A hollow sphere of inner radius 8.69 cm and outer radius 9.99 cm floats half-submerged in a liquid of density 927.00 kg/m3. (a) What is the

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A hollow sphere of inner radius 8.69 cm and outer radius 9.99 cm floats half-submerged in a liquid of density 927.00 kg/m3. (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made.

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Thu Cúc 4 years 2021-07-28T05:47:26+00:00 1 Answers 16 views 0

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  1. Vodka
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    2021-07-28T05:48:50+00:00
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    Answer:

    a) m_{sph} = 3.768\,kg, b) \rho_{sph} = 4897.795\,\frac{kg}{m^{3}}

    Explanation:

    According to the Archimedes’ Principle, the buoyancy force is equal the weight of the displaced fluid. Then, this equation of equilibrium is constructed by the Newton’s Laws:

    \Sigma F = \rho_{w}\cdot V_{disp}\cdot g -\rho_{sph}\cdot V_{sph}\cdot g = 0

    After some algebraic handling:

    \rho_{w}\cdot V_{disp} = \rho_{sph}\cdot V_{sph}

    a) The mass of the sphere is:

    m_{sph} = \rho_{w}\cdot V_{disp}

    m_{sph} = (927\,\frac{kg}{m^{3}} )\cdot \left[\frac{4}{3}\pi\cdot (0.099\,m)^{3} \right]

    m_{sph} = 3.768\,kg

    b) The density of the sphere is:

    \rho_{sph} = \frac{m_{sph}}{V_{sph}}

    \rho_{sph} = \frac{3.768\,kg}{\frac{4}{3}\pi\cdot [(0.099\,m)^{3}-(0.087\,m)^{3}]}

    \rho_{sph} = 4897.795\,\frac{kg}{m^{3}}

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