Nowadays, there are many devices using batteries as power supply, and the amount of batteries used is very large. Some of them are rechargeable batteries, such as battery car batteries and automobile batteries. These rechargeable batteries all have one feature: the newly purchased batteries can be used for a long time after charging. After long-term use, they will be fully charged, and the power will soon run out. Please use Ohm's law of the whole circuit for analysis! Thank you for your explanation!

Nowadays, there are many devices using batteries as power supply, and the amount of batteries used is very large. Some of them are rechargeable batteries, such as battery car batteries and automobile batteries. These rechargeable batteries all have one feature: the newly purchased batteries can be used for a long time after charging. After long-term use, they will be fully charged, and the power will soon run out. Please use Ohm's law of the whole circuit for analysis! Thank you for your explanation!

The reason is that the internal resistance of the rechargeable battery increases. When the external circuit is connected, the power consumption of the battery increases
The reason why the internal resistance of the battery increases is due to the chemicals inside the battery. I think it's a matter of purity

If the distance from the moving point P to the fixed point F (1,0) on the plane is 1 greater than the distance from P to the Y axis, then the trajectory equation of the moving point P is ()
A. Y2 = 2xb. Y2 = 4xc. Y2 = 2x or y = 0 x ≤ 0 D. y2 = 4x or y = 0 x ≤ 0

Let P (x, y), the distance from P to the fixed point F (1,0) be (x − 1) 2 + Y2, and the distance from P to the y-axis be | x |. When x ≤ 0, the trajectory of P is y = 0 (x ≤ 0); when x > 0, the distance from P to the fixed point F (1,0) is greater than that from P to the y-axis by 1. The equation: (x − 1) 2 + Y2 - | x | = 1 is listed, and y2 = 4x & nbsp; Then the trajectory equation of the moving point P is y 2 = 4x or y = 0 x ≤ 0