Your boss asks you to design a drone that begins its flight near the surface and rises to 9600 m. At the surface it will fly through air hav

Question

Your boss asks you to design a drone that begins its flight near the surface and rises to 9600 m. At the surface it will fly through air having a density of 1.23 kg per cubic meter and at its highest altitude the air density will become 0.62 kg per cubic meter. If the flight velocity near sea level is 45 mph, then how fast will in need to go at its highest altitude to maintain the same lift. Assume the coefficient of lift remains constant.

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Kim Cúc 4 years 2021-09-03T05:45:58+00:00 1 Answers 3 views 0

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  1. mocmien2
    0
    2021-09-03T05:47:54+00:00
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    Answer:

    63.38\ \text{mph}

    Explanation:

    L = Lift force

    \rho = Density of air

    A = Surface area

    v = Velocity

    v_1 = 45 mph

    \rho_1=1.23\ \text{kg/m}^3

    \rho_2=0.62\ \text{kg/m}^3

    Coefficient of lift is given by

    CL=\dfrac{2L}{\rho v^2A}\\\Rightarrow \rho=\dfrac{2L}{CL v^2A}

    So

    \rho\propto \dfrac{1}{v^2}

    \dfrac{\rho_1}{\rho_2}=\dfrac{v_2^2}{v_1^2}\\\Rightarrow v_2=\sqrt{\dfrac{\rho_1}{\rho_2}}\times v_1\\\Rightarrow v_2=\sqrt{\dfrac{1.23}{0.62}}\times 45\\\Rightarrow v_2=63.38\ \text{mph}

    The velocity at the required altitude should be 63.38\ \text{mph} to maintain the same lift.

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