Respuesta :
Answer:
6.5 cm
Explanation:
Given that the edge dimension of a cube, a = 18.7 cm
Density of the cube, [tex]\rho_c= 653 kg/m^3[/tex]
Volume of the cube, [tex]V_c= a^3= 18.7^3 cm^3=18.7^3\times 10^{-6}m^3[/tex]
Mass of the cube. [tex]m_c= \rho_c \times V_c[/tex]
[tex]=653\times 18.7^3\times 10^{-6}[/tex] kg
Gravitational force action on the cube in the downward direction is
[tex]F_g=m_cg=653\times 18.7^3\times 10^{-6}\times g \;N \cdots(i)[/tex]
where g is the acceleration due to gravity.
As the cube floats on the water, let the distance from the horizontal top surface of the cube to the water level is x cm.
So, the distance from the horizontal bottom surface of the cube to the water level= (18.7 - x) cm.
Volume of water replaced, [tex]V_w=[/tex] volume of a portion of the
cube inside the water.
So, [tex]V_w= 18.7\times 18.7\times(18.7 - x) cm^3= 18.7^2\times(18.7 - x)\times 10^{-6}m^3[/tex]
As the density of water, [tex]\rho_w= 1000 kg/m^3[/tex]
So, the mass of the water displaced, [tex]m_w= \rho_w\times V_w[/tex]
[tex]m_w=1000\times 18.7^2\times(18.7 - x)\times 10^{-6}[/tex] kg
The buoyancy force acting on the cube in the upward direction,
[tex]F_b= m_w\times g[/tex]
So, [tex]F_b=1000\times 18.7^2\times(18.7 - x)\times 10^{-6}\times g \; N \cdots(ii)[/tex]
For the floating condition, the gravitational force acting on the cube in the downward direction must be equal to the buoyancy force acting on the cube in the upward direction.
So, equation equations (i) and (ii), we have
[tex]F_b=F_g \\\\\Rightarrow 1000\times 18.7^2\times(18.7 - x)\times 10^{-6}\times g=653\times 18.7^3\times 10^{-6}\times g \\\\[/tex]
[tex]\Rightarrow 18.7-x=\frac{653\times 18.7^3\times 10^{-6}\times g}{1000\times 18.7^2\times 10^{-6}\times g} \\\\\Rightarrow 18.7-x=\frac {653\times 18.7}{1000} \\\\[/tex]
[tex]\Rightarrow 18.7-x=12.2 \\\\\Rightarrow x= 18.7-12.2 \\\\[/tex]
[tex]\Rightarrow x= 6.5[/tex] cm
Hence, the distance from the horizontal top surface of the cube to the water level is 6.5 cm.
