An adiabatic closed system is accelerated from 0 m/s to 34 m/s. Determine the specific energy change of this system, in kJ/kg.

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

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .