Respuesta :
Answer:
(a). The mass of displaced water is [tex]41.3\times10^{-3}\ kg[/tex]
(b). The volume of the bone is [tex]41.3\times10^{-6}\ m^3[/tex]
(c). The average density is 1089.58 kg/m³.
Explanation:
Given that,
Mass of bone in air = 45.0 g
Mass of bone in water = 3.60 g
(a). We need to calculate the mass of displaced water
Using buoyant force
The buoyant force equal to the difference in apparent weight in air and water
[tex]F_{b}=m_{dw}g[/tex]
[tex]m_{dw}g=m_{a}g-m_{w}g[/tex]
Put the value into the formula
[tex]m_{dw}g=45\times10^{-3}\times9.8-3.60\times10^{-3}\times9.8[/tex]
[tex]m_{dw}g=0.405[/tex]
[tex]m_{dw}=\dfrac{0.405}{9.8}[/tex]
[tex]m_{dw}=0.0413\ kg[/tex]
[tex]m_{dw}=41.3\times10^{-3}\ kg[/tex]
[tex]m_{dw}=41.3\ g[/tex]
(b). We need to calculate the volume of the bone
Using formula of volume
[tex]V_{dw}=\dfrac{m_{dw}}{\rho_{w}}[/tex]
We know that,
[tex]V_{dw}=V_{b}[/tex]
Put the value into the formula
[tex]V_{dw}=\dfrac{41.3\times10^{-3}}{1000}[/tex]
[tex]V_{dw}=41.3\times10^{-6}\ m^3[/tex]
(c). We need to calculate the average density
Using formula of density
[tex]\rho_{b}=\dfrac{m_{a}}{V_{b}}[/tex]
Put the value into the formula
[tex]\rho_{b}=\dfrac{45.0\times10^{-3}}{41.3\times10^{-6}}[/tex]
[tex]\rho_{b}=1089.58\ kg/m^3[/tex]
Hence, (a). The mass of displaced water is [tex]41.3\times10^{-3}\ kg[/tex]
(b). The volume of the bone is [tex]41.3\times10^{-6}\ m^3[/tex]
(c). The average density is 1089.58 kg/m³.
