Respuesta :

P=∆E/∆t. ∆E=P×∆t
Lightbulb: ∆E=60×(24×3600) the unit of the time has to be seconds
Stereo: ∆E= 50×(2×3600)

Answer:

The lightbulb uses 4,104,000 J more than the stereo.

Explanation: