If the energy stored in the fully charged battery is used to lift the battery with 100-percent efficiency, what height is attained? Assume that the acceleration due to gravity is 9.8 m/s2m/s2 and is constant with height

h = 32059.37 m

Explanation:

Assuming the missing in formation as

A certain lead-acid storage battery has a mass of 33 kg. Starting from a fully charged state, it can supply 6 A for 20 hours with a terminal voltage of 24 V before it is totally discharged.

Now, Applying energy conservation (Electrical to potential)

Electrical Energy E = I*V*t

I = correct, V = voltage, t = time of flow of current

E = 6*24*20*60*60.

E = 10368 KJ

Now this energy is used to lift the battery with 100% efficiency

Hence,

electrical energy E = potential energy P

P = mgh

m=mass of the battery, g = the acceleration due to gravity is 9.8 m/s^2

h = height

mgh = 10368 kJ

33*9.8*h = 10368*1000

h = 10368*1000 / (33*9.8)

h = 32059.37 m