When an incandescent light bulb is turned on, its thin wire filament heats up quickly. as the temperature of this wire filament increases, its electrical

here is it answer

Because the lamp filament is physically very small and dissipates a relatively large amount of power (say, 60W for a typical incandescent bulb), while the conductors which feed it are relatively large and of negligible resistance, so they dissipate very little power, for a relatively large conductor area. How hot a component gets from resistive heating is proportional to its power dissipation, and inversely proportional to its area/size.

Thus if you dissipate, say 60W of power in a tiny lamp filament, it will get hot enough to become (surprise !!) incandescent, while the conductors which supply it (which dissipate perhaps a fraction of a watt for many feet of conductor length) will experience no significant temperature rise.

In fact, applicable electrical codes mandate that circuit conductors be sized large enough and provided with appropriate overcurrent protection, so that no significant conductor heating can take place.

a) electrical > light, sound, and thermal< < < <

) the mechanical energy of the firefighter moving.< < < < a) electrical > light, sound, and thermal< < < <

c) both a and b.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6