more voltage or amps
The minimum speed of 6v motor to produce electricity is 15 Mph.
6V dc motor
A 6v motor might not run at all on a 4.5v battery. Whether or not it does, it's likely to discharge the battery rather rapidly. A series resistor will help the battery last longer, but will make it even less likely that the motor will run at all. You're better off taking some time to find either a 4.5v motor or a 6v battery.
IC 7806 is the voltage regulator that gives 6V dc voltage from 230V ac power supply. It is used to give power supply to the IC which operates on 6V.
If the machine is designed to run on 6v and you use two 6v batteries instead of one, then . . . -- If you connect the two batteries in parallel, then they will last twice as long between recharges as one battery does. -- If you connect the batteries in series, then they will definitely overheat the motor, and possibly destroy it.
Well first, more pistons means more exploding. That means more power is transfered to the wheels.
No, underpowering will not "power" the device, it will not run. You must get an adapter that is 6V and (2A or higher) will be ok as well.
Not always. Wattage is a measure of power consumption, not motor power. A more powerful motor may have higher wattage, but efficiency and design also play a role in motor performance.
Use an "Inverter". They make a lot of inverters for 12v but few for 6v. You could use 2 6v batteries hooked in series though. Just make sure the inverter is big enough to power what you wish. A 300 watt one will power a electric razor, or small electronics fine. A 1000 watt would power a computer or a laptop also, etc.
1.25 should have more power but it depends on the design.
The peak-to-peak (P-P) value of a 6V square wave is the difference between its maximum and minimum voltage levels. For a square wave that oscillates between +3V and -3V, the peak-to-peak voltage would be 6V (3V - (-3V) = 6V). If the square wave oscillates between 0V and 6V, the peak-to-peak voltage would also be 6V (6V - 0V = 6V). Therefore, regardless of the specific levels, a 6V square wave has a peak-to-peak voltage of 6V.
It is: -10v+6v = -4v