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

e) none of the above answers are correct