A rock of mass 200kg is dropped from a height of 200m. What is the Kinetic energy at:
2 answers:
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Answer:
Δe=0.578 kJ/kg
Explanation:
Given data
Velocity v₁=0 m/s
Velocity v₂=34 m/s
to find
Specific energy change Δe
Solution
The specific energy change is simply determined from change in velocity
Δe=(v₂²-v₁²)/2
Put the given values to find the specific energy change
Δe=0.578 kJ/kg
To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.
Energy in a capacitor can be defined as
Here,
V = Potential difference across the capacitor plates
Q = Charge stored on the capacitor plates
At the same time capacitance can be defined as,
Here,
Vacuum permittivity constant
A = Area
d = Distance
Replacing with our values we have that,
PART A) Energy stored in the capacitor is
PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore
Now if we have that the capacitance is
Then the energy stored is
PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore