Respuesta :
Answer:
10.28 Kg
Explanation:
Given:
[tex]M_{iron} =7.70 Kg\\\rho_{iron}=7.86 x 10^-3 kg/m^3\\\rho_{silver}=10.50 x 10^3 kg/m^3[/tex]
we know that
[tex]\rho=\frac{m}{V}\\V=\frac{m}{\rho}[/tex]
So,
[tex]\frac{m_{iron}}{\rho_{iron}} =\frac{m_{silver}}{\rho_{silver}}\\ \frac{7.70}{7.86*10^3} =\frac{m_{silver}}{10.50*10^3}\\ m_{silver}=10.28 kg[/tex]
The mass of the silver sculpture is 10.28 kg.
How do you calculate the mass?
Given that the mass of iron is 7.70 kg and the density of the iron is 7.86 ✕ 103 kg/m3. The density of the silver is 10.50 ✕ 103 kg/m3.
The density of the iron is given below.
[tex]\rho_i = \dfrac {m_i}{V_i}[/tex]
Where m_i is the mass and V_i is the volume of the iron.
[tex]\rho_i = \dfrac {7.70}{7.86\times 10^3}[/tex]
The density of the silver is given below.
[tex]\rho_s = \dfrac {m_s}{V_s}[/tex]
Where m_s is the mass and V_s is the volume of the silver.
[tex]\rho_s = \dfrac {m_s}{10.5\times 10^3}[/tex]
The sculpture is identical to the iron sculpture, hence their density will be equal.
[tex]\rho_i =\rho_s[/tex]
[tex]\dfrac {7.70}{7.86\times 10^3} = \dfrac {m_s}{10.5 \times 10^3}[/tex]
[tex]m_s = 10.28 \;\rm kg[/tex]
hence we can conclude that the mass of the silver sculpture is 10.28 kg.
To know more about the density, follow the link given below.
https://brainly.com/question/952755.
