The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by a 12.0 - V battery, what total energy in Joules is delivered to the lightbulb filament during 2.00 s

E = 20.03 J

Explanation:

Given that,

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,

Voltage, V = 12 V

We need to find the energy delivered to the lightbulb filament during 2.00 s.

The energy delivered is given by :

. ....(1)

As,

As per Ohm's law, V = IR

Using formula (1).

So, the energy delivered to the lightbulb filament is 20.03 J.

answer; how far out of place which the object finished relative to the starting location;

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: