbler. This is called shielding. Here ...


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Well, assuming they are bright enough to affect the other light, then, it will depend upon if they have exactly identical timing or not. Theoretically, with identical timing and mechanism, they should blink. Even a tiny fraction of a second difference will be important. If one light switches on a fraction earlier, the other one will stay switched off. The ...


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Assuming that each sensor sees only the light from the other night light, and assuming that each night lights is bright enough to reliably trigger the other's sensor, then you have discovered a configuration that computer engineers call a "flip-flop". https://en.wikipedia.org/wiki/Flip-flop_%28electronics%29 It has another name, "bistable multivibrator." "...


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Interesting question. What you have arrived on is a classic conundrum. Let me make my argument clear through a few scenarios. First, is the light from the nightlight bright enough to affect itself? I mean, if the light is turned on, won't the environment become bright again and hence and trigger a change? I am assuming it does not work that way. Second, ...


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The picture is correct. By the passive sign convention, the reference direction for current is into the positive labeled terminal of the circuit element and thus the circuit element is absorbs (not necessarily dissipates) power when the product of the voltage across and current through is positive. However, the reference direction for $I_S$ is out of the ...



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