In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2 ? Take the density of air to be 1.3 kg m–3.
Speed of wind on the upper surface of the wing, V1 = 70 m / s
Speed of wind on the lower surface of the wing, V2 = 63 m / s
Area of the wing, A = 2.5 m2
Density of air, ρ = 1.3 kg m–3
According to Bernoulli’s theorem, we have the relation :
\(P_1 + \frac{1 }{ 2} ρV_1^2 = P_2 + \frac{1 }{ 2} ρV_2^2\)
\(P_2 - P_1 = \frac{1 }{ 2} ρ(V_1^2 - V_2^2)\)
Where,
P1 = Pressure on the upper surface of the wing
P2 = Pressure on the lower surface of the wing
The pressure difference between the the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing = (P2 - P1)A
= \(\frac{1 }{ 2} ρ(V_1^2 - V_2^2)A \)
\(= \frac{1 }{ 2} 1.3((70)^2 - (63)^2) × 2.5 \)
= 1512.87
= 1.51 × 103 N
Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.
Give reasons for the following.
(i) King Tut’s body has been subjected to repeated scrutiny.
(ii) Howard Carter’s investigation was resented.
(iii) Carter had to chisel away the solidified resins to raise the king’s remains.
(iv) Tut’s body was buried along with gilded treasures.
(v) The boy king changed his name from Tutankhaten to Tutankhamun.
Find the mean deviation about the median for the data
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |