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Any force greater than zero can stop the car. Only it will take longer and the distance moved by it by the time it stops also will be greater. If the force is larger these parameters (time to stop and distance traveled before stopping) will decreasing. Theoretically, infinite force is required to stop it instantaneously.

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Depending on how the energy is stored, it is certainly possible to transfer a lot in a short time. Here are several examples of delivering 1000 J in a short time: A bank of capacitors adding up to 200 mF charged to 100 V holds 1000 J. The right type of capacitor is capable of discharging in 1 ms. Since it may take too long to discharge a capcitor all the ...

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You can deliver that with a rather small capacitor bank. 1MW is 1000V and 1000A of current into a 1 Ohm load. That's fairly trivial to switch with a few dozen bipolar transistors. To store 1000J of energy, you would need a capacitor of 2*1000J/(1000V^2)=2000uF. That's about $2000 worth of electronics components. 1 The obvious example of this is the laser system at the National Ignition Facility. The energy delivered to the target is only a few kJ, but it's delivered in about a picosecond so the power during that time is around 500 terawatts - that's$5 \times 10^{14}$W. The power is so great that it heats a hydrogen pellet enough to make it undergo nuclear fusion. 0 Definitely yes. Thinking of energy in terms of "work done" might help. Just take an example: a big Ferrari with$10^3$kW of power,wont do too much if its engine is turn on for just a few milliseconds seconds and then immediately turned of. Hardly any petrol is used and there is no appreciable increase of kinetic energy. 2 No electrons are generated. They are just set in motion. When the wings of a turbine are set in motion, the axis combined with the generator spins. If (very simply speaking to make the logic clear) you place a magnet on this spinning axis and hold a coil of wires close, this moving magnet causes inductive forces on the wire and the electrons start moving. ... 11 If you express power loss in a power line as$V^2/R$, the$V$in that expression is the voltage difference between the two ends of the power line, not the voltage difference between the power line and ground. To supply a fixed amount of power$P_L$to a load, if the voltage at the load$V_L$is larger, the current$I=P_L/V_L\$ can be smaller. If the power ...

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The short answer is that, if steeped in water, the optical power decreases, possibly a great deal if your lens is a crown glass (like N-BK7) with a refractive index of 1.5 or so. It doesn't matter whether the lens is concave or convex: the power is scaled down (made less negative if the lens is diverging) by the same scaling constant: ...

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