Solution :
Speed of the air craft,
= 262 m/s
Fuel burns at the rate of,
= 3.92 kg/s
Rate at which the engine takes in air,
= 85.9 kg/s
Speed of the exhaust gas that are ejected relative to the aircraft,
=921 m/s
Therefore, the upward thrust of the jet engine is given by

F = 85.9(921 - 262) + (3.92 x 921)
= 4862635.79 + 3610.32
= 
Therefore thrust of the jet engine is
.
Answer:
The linear charge density is 5.19 X 10⁻⁶ C/m
Explanation:
The potential difference between two cylinders, is given as
V = (λ/2πε)ln(b/a)
where;
λ is the line charge density on the power line.
b is the distance between the power line = 1 m
a is the radius of the wire = 1.5 cm = 0.015 m
ε is the permittivity of free space = 8.9 X 10⁻¹² C
V*2πε = λ* ln(b/a)
3900 *(2π*8.9 x10⁻¹²)= λ *ln(1/0.015)
2.1812 X 10⁻⁷ = 4.1997* λ
λ = 5.19 X 10⁻⁶ C/m
Therefore, the linear charge density is 5.19 X 10⁻⁶ C/m
The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.
so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.
airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.
the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h
the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.
if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.
so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.
the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.