Duracell batteries are an example of an electrochemical cell that is powered between the reaction of Magnesium and Zinc, occurring in basic conditions (alkaline battery). This type of reaction has a precise output of 1.5 volts, and looks like this:
Zn + 2MnO2 ➡️ ZnO + Mn2O3
It’s not rechargeable.
Golf Cart Batteries are an example of an electrochemical cell that is powered by the reaction between Lead and Sulfuric Acid (Lead-Acid battery). This type of reaction occurs on larger scales than an alkaline battery, and thus can generate a variety of powers depending on how many instruments are present within the battery. The reaction looks like this:
PbO2 + Pb + 2H2SO4 ➡️ 2PbSO4 + H2O
This is a rechargeable cell, but is rather prone to discharging by the environment and surroundings of the battery.
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula 
E = 1461.95 N/C
c) The electric field E is calculated as:
E = 239.76 N/C
Potential and kinetic energy are the two types of energy, but they do get separated into subgroups, for which I do not know. Hope that helps.
Answer:
q = 400 nC
the correct answer is b
Explanation:
The expression for the electric potential of a point charge is
V = k q / r
they ask us for the electrical charge
q = V r / k
let's calculate
Q = 600 6.0 / 9 10⁹
Q = 4 10⁻⁷ C
let's reduce to nC
Q = 4 10⁻⁷ C (10⁹ nC / 1C)
q = 4 10² nC = 400 nC
the correct answer is b
Traslate
La expresión para el potencial eléctrico de una carga puntual es
V = k q/r
nos piden la carga eléctrica
q= V r /k
calculemos
Q= 600 6,0 / 9 10⁹
Q= 4 10⁻⁷ C
reduzcamos a nC
Q = 4 10⁻⁷ C(10⁹ nC/1C )
q = 4 10² nC = 400 nC
la respuesta correcta es b
Im not sure if this is physics or mathematics. but if 300 msec then per minute this will equal to 300 X 60 sec =18000 m per minute
