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)
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: