An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent resistance 5.20 (assumed constant).
(a) What is the potential difference across the headlight bulbs when they are the only load on the battery?
1 V
(b) What is the potential difference across the headlight bulbs when the starter motor is operated, taking an additional 35.0 A from the battery?

(a)

(b)

Explanation:

Electric Circuits

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

(a) The electromagnetic force of the battery is and its internal resistance is . Knowing the equivalent resistance of the headlights is , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

Solving for i

i=2.28\ A

The potential difference across the headlight  bulbs is

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

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when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.