Density of Geometric Objects: Videos & Practice Problems

Density of Geometric Objects Example

To determine the mass of a cube of silver measuring 0.56 meters on each side, we start by recognizing the density of silver, which is 10.5 grams per cubic centimeter. The first step is to convert the dimensions of the cube from meters to centimeters, as the density is given in grams per cubic centimeter. Since 1 meter equals 100 centimeters, the side length of the cube in centimeters is:

0.56 \, \(\text{m}\) \(\times\) 100 \, \(\text{cm/m}\) = 56 \, \(\text{cm}\)

Next, we calculate the volume of the cube using the formula for volume:

V = \(\text{length}\)^3 = (56 \, \(\text{cm}\))^3 = 175616 \, \(\text{cm}\)^3

Now that we have the volume, we can use the density to find the mass. The formula relating mass, density, and volume is:

m = \(\text{density}\) \(\times\) \(\text{volume}\)

Substituting the known values:

m = 10.5 \, \(\text{g/cm}\)^3 \(\times\) 175616 \, \(\text{cm}\)^3 = 1843.968 \, \(\text{g}\)

To convert grams to kilograms, we use the conversion factor where 1 kilogram equals 1000 grams:

m = \(\frac{1843.968 \, \text{g}\)}{1000 \, \(\text{g/kg}\)} = 1.843968 \, \(\text{kg}\)

Considering significant figures, the original measurement of 0.56 meters has 2 significant figures, while the density of silver has 3 significant figures. Therefore, we round our final answer to 2 significant figures, resulting in:

m \(\approx\) 1.8 \(\times\) 10^3 \, \(\text{kg}\) \(\text{ or }\) 1800 \, \(\text{kg}\)

This example illustrates the importance of dimensional analysis and unit conversion in solving problems involving density and volume. By carefully managing units and significant figures, we can arrive at a precise and accurate answer.

A copper wire (density = 8.96 g/cm³) has a diameter of 0.32 mm. If a sample of this copper wire has a mass of 21.7 g, how long is the wire?

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• B.
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Answer: (B)

If the density of a certain spherical atomic nucleus is 1.0 x 10¹⁴ g/cm³ and its mass is 3.5 x 10^-23 g, what is the radius in angstroms? (Å= 10⁻¹⁰ m)

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Answer: (C)

Here's what students ask on this topic:

How do you calculate the density of a sphere given its mass and radius?

To calculate the density of a sphere, you first need to find its volume using the formula $V=\frac{4}{3}\pi r^3$, where $r$ is the radius. Once you have the volume, use the density formula $\rho =\frac{m}{V}$, where $m$ is the mass of the sphere. Substitute the volume into this equation to get $\rho =\frac{m}{\frac{4}{3}}$. This allows you to relate the mass and radius to find the density of the sphere.

What is the formula for the volume of a cube and how is it used to find density?

The volume of a cube is calculated using the formula $V=a^3$, where $a$ is the length of one edge of the cube. Since all edges are equal, cubing the edge length gives the volume. To find the density, use the formula $\rho =\frac{m}{V}$, where $m$ is the mass of the cube. Substitute the volume into the density formula to get $\rho =\frac{m}{a^3}$. This connects the cube's mass and edge length to its density.

How do you determine the density of a cylinder if you know its mass, radius, and height?

To determine the density of a cylinder, first calculate its volume using the formula $V=\pi r^{2}h$, where $r$ is the radius and $h$ is the height of the cylinder. Then, use the density formula $\rho =\frac{m}{V}$, where $m$ is the mass. Substituting the volume gives $\rho =\frac{m}{\pi }$. This formula relates the mass, radius, and height to the density of the cylinder.

Why is it important to understand the volume formulas of geometric objects when calculating density?

Understanding the volume formulas of geometric objects is essential when calculating density because density is defined as mass divided by volume. Each geometric shape—such as spheres, cubes, and cylinders—has a unique volume formula that depends on its dimensions. Without knowing the correct volume, you cannot accurately calculate density. For example, the volume of a sphere depends on its radius cubed, while a cube's volume depends on the cube of its edge length, and a cylinder's volume depends on both its radius squared and height. Knowing these formulas allows you to convert physical measurements into volume, which, combined with mass, gives you the density. This understanding is crucial in chemistry and physics for identifying materials and understanding their properties.

Can you explain how to relate mass, volume, and density for different geometric shapes?

Mass, volume, and density are related by the formula $\rho =\frac{m}{V}$, where $\rho$ is density, $m$ is mass, and $V$ is volume. For different geometric shapes, volume is calculated differently: a sphere's volume is $\frac{4}{3}\pi r^3$, a cube's volume is $a^3$, and a cylinder's volume is $\pi r^{2}h$. By measuring the mass and using the appropriate volume formula, you can calculate the density for any geometric object. This relationship helps in identifying materials and understanding their physical properties.