density homework\n1. consider two metal cubes, cube 1 and cube 2. (make sure you fully understand these…
density homework\n1. consider two metal cubes, cube 1 and cube 2. (make sure you fully understand these before commencing.)\na. if the two cubes have the same mass, but cube 1 has half the volume as cube 2, which has a greater density?\nb. the two cubes are both solid copper, but cube 1 has twice the volume as cube 2. which has a greater density?\nc. cube 1 has a mass of 5 kg and cube 2 has a mass of 20 kg. if they occupy the same volume, which has a greater density?\nd. cube 1 has a greater density than cube 2. which occupies a larger volume: a 2 - kg block of cube 1 or a 2 - kg block of cube 2?\n2. a 12.0 cm³ bar of lead metal has a mass of 136 g. what is the density of lead?\n3. liquid mercury has a density of 13.546 g/cm³. what is the mass of 5.0 cm³ of mercury?
Answer
Explanation:
Step1: Recall density formula
The density formula is $\rho=\frac{m}{V}$, where $\rho$ is density, $m$ is mass, and $V$ is volume.
a.
Step1: Analyze given conditions
Given $m_1 = m_2$ and $V_1=\frac{1}{2}V_2$. Using the density formula $\rho=\frac{m}{V}$, for cube 1, $\rho_1=\frac{m_1}{V_1}$, and for cube 2, $\rho_2=\frac{m_2}{V_2}$. Substituting $m_1 = m_2$ and $V_1=\frac{1}{2}V_2$ into the density - expressions, we get $\rho_1=\frac{m_1}{\frac{1}{2}V_2}=2\frac{m_1}{V_2}$ and $\rho_2=\frac{m_2}{V_2}$. Since $m_1 = m_2$, $\rho_1>\rho_2$. So cube 1 has a greater density.
b.
Step1: Recall property of same - material
The density of a homogeneous material is a characteristic property and does not depend on the volume. Since both cubes are solid copper (same material), their densities are the same.
c.
Step1: Analyze given conditions
Given $V_1 = V_2$, $m_1 = 5kg$ and $m_2 = 20kg$. Using the density formula $\rho=\frac{m}{V}$, for cube 1, $\rho_1=\frac{m_1}{V_1}$, and for cube 2, $\rho_2=\frac{m_2}{V_2}$. Substituting $V_1 = V_2$, $m_1 = 5kg$ and $m_2 = 20kg$ into the density - expressions, we know that $\rho_2=\frac{20}{V_2}$ and $\rho_1=\frac{5}{V_1}$. Since $V_1 = V_2$, $\rho_2>\rho_1$. So cube 2 has a greater density.
d.
Step1: Rearrange density formula for volume
From $\rho=\frac{m}{V}$, we can get $V=\frac{m}{\rho}$. Given $m_1 = m_2=2kg$ and $\rho_1>\rho_2$. Substituting into the volume formula, $V_1=\frac{m_1}{\rho_1}$ and $V_2=\frac{m_2}{\rho_2}$. Since $m_1 = m_2$ and $\rho_1>\rho_2$, $V_2>V_1$. So the 2 - kg block of cube 2 occupies a larger volume.
2.
Step1: Apply density formula
Given $m = 116g$ and $V = 12.0cm^3$. Using the density formula $\rho=\frac{m}{V}$, we substitute the values: $\rho=\frac{116g}{12.0cm^3}\approx9.67g/cm^3$.
3.
Step1: Rearrange density formula for mass
From $\rho=\frac{m}{V}$, we can get $m=\rho V$. Given $\rho = 13.546g/cm^3$ and $V = 5.0cm^3$. Substitute the values: $m=(13.546g/cm^3)\times5.0cm^3 = 67.73g$.
Answer:
a. Cube 1 b. They have the same density c. Cube 2 d. The 2 - kg block of cube 2 2. $9.67g/cm^3$ 3. $67.73g$