Answer:
223.25
Explanation:
The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.
In the context, the thermal conductivity of the material is given as
And it is given that :
1 Btu = 1055 J
1 ft = 0.3048 m
We know that 1 h = 3600 s
So the thermal conductivity of the material in is :
Thermal conductivity :
= 223.25
Answer:
- the effective lift area for the aircraft is 8.30 m²
- the required engine thrust is 1275 N
- required power is 79.7 kW
Explanation:
Given the data in the question;
Speed V = 225 km/hr = 62.5 m/s
The lift coefficient CL = 0.45
drag coefficient CD = 0.065
mass = 900 kg
g = 9.81 m/s²
a) the effective lift area for the aircraft
we know that for a steady level flight, weight = lift and thrust = drag
Using the equation for the lift force
F = C
ρV²A = W
we substitute
0.45 × × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )
1081.05 × A = 8829
A = 8829 / 1081.05
A = 8.30 m²
Therefore, the effective lift area for the aircraft is 8.30 m²
b) the required engine thrust and power to maintain level flight.
we use the expression for drag force
F = T = C
ρV²A
we substitute
= 0.065 × × 1.21 × ( 62.5 )² × 8.30
T = 1275 N
Since drag and thrust force are the same,
Therefore, the required engine thrust is 1275 N
Power required;
P = TV
p = 1275 × 62.5
p = 79687.5 W
p = ( 79687.5 / 1000 )kW
p = 79.7 kW
Therefore, required power is 79.7 kW
Answer:
b. 29.2 rev/min
Explanation:
where I₀ = initial moment of inertia = moment of inertia of the disk +
moment of inertia of the cylinder and ω₀ = initial angular velocity =
30.0 rev/min.
where If = final moment of inertia = moment of the inertia of the solid
disk + moment of inertia of the clay flattened on a disk, and ωf = final
angular velocity.
⇒ Lo =Lf = If*ωf