In this exercise we have to calculate the drag coefficient and the power it will have
A) [tex]F_D= 950.166 N[/tex]
B) [tex]P= 50.971 Kw[/tex]
Using aerodynamic concepts we will have:
[tex]F_D[/tex]: Drag force on a body
This can be write as:
[tex]F_D= C_D(1/2)(\rho)(V^2)(A)[/tex]
Where:
- [tex]C_D[/tex]: Drag coefficient equal to [tex]0.290[/tex]
- [tex]\rho[/tex]: Density equal to [tex]0.002378[/tex]
- [tex]V[/tex]: Velocity equal to [tex]120[/tex]
- [tex]A[/tex]: Area (frontal) equal to [tex]20 \ ft^2[/tex]
We need to convert some units, this way we will have how:
For density we have:
[tex]0.002378 lug/ ft^3= 1.2255 \ Kg/m^3[/tex]
For velocity we have:
[tex]120 \ mph= 53.645 m/sec[/tex]
For Area we have:
[tex]20 ft^2 = 185. 806 m^2[/tex]
After the conversions, performing the calculations becomes simple:
[tex]F_d = (0.29)(0.50)(1.2255)(53.645)(185806)\\F_D= 950.166 N[/tex]
Now, power required to overcome drag force, will be;
[tex]P= F_D V\\= (950166)(53645)= 50.971 Kw[/tex]
See more about aerodynamic at brainly.com/question/3800219
