A cube has a drag coefficient of 0.8. What would be the terminal velocity (in m/s) of a jumbo sugar cube 5-cm per side in air. The density o

Question

A cube has a drag coefficient of 0.8. What would be the terminal velocity (in m/s) of a jumbo sugar cube 5-cm per side in air. The density of air is 1.2 kg/m3 and the density of sugar is 1600 kg/m3. Round your answer to the nearest tenth.

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Thành Công 1 month 2021-08-13T20:33:45+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-13T20:35:20+00:00

    Answer:

    The terminal velocity of the sugar would be 40.4 m/s

    Explanation:

    Given L as the length of the cube = 0.05 m

    Area of the sugar = (0.05 m)^{2} = 0.0025 m^{2}

    The volume of the cube = L^{3} = (0.05 m)^{3} = 0.000125 m^{3}

    Mass of the sugar = Density x volume  =1600 kg/m^{3} x 0.000125

    The terminal velocity of a body can be calculated with the expression below;

    V =\sqrt{(2mg)/pAC}

    Where;

    m is the mass of the body = 0.2 kg

    g is the acceleration due to gravity = 9.81 m/s^{2}

    p is the density of the fluid (air) = 1.2 kg/m^{3}

    A is the area of the body = 0.0025 m^{2}

    Given L as the length of the cube = 0.05 m

    Volume of the cube = L^{3} = 0.05 m^{3} = 0.000125 m^{3}

    Substituting the values in the equation, we have;

    V =\sqrt{(2*0.2*9.8)/1.2*0.0025*0.8}

    V = \sqrt{3.92/0.0024}

    V = \sqrt{1633.33}

    V = 40.4 m/s to the nearest tenth

    Therefore the terminal velocity of the sugar would be

    40.4 m/s

    0
    2021-08-13T20:35:44+00:00

    Answer:

    Explanation:

    Formula for Terminal Velocity is given as :

    • v = \sqrt{(}2 x mass x acceleration (g) / ( density x area x drag coefficient) )

    Mass of sugar =  Density x Volume

    Volume of cube = 5cm x 5cm x 5cm = 125cm^{3}  = 0.000125m^{3}

    Mass of sugar cube = 1600 kg/m^{3} x 0.000125m^{3} = 0.2 kg

    Area = 5cm x 5 cm = 25 cm^{2} = 0.0025 m^{2}

    V = \sqrt{(} 2 x 0.2 x 9.81 )/ (1600 x 0.0025 x 0.8)

    V = \sqrt{(} 3.924/32) = \sqrt{0.1226}

    V= 0.350 m/s

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