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What you can do is find out the illuminance (lux=lm/m2) of a magnitude 6 star, which is the limit of human vision of light spots and see what luminous flux (lm) creates that illuminance at your certain distance from the spot, assuming that the light disperses uniformly in a semisphere (unless you have better information on the light distribution). From the ...


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A 1 mW red (diode) laser will make a dot visible many meters away, is cheap, widely, legally available, not dangerous and uses little electricity.


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I recently came across this information which seems quite relevant here: When you are comparing equal disk area, then yes staggering them is better than coaxial (in the same column) by roughly 20-30%. But consider the configuration shown below. For a given fram size, when you stagger them you have less room around the circumference of the vehicle's frame ...


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Amps travel in a straight line and so must travel inside the wire. Volts travel around the amps and usually outside the wire. So amps will generate heat - because of the atoms and valence electrons create degrees of resistance - while volts , generally, will not. But if you use thick enough wire you will not notice the heat increase.


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When the rocket sits on the launch pad and the engine fires up, an enormous amount of energy is expended moving the exhaust gases and heating the air. No work is done to move the rocket, but that doesn't mean no energy is expended. But the kinetic / potential energy of the rocket are not changing (actually the potential energy of the rocket becomes less ...


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Useful work=0, power DRIVING the engine =0, ALL useful power lost in the form of wasteful energy, example: exhaust gases, heat, sound and all that..


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In your first example, a rocket on a launchpad that has just started its engines, the equation is exactly correct. The motor is generating huge force, but it has not yet moved the vehicle, so no work or power is being generated. In reality, that moment of no work or power lasts only a tiny fraction of a second. The power moves the rocket and then you can ...


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The dB scale is logarithmic, so when you have two dB levels, their difference is their ratio. Going from -15dB to -7dB is an 8 dB step. It's that simple. Doing the math more explicitly (and showing where you equations come in): If a signal has a dB value $d$ then its intensity is $10^{d/20}$ (that is just inverting your expression for dB value given sound ...


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First of all, I'll write the energy with the letter $U$ to not confuse with the electric field $E$. By definition of power, $P=\frac{dU}{dt}$. Now, where is this energy in the equation coming from? $$U=\int \vec{F}\cdot \vec{dl} = q \int \vec{E}\cdot \vec{dl} = qV$$ This is the amount of energy gained by the charge when moving across an electric potential ...


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First equation: power is defined as energy per unit time. Second equation: if a current flows through a circuit, the power dissipated is the product of voltage and current. This is because the voltage describes the energy each electron is given to traverse the circuit, and the current describes the number of electrons that travel the circuit per unit time. ...


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Define the efficiency of a heat engine as \begin{equation} \eta = \frac{\text{Net work out}}{\text{Heat in}} \end{equation} The Second Law of Thermodynamics tells us that the efficiency can't be greater than the Carnot efficiency: \begin{equation} \eta \leq 1 - \frac{T_L}{T_H} \end{equation} where the $T$'s must be in absolute units (e.g. Kelvin) for the ...


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Since both you and your teachers are stumped I will give some pointers. Pointer 1 - light is an electromagnetic wave. The energy flow is given by the Poynting vector. In vacuum (or air) this is $$\vec S = \vec E \times \vec H$$ Conveniently, for plane waves the time averaged Pointing vector is (see wiki ) $$\langle S \rangle = \frac12 \epsilon_0 c E^2$$ ...


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should the voltage be the input voltage or the voltage across that specific resistor The latter. generally when you want to know something about a particular component, you work with the conditions applying at the boundary of that specific component. I have a desk lamp with a LED. The other end of the lamp plugs into a 240 V AC power outlet. If I want ...


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The symbol kW(e) or kW${}_e$ refers to the "kilowatt electrical". It is the part of the power that is actually used by the devices connected to the power station or the grid, effectively the average of $P = U\cdot I$. The "kilowatt electrical" should be contrasted with "kilowatt thermal" i.e. kW(th) which represents the power including the thermal losses. So ...



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