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In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are $$70$$ m s$$^{-1}$$ and $$65$$ m s$$^{-1}$$ respectively. If the wing area is $$2$$ m$$^2$$, the lift of the wing is _______ N. (Given density of air $$= 1.2$$ kg m$$^{-3}$$)
Correct Answer: 810
Lift = $$\frac{1}{2}\rho(v_1^2 - v_2^2) \times A = \frac{1}{2}(1.2)(70^2 - 65^2)(2) = 1.2(4900-4225) = 1.2 \times 675 = 810$$ N.
The answer is $$\boxed{810}$$.
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