Question

A medium-sized jet has a 3.8-m-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to the cylindrical fuselage, and aerodynamic shaping gives it a drag coefficient of 0.37. How much thrust must the jets engines provide to cruise at 260 m/s at an altitude where the air density is 1.7 kg/m^3? Express your answer with the appropriate units.

Answer 1

Answer:

$2.4\times 10^{5} N$

Explanation:

The force of drag is given by

$F= 0.5 C\rho Av^{2}$

C is coefficient of drag

$\rho$is density of air

A is area which is given by $\frac {\pi d^{2}}{4}$

V is the velocity

$F= 0.5*0.37*1.7\times \frac {\pi 3.8^{2}}{4}\times 260^{2}=2.4\times 10^{5} N$

Answer 2

Answer:

The thrust  is $F_{thrust}=240.87\rm kN$

Explanation:

Given information:

D=$3.80\rm m$

$C=0.37$

$\rho= \rm 1.7kg/m^3$

$v=\rm 260m/s$

$m=85000kg$

$F_{thrust}=?$

For equilibrium the net force will be zero ,so the thrust must be equal in magnitude to the drag.

$F_{thrust}=F_{drag}=\frac{1}{2}\rho CAv^2$

$A=\frac{\pi\times D^2}{4}=\frac{\pi\times 3.8^2}{4}=11.33\rm m^2$

On substitution,

$F_{thrust}=F_{drag}=\frac{1}{2}\rho CAv^2=\frac{1}{2}\times 1.7\times0.37\times11.33\times (260)^2=240.87\rm kN$

$F_{thrust}=240.87\rm kN$

Hence,the thrust  is $F_{thrust}=240.87\rm kN$.

https://brainly.com/question/12774964?referrer=searchResults