The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.

a. what is the forward thrust of this engine?
b. what are the magnitude and direction (i. e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?
c. what are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?

T = 5416.67 N

T = -2083.5 N

T = 0

Explanation:

Forward thrust has positive values and reverse thrust has negative values.

part a

Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

The thrust force T = 5416.67 N

part b

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:

The thrust force T = -2083.5 N reverse direction

part c

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.

T = 0 N

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

answer:

try looking it up that is always

explanation: