Write the density formula. 

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

The density of air is: \(\displaystyle 1.29\:\frac{kg}{m^3}\)

Find the volume of the freezer in meters. Converting the dimensions from \(\displaystyle cm\) to \(\displaystyle m\) gives.

\(\displaystyle V_{freezer}=(0.6)(0.85)(1)= 0.51 \textup{m}^3\)

Solve for the mass of the air.

\(\displaystyle m=\rho V=(1.29\frac{kg}{m^3})(0.51 \textup{m}^3)= 0.6579 \textup{kg}\)

To answer this question, it's best to start by drawing a free body diagram so that we can identify the forces acting on the cube. Since the cube is submerged in water but not accelerating, we know that we are dealing with a situation in which there is no net force acting on the cube. In the upward direction we have the force caused by the tension in the string as well as the buoyant force of the displaced water. In the downward direction we have the weight of the cube itself.

\(\displaystyle \Sigma F_{y}=0=T+F_{B}-mg\)

\(\displaystyle mg=T+F_{B}\)

\(\displaystyle m=\frac{T+F_{B}}{g}=\frac{T}{g}+\frac{F_{B}}{g}\)

In order to calculate the buoyant force, we'll need to calculate the volume of the cube in order to determine the amount of water displaced. Since the cube is fully submerged, the volume of the cube will be exactly equal to the volume of displaced water.

\(\displaystyle F_{B}=\rho _{water}V_{cube}g\)

\(\displaystyle V_{cube}=(8cm)^{3}=512cm^{3}\)

\(\displaystyle F_{B}=(1* 10^{-3} \frac{kg}{cm^{3}})(512 cm^{3})g=(0.512kg)g\)

Next, plug this value into the above equation for the mass of the cube.

\(\displaystyle m=\frac{T}{g}+\frac{F_{B}}{g}=\frac{15 N}{9.8 \frac{m}{s^{2}}}+\frac{(0.512kg)g}{g}\)

\(\displaystyle m=1.53\:kg+0.512kg=2.04\:kg\)

Now that we've found the mass of the cube, we can divide this value by its volume in order to obtain its density.

\(\displaystyle \rho _{cube}=\frac{m}{V}=\frac{2.04 kg}{512 cm^{3}}=0.00398\frac{kg}{cm^{3}}\)

And lastly, since we have the density of the cube, we can calculate its specific gravity by dividing its density by the density of water.

\(\displaystyle SG=\frac{\rho _{cube}}{\rho _{water}}=\frac{0.00398 \frac{kg}{cm^{3}}}{0.001 \frac{kg}{cm^{3}}}=3.989\)

Use the information given to find the density of the raft. Since the raft is at rest, the forces on it add to zero. Therefore the magnitude of the gravity force is equal to the magnitude of the buoyant force:

\(\displaystyle mg=F_B=\rho V g\)

The mass of the raft is equal to its density times its volume. We start using subscripts to keep track of which densities and volumes we mean:

\(\displaystyle \rho_R V_Rg=\rho _{H_2O}\, V_{submerged}\,g\)

The \(\displaystyle g\)'s cancel, and the \(\displaystyle V{submerged}\) is \(\displaystyle \frac{9}{10}\) of the raft's total volume:
\(\displaystyle \rho_R\,V_R=\rho_{H_2O}\,0.9V_R\)

\(\displaystyle \rho_R=0.9\rho_{H_2O}=900\frac{kg}{m^3}\)

Setting the magnitudes equal again at the Great Salt Lake:

\(\displaystyle \rho_R V_Rg=\rho _{GSL}\, V_{submerged}\,g\). Cancel the \(\displaystyle g\)'s and solve for the ratio of submerged volume to total volume of the raft:

\(\displaystyle \frac{V_{submerged}}{V_R} =\frac{\rho _{R} }{\rho_{GSL}}=\frac{900}{1100}=\frac{82}{100}\)

Specific gravity is a ratio of a material's density to that of the density of water. Density units can be in any system or scale, but when dividing one unit by itself, the number becomes dimensionless. 

Using density equation:

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

For a sphere:

\(\displaystyle V=\frac{4\pi r^3}{3}\)

If diameter is \(\displaystyle 74.68mm\), then radius is \(\displaystyle 37.34cm\).

Converting \(\displaystyle mm\) to \(\displaystyle cm\) and plugging in values:

\(\displaystyle \rho=\frac{145}{\frac{4\pi*3.734^3}{3}}\)

\(\displaystyle \rho=.665\frac{g}{cm^3}\)

This is less than the density of water, so it will float.

In this question, we're presented with a scenario in which an object of a given density is floating in water. Based on the information given, we're asked to solve for the volume of the object that is above the surface of the water while floating.

To begin, it is useful if we consider a free-body diagram of the object floating on water. In such a diagram, we need not concern ourselves with the horizontal forces, since these will cancel each other out. However, the vertical forces will provide us with the information we need.

Acting in the downward direction, we have the weight of the object, \(\displaystyle mg\). In the upward direction, we have the buoyant force caused by the displacement of water by the object. This upward buoyant force is equivalent to \(\displaystyle \rho _{water}V_{water}g\). Thus, we can write an equation for the vertical forces.

\(\displaystyle \Sigma F_{y}=\rho _{water}V_{water}g-m_{object}g\)

Furthermore, recall that the object is floating. Since it is floating, we know that it is not accelerating in any direction, and thus the net force on it must be zero. Thus, we can set the above expression equal to zero.

\(\displaystyle \Sigma F_{y}=\rho _{water}V_{water}g-m_{object}g=0\)

Next, we can rearrange the equation to obtain the following:

\(\displaystyle \rho _{water}V_{water}g=m_{object}g\)

\(\displaystyle \rho _{water}V_{water}=m_{object}\)

Next, it's important for us to recognize that we can rewrite the expression for the mass of the object. Since density is equal to mass divided by volume, we can rearrange this in terms of density and volume as follows:

\(\displaystyle \rho_{object}=\frac{m_{object}}{V_{object}}\)

\(\displaystyle m_{object}=\rho _{object}V_{object}\)

\(\displaystyle \rho _{water}V_{water}=\rho _{object}V_{object}\)

We can then make one final rearrangement

\(\displaystyle \frac{V_{water}}{V_{object}}=\frac{\rho _{object}}{\rho _{water}}\)

Now let's think about this for a moment. The volume of water that has been displaced by the object is also the volume of the object that is submerged in the water. Thus, the \(\displaystyle \frac{V_{water}}{V_{object}}\) term shown above gives us the fraction of the object that is under the water. Therefore, the remaining fraction of the object must be the portion of it that is above the water. If we take \(\displaystyle 1\) as representing the entire volume of the object, we can solve for the fraction of the object that is above the water like so:

\(\displaystyle 1-\frac{\rho _{object}}{\rho _{water}}=Volume\: of\: object\: above\: water\)

\(\displaystyle 1-\frac{0.65\: \frac{g}{cm^{3}}}{1.0\: \frac{g}{cm^{3}}}=1-0.65=0.35\)

Thus, \(\displaystyle 35\: \%\) of the object's volume is above the water.

We can start with Newton's 2nd law for this problem:

\(\displaystyle F_{net} = ma\)

Upon release, we only have two forces, gravity and buoyancy:

\(\displaystyle F_g =mg\)

\(\displaystyle F_b = m_{w,displaced}g\)

Since the ball is accelerating upward, let's designate an upward force as positive. Then, the first expression becomes:

\(\displaystyle m_{w,displaced}g-m_bg=m_ba\)

Now we need to determine the mass of water displaced. To do this, we will need to choose an arbitrary volume for the sphere. To keep things simple, let's just make it \(\displaystyle V_b=1L\).

Therefore we can say:

\(\displaystyle m_b = \rho_b V_b\)

Substituting this into our last expression, we get:

\(\displaystyle m_wg-\rho_bV_bg=\rho_bV_ba\)

Rearranging for density:

\(\displaystyle m_wg=\rho_bV_b(a+g)\)

\(\displaystyle \rho_b = \frac{m_wg}{V_b(a+g)}\)

We have values for each of these except for the mass of water displaced. Since the ball has a volume of \(\displaystyle 1L\), we know that \(\displaystyle 1L\) of water is also displaced. Then we can say:

\(\displaystyle m_{w,displaced} = \rho_w\cdot V_w\)

\(\displaystyle m_{w,displaced} = \left ( 1.0\frac{kg}{L}\right )(1L) = 1kg\)

Now plugging in values to the density expression, we get:

\(\displaystyle \rho_b = \frac{(1kg)\left ( 10\frac{m}{s^2}\right )}{1L(4.5\frac{m}{s^2}+10\frac{m}{s^2})}\)

\(\displaystyle \rho_b = 0.69\frac{kg}{L}\)

We know that the weight of the object in the air is \(\displaystyle 21 N\). This means that the buoyant force will be \(\displaystyle 21 N\) subtracted by the cube's weight in water (\(\displaystyle 7 N\)). This gives us a buoyant force of \(\displaystyle 14 N\).

The ratio of the weight of the object to the buoyant force in water will give us the specific gravity of the cube in the following equation:

\(\displaystyle SG=\frac{Weight}{F_{b}}=\frac{21N}{14N}=1.5\)

Therefore the correct answer is \(\displaystyle 1.5\)

Since the block is submerged by \(\displaystyle 75%\) percent we know that it has a density that is \(\displaystyle 75%\) percent that of water. We know that the density of water is \(\displaystyle 1 \frac{kg}{L}\) so \(\displaystyle 75%\) percent of that will give us the answer of:

\(\displaystyle .75\frac{kg}{L}\)

Equation of density is \(\displaystyle p=\frac{m}{V}\) so multiplying the volume by three would decrease the density by \(\displaystyle \frac{1}{3}\).