PLEASE HELP I WILL MARK BRAINIEST!!!!!!!!! POSSIBLE ANSWERS BELOW!
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Answer:
0·95
Explanation:
Given the combined mass of the rider and the bike = 100 kg
Percent slope = 12%
∴ Slope = 0·12
Terminal speed = 15 m/s
Frontal area = 0·9 m²
Let the slope angle be β
tanβ = 0·12
As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions
The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached
From the diagram m × g × sinβ = drag force
Drag force = 0·5 × d × × v² × A
where d is the density of the fluid through which it flows
is the drag coefficient
v is the speed of the object relative to the fluid
A is the cross sectional area
As tanβ = 0·12
∴ sinβ = 0·119
Let the fluid in this case be air and density of air d = 1·21 kg/m³
m × g × sinβ = 0·5 × d × × v² × A
100 × 9·8 ×0·119 = 0·5 × 1·21 × × 15² × 0·9
∴ ≈ 0·95
∴ Drag coefficient is approximately 0·95
To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables
Mathematically this can be determined as
Where
Temperature at inlet of turbine
Temperature at exit of turbine
Pressure at exit of turbine
Pressure at exit of turbine
The steady flow Energy equation for an open system is given as follows:
Where,
m = mass
= mass at inlet
= Mass at outlet
= Enthalpy at inlet
= Enthalpy at outlet
W = Work done
Q = Heat transferred
= Velocity at inlet
= Velocity at outlet
= Height at inlet
= Height at outlet
For the insulated system with neglecting kinetic and potential energy effects
Using the relation T-P we can find the final temperature:
From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK
So:
Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg