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0

Adding mass will increase inertia, but will also lower the rotating speed. The "power" are those 2.2 kW so there's not much mechanical cheating if you don't modify tension/current supply, that would change the working diagram of the motor


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My first inclination would be to put a better motor controller that would regulate first current, then speed. By this method when the bit is loaded, the controller would source more current to the drive to maintain a set velocity. But you suggested perhaps doing something to the flywheel/pulley, and yes indeed there is a simpler and cheaper solution, and ...


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We do, but not everywhere. The typical plant is the thermonuclear type, using plutonium. to create the kind of power for the ISS would be about $500M. the solar array is like $10M. there is the danger as well. the probes that use Pu are unmanned and of zero risk to people. if the iss lost orbit there would be a big mess.


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Power is defined as the derivative of the work done onto something with respect to the time, namely $P=\dot{W}$. Integrating over the duration of the time interval gives you the work done by the engine on the car: $$ W=\int_0^t dt' P(t') $$ which in turn is the differential form associated to the force vector field, that is $$ dW = F_xdx + F_ydy + F_zdz $$ ...


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If resistor is an ohmic one, on applying a higher volt battery, power will increase as power for a resistor is $V^2/r$. If it is non ohmic one, the answer may depend upon other factors as well.


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No. a meltdown of uranium would sink in lead due to uranium's higher density. that is the whole problem. there is really nothing that can stop it once it gets going. the only thing at that point is the dispersal of the fuel to lower the thermal neutron density that is causing the heat. this will happen when enough of the rock below mixes with the uranium. if ...


1

Lockheed Martin makes an LH3 and LH8 that are thermoelectric nuclear power generators. You can by new or used. You have to get the fuel from DoE or the Russians.


6

In principle, the drop in the Gibbs energy when the uranium gets converted to the fission products is available for doing useful work. While a steam engine will not come close to the maximum possible efficiency attainable (which is very close to 100%), a thermoelectric device will have much worse performance, as pointed out in detail in the other answers. ...


17

my question is about whether it's possible in principle The answer is yes. and whether anyone tried it. The answer is by all chances, no. So, how come? The effect The thermoelectric effect for electricity generation (called the Seebeck effect) is the phenomenon that a voltage is generated at a temperature different across the ends of a ...


43

The efficiency of a thermoelectric generator is around 5 - 8%. The efficiency of a large steam turbine power plant aproaches 40%. In fact the thermodynamic efficiency of a large steam turbine power plant is over 90%, so it's about as efficient as anything could be. The maximum possible efficiency of a steam driven engine is given by the idealised model ...


0

This is an interesting question whose answer is not entirely obvious. There are many choices of filament dimension which can satisfy the power equations, some of them with bigger diamters and some of them with smaller diamters. However, the real constraint is that all filaments must operate at the same temperature. To make an effective light bulb, the ...


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Power: $$P=VI$$ Ohm's law: $$V=RI$$ Resistance: $$R=\rho \frac{L}{A} $$ These should be what you are looking for.


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If energy is lost, it must go somewhere. So, there won't be any "interference" where the beams overlap which saps the power. Unless, and this does happen, the lasers are powerful enough to ionize the air, in which case the light is blocked at that point and converted to heat, which will never make it to the surface. See this article on laser-induced ...


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You're exactly in trouble! Wh is a symbol of ENERGY, not the POWER!


1

The unit Wh, shown on the battery, is Watt hours, which is the total energy (Joules), not the power.


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The unit langley is amount of energy distributed over an area found in solar radiation, named after Samuel Pierpont Langley. The conversion is straight-forward: $$ 1\frac{\rm ly}{\rm min}=697.3\,\frac{\rm W}{\rm m^2} $$ So you'll have to do some more unit conversions to get the ly/day to W/m$^2$ conversion. See also this site for other unit conversions ...


0

If I have a source that can give an output of 10,000 Volts with some unknown current value, now, can I make that 10,000 Volts to 50 numbers(need not be 50 exactly but may be more than 30) of 200 volts with the same current as previously. Yes, you can. What you have is effectively a constant current source with a maximum terminal voltage. Common ...


0

normal "incoherent" light (white light from a bulb for example) is a world apart from coherent laser (Light Amplification by Stimulated Emission of Radiation) light white light is made up of many frequencies (colors) of light, moving outward in all directions from a point source laser light is made up of ONE color (read frequency) of light, moving in ONE ...



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