The rotor is situated inside the stator and is mounted on the AC motor's shaft. It is the rotating part of the AC motor. And while we know this, the major function of the rotor and the stator is helping the motor shaft rotate.
Complete question:
An airplane is cruising at a velocity of 800 km/h in air whose density is 0.526 kg/m³ . The airplane has a wing planform area of 90 m² . The lift and drag coefficients on cruising conditions are estimated to be 2.0 and 0.06, respectively. The power that needs to be supplied to provide enough trust to overcome wing drag is
Answer:
The power that needs to be supplied to provide enough trust to overcome wing drag is 15,600 kW.
(C) 15,600 kW
Explanation:
Given;
velocity of the airplane, v = 800 km/h = 222.22 m/s
density of air, ρ = 0.526 kg/m³
wing planform area, A = 90 m²
lift coefficients, CL = 2.0
drag coefficients, CD = 0.06
Power supplied = FD* V
This is approximately 15,600 kW.
Therefore, The power that needs to be supplied to provide enough trust to overcome wing drag is 15,600 kW.
The correct option is C
Answer: downward velocity = 6.9×10^-4 cm/s
Explanation: Given that the
Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m
Where radius r = 2.5 × 10^-5 m
Density = 1200 kg/m^3
Area of a sphere = 4πr^2
A = 4 × π× (2.5 × 10^-5)^2
A = 7.8 × 10^-9 m^2
Volume V = 4/3πr^3
V = 4/3 × π × (2.5 × 10^-5)^3
V = 6.5 × 10^-14 m^3
Since density = mass/ volume
Make mass the subject of formula
Mass = density × volume
Mass = 1200 × 6.5 × 10^-14
Mass M = 7.9 × 10^-11 kg
Using the formula
V = sqrt( 2Mg/ pCA)
Where
g = 9.81 m/s^2
M = mass = 7.9 × 10^-11 kg
p = density = 1200 kg/m3
C = drag coefficient = 24
A = area = 7.8 × 10^-9m^2
V = terminal velocity
Substitute all the parameters into the formula
V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]
V = sqrt[ 1.54 × 10^-9/2.25×10-4]
V = 6.9×10^-6 m/s
V = 6.9 × 10^-4 cm/s