an object of volume 1m3 and density of 500kg/m3 float in water what volume of water is displaced​

Question

an object of volume 1m3 and density of 500kg/m3 float in water what volume of water is displaced​

in progress 0
Dulcie 4 years 2021-07-19T01:20:40+00:00 1 Answers 27 views 0

Answers ( )

  1. haidang
    0
    2021-07-19T01:22:05+00:00
    Reply

    Answer:

    0.5\; \rm m^{3}.

    Explanation:

    Calculate the mass of this object:

    \begin{aligned}& m(\text{object}) \\ &= \rho(\text{object}) \cdot V(\text{object}) \\ &= 500\; \rm kg \cdot m^{-3} \times 1\; \rm m^{3} = 500\; \rm kg\end{aligned}.

    Multiple the mass of this object by the gravitational field strength, g, to find the weight of this object: W(\text{object}) = m(\text{object}) \cdot g.

    Since this object is floating in water, the buoyancy force on it should be equal to its weight:

    F(\text{buoyancy}) = W(\text{object}) = m(\text{object}) \cdot g.

    By Archimedes’ Principle, the weight of the water that this object displaces would be equal to F(\text{buoyancy}), the size of buoyancy force on this object,

    Hence, the weight of water displaced would be W(\text{object}) = m(\text{object}) \cdot g.

    Divide this weight by g to find the mass of water displaced:

    \begin{aligned} & m(\text{water displaced}) \\ &= \frac{W(\text{water displaced})}{g} \\ &= \frac{m(\text{object}) \cdot g}{g} = m(\text{object}) = 500\; \rm kg\end{aligned}.

    Assume that the density of water is \rho(\text{water}) = 1000\; \rm kg \cdot m^{-3}. The volume of water displaced would be:

    \begin{aligned}& V(\text{water displaced}) \\ &= \frac{m(\text{water displaced})}{\rho(\text{water})} \\ &= \frac{500\; \rm kg}{1000\; \rm kg \cdot m^{-3}} = 0.5\; \rm m^{3}\end{aligned}.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )