# Tag Info

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For turbulent flow, the friction goes up as the square of the velocity. So the force (pressure) goes as $n^2$. And you are moving $n$ times more liquid. The velocity (or volume flow) increases by $n$. The two factors combine to give the cube law since power = force times velocity, or pressure times flow rate.

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Let's assume that the power company is supplying a neighborhood with 1000 A of current at 120 V. Since P = IV, the neighborhood is receiving 120 kW of power, which is the "load" seen by the power company. To maximize efficiency, the power company wants to minimize the losses involved with transmitting power to the neighborhood, which occur due to ...

0

As your first equation is correct, there is no difference in using $RI^2$ or $U^2/R$ to calculate the power $P = U \cdot I$. My best guess to their choise is that the example gives you the resistance and the current, so this is the simplest way to calculate the power since you don't have to calculate the voltage in a middle step. Furthermore, it is often ...

2

The short answer This is not 100% true since it assumes DC transmission, but it gives the simplest form of the idea: even if the transmission lines are themselves at high voltages, that doesn't directly mean anything, since voltages are not defined relative to anything special (they're defined relative to some other line which is in parallel with your ...

2

There are two different $V$'s here. Suppose the power station outputs at 10,000 V. By the time the wire makes it to your house, this may have dropped to, say, 9,000 V. The $V$ in the first equation refers to the voltage difference you can use, which is 9,000 V (between the wire you receive and ground). The $V$ in the second equation refers to how much ...

0

Voltage is a measure of the electric potential difference across two points in a circuit. It may be considered the work done to transport an electric charge. Power lines are made of thick easily conductive material in order to minimize resistance and power loss to heat. But resistance within power lines is fixed, and power is delivered through the line ...

0

It's true that a 600 hp engine does not necessarily have to be generating that amount of power in order to get from point A to point B. But even though you manage the throttles so that both cars accelerate at the same rate, the larger engine has greater internal cylinder displacement, and therefore consumes more air and fuel per engine revolution than the ...

0

Unfortunately, I see a premise here that is difficult to prove true or false. Mainly because it is almost impossible to use a 600hp engine in a way similar to an 80hp engine and compare them directly. Any car that has a 600hp engine will have other components to support it. Compared with an 80hp car, that might be a larger, more robust transmission, ...

1

Oh yes, particularly if we ever resurrect Project Orion, which works by tossing nukes out the back of the spacecraft and setting them off. https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion) It does, of course, take a pretty good buffer plate to absorb the momentum impulse. Using Tsar Bomba https://en.wikipedia.org/wiki/Tsar_Bomba as an upper ...

11

What is the size/scale of a wood fire that is producing 1kW? Based on what my understandings re energy content of fuels and combustion processes, about 0.5 to 1 cubic inch per minute of typically dry wood in an open fire. Based on an utterly superb 80 page Wood Fuels handbook which I discovered along the way - about the same, rather to my surprise. ...

23

Let's work this out from the Stephan Boltzmann law. What color is a fire? If you look at color charts for black body radiators at various temperatures, I estimate it to be about 1000K. (Be careful: some flames are colored by strong emission spectra, making their light very different from a blackbody radiator's color). Glancing around the web from various ...

-2

During summer due to heat, the intra molecular space increases and the substance expands. As the substance expands, the weight of the conductor increases resulting in the increased sag.

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The acceleration seems quite high - but assuming it is correct, you need to take into account that your velocity keeps increasing. The force might be constant at 8.2 kN (please, drop the extra digits. You are estimating the mass of the car: do you know how much the passengers weigh?) but as the velocity changes (from acceleration) so will the power. In ...

0

What is the acceleration of the space ship, if you turn off the earth's gravitational force? The acceleration of the ship is $$a = \frac{F + G}{m}$$ where $F$ is thrust of the engine (force), $m$ is mass of the rocket and $G=-mg$ is gravity force. If thrust is such as to exactly cancel gravity, it is $F=mg$ so the acceleration with no gravity would ...

0

As @Floris said, rocket motors provide thrust (reaction force), not power. The thrust is just the mass times velocity of exhaust, per second. The power is the mass times velocity squared, per second (over two). Example, shooting a rifle bullet has low reaction force, but high energy. Shooting a bowling ball with the same reaction force takes a lot less ...

2

If you really had a "constant power" engine, and all that power was transferred to your rocket which does not lose mass, it would result in a linear increase in the kinetic energy. And since the kinetic energy $E=\frac12 m v^2$, you can find the velocity at a given time from $$P\cdot t = \frac12 m v^2\\ v = \sqrt{\frac{2 \cdot P \cdot t}{m}}$$ If you ...

0

Yes, people use thermoelectrics as part of very-small-scale nuclear power generation systems, mainly in spacecraft: See Radioisotope Thermoelectric Generator. People do not normally call these things "nuclear power plants", but they are definitely a type of nuclear power generation.

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