The tungsten filament of a light bulb has an operating temperature of about 2100 K. If the emitting area of the filament is 78 cm2, and its emissivity is 0.65, what is the power output of the light bulb

80cm/s.

24cm/s.

53.3cm/s.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

490 n

f=125n a=2.5m/s2