Specific gravity is the density of the substance over the density of water.

\(\displaystyle SG= \rho _{substance}/\rho _{water}\)

\(\displaystyle \rho _{water}= 1000 kg/m^{3}\)

Density is given by mass over unit volume.

\(\displaystyle \rho =m/V\)

A boat with a mass of 6000kg and a volume of 10m³ will have a density of 600 kg/m³.

\(\displaystyle \rho=\frac{6000kg}{10m^3}=600\frac{kg}{m^3}\)

\(\displaystyle SG=\frac{600\frac{kg}{m^3}}{1000\frac{kg}{m^3}}=0.6\)

If the object is at rest, the net force on it must equal 0. At any point in the liquid, the total downwards force is described as the difference between the gravitational force and bouyant force: \(\displaystyle F_{net} = F_{g} - F_{b}\). When the object is not accelerating, this reduces to \(\displaystyle F_{g} = F_{b}\).

Plugging in the equations for these forces respectively yields \(\displaystyle m_{object}g = \rho_{water}V_{displaced}g\).

Since mass is the product of density and volume, \(\displaystyle \rho_{object}V_{object}g = \rho_{water}V_{displaced}g\).

Note that in this case the object's volume equals the volume of liquid displaced, since the object is completely submerged. So, the volumes on either side cancel, as does gravity, leaving just \(\displaystyle \rho_{object} = \rho_{water}\).

The density of ice is less than that of water (density of ice = 0.92 g/cm³, density of water = 1 g/cm³). Ice will only sink in liquids that are less dense than 0.92 g/cm³. Since ice will only sink in liquids that are less dense than water, the unknown liquid must have a lower density than that of water.

The slope of the solid/liquid phase transition line can predict the comparative density of the solution in each section. A negative slope indicates that the solid phase is less dense than the liquid phase. A positive slope indicates that the solid is more dense. In the above phase diagram, the slope is negative, indicating that the solid is less dense than the liquid. 

To solve, we can equate the volume of the gasoline displaced to the volume of the portion of the object that is submerged.

\(\displaystyle V_{gas}=V_{obj}\)

\(\displaystyle V=\frac{m}{\rho}\)

We know that of the object is submerged, thus of the object's total volume will equal the volume of water displaced.

\(\displaystyle \frac{m_{gas}}{\rho_{gas}}=0.33*\frac{m_{obj}}{\rho_{obj}}\)

\(\displaystyle \frac{\rho_{obj}}{\rho_{gas}}=0.33\frac{m_{obj}}{m_{gas}}\)

The displaced masses are equal.

\(\displaystyle \frac{\rho_{obj}}{\rho_{gas}}=0.33\)

 \(\displaystyle \rho_{obj}=(0.33)(0.8\frac{g}{mL})=0.264\frac{g}{mL}\)

The formula for density is:

\(\displaystyle \rho=\frac{\text{mass}}{\text{volume}}=\frac{m}{V}\)

Specific gravity is the density of a material relative to the density of water.

\(\displaystyle SG_{oil}=\frac{\rho_{oil}}{\rho_{water}}\)

We are given the specific gravity of the oil and the density of water, allowing us to calculate the density of the oil.

\(\displaystyle 0.882=\frac{\rho_{oil}}{1.00\frac{g}{cm^3}}\)

\(\displaystyle \rho_{oil}=(0.882)(1.00\frac{g}{cm^3})=0.882\frac{g}{cm^3}\)

Returning to the equation for density, we can use the density of the oil to find the mass on one liter.

\(\displaystyle \rho_{oil}=\frac{m}{V}\)

\(\displaystyle 0.882\frac{g}{cm^3}=\frac{m}{1000cm^3}\)

\(\displaystyle m=(0.882\frac{g}{cm^3})(1000cm^3)=882g\)

We can solve this question by using a density ratio, given by the equation:

\(\displaystyle X=\frac{\rho_{iceberg}}{\rho_{saltwater}}\)

We know the density of ice, and we can find the ratio of the densities by using buoyancy. If the iceberg were completely submerged, then we could conclude that the density of the ice was equal to the density of the salt water. Since the iceberg is 90% submerged, we can conclude that the density of the ice is 90% of the density of the salt water.

\(\displaystyle \frac{\rho_{iceberg}}{\rho_{saltwater}}=0.9\)

Use this ratio and the density of ice to solve for the density of the salt water.

\(\displaystyle 0.9=\frac{0.917\frac{g}{mL}}{\rho_{saltwater}}\)

\(\displaystyle \rho_{saltwater}=1.019\frac{g}{mL}\)

Density is a measure of concentration, defined as mass per unit volume.

\(\displaystyle \rho=\frac{\text{mass}}{\text{volume}}\)

Since the stock solutions contain different concentrations, the density of each solution must be different. The solution with the highest concentration will contain the highest density, and the solution with lowest concentration will contain the lowest density. The order of increasing density is solution A, solution B, and solution C. Solution C is the most concentrated and will contain the greatest amount of mass per unit volume, followed by solution B and solution A.

\(\displaystyle \rho_A< \rho_B< \rho_C\)

Remember that concentration can be defined in various ways: molarity, molality, mass percent, density, etc. Increasing concentration means you are increasing all of these quantities. Also remember that the density of the compound remains the same in each solution; only the solutions, as a whole, have different densities.

To answer this question, we use F_b = Vgρ, where V is volume, g is gravity, and ρ is the density of the fluid.

In this case we can use the density of blood, since it is given in the passage as 1060kg/m³. Since the density is in cubic meters, we have to make sure that the volume is also converted to cubic meters.

(4.0 * 10^-11)x (1 * 10^-6m/ 1cm) = 4 * 10^-17m³

F_b = (4.0 x 10^-17m³) (10 m/s²) (1.06 x 10³kg/m³) = 4 * 10^-13N

As temperature increases, density decreases.

In a free body diagram of each balloon, the only forces acting in the vertical direction are gravity and buoyant force.

\(\displaystyle F_g =mg= (\rho_{obj})(V_{obj})g\)

\(\displaystyle F_B = (\rho_{medium})(V_{obj})g\)

These forces act in opposite directions, with the buoyant force pulling up and gravity pulling down.

As gas B is heated to the higher temperature, the density decreases. Gas B already has a lower density than gas A; heating it simply intensifies this discrepancy. As the density decreases, the force of gravity on the balloon decreases as well, allowing a greater net force in the upward direction due to the buoyant force.

The net upward force experienced by balloon B at the highest temperature will be greater than any of the other scenarios listed, resulting in the greatest upward acceleration.