Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
<u>Determine the Thrust developed</u>
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : <em>calculate the area of the duct </em>
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
<em>next : calculate the velocity of propeller</em>
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
<em>Finally determine the thrust developed </em>
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
The process is called neutralization.
Hope this helps!
D is your answer hope this helps
Answer:
(a) 1294.66 m
(b) 88.44°
Explanation:
d1 = 580 m North
d2 = 530 m North east
d3 = 480 m North west
(a) Write the displacements in vector forms





The resultant displacement is given by



magnitude of the displacement

d = 1294.66 m
(b) Let θ be the angle from + X axis direction in counter clockwise

θ = 88.44°