A certain lead-acid storage battery has a mass of 30 kg. starting from a fully charged state, it can supply 5 a for 24 hours wit
h a terminal voltage of 12 v before it is totally discharged.
a. 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/s2 and is constant with height. b if the stored energy is used to accelerate the battery with 100 percent efficiency, what velocity is attained?
c. gasoline contains about 4.5 x 107 j/kg. compare this with the energy content per unit mass for the fully charged battery.
a. 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/s2 and is constant with height. b if the stored energy is used to accelerate the battery with 100 percent efficiency, what velocity is attained?
c. gasoline contains about 4.5 x 107 j/kg. compare this with the energy content per unit mass for the fully charged battery.
1 answer:
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Answer:
v₂ = 0.56 m / s
Explanation:
This exercise can be done using Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Where points 1 and 2 are on the surface of the glass and the top of the straw
The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0
The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m
We substitute in the equation
+ ρ g y₁ =
+ ½ ρ v₂² + ρ g y₂
½ v₂² = g (y₂-y₁)
v₂ = √ 2 g (y₂-y₁)
Let's calculate
v₂ = √ (2 9.8 1.6 10⁻²)
v₂ = 0.56 m / s