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Determine the order-of-magnitude difference in the sizes of the radii for: a. The solar system \(\left(10^{12}\right.\) meters) compared with Earth \(\left(10^{7}\right.\) meters) b. Protons \(\left(10^{-15}\right.\) meter) compared with the Milky Way \(\left(10^{21}\right.\) meters) c. Atoms \(\left(10^{-10}\right.\) meter) compared with neutrons \(\left(10^{-15}\right.\) meter)

Short Answer

Expert verified
a. 5, b. 36, c. 5

Step by step solution

01

Identify the magnitudes of the radii

First, write down the given magnitudes for the radii: a. Solar system: \(10^{12}\) meters Earth: \(10^{7}\) meters b. Protons: \(10^{-15}\) meter Milky Way: \(10^{21}\) meters c. Atoms: \(10^{-10}\) meter Neutrons: \(10^{-15}\) meter
02

Calculate the order-of-magnitude difference for part (a)

To find the order-of-magnitude difference for part (a), divide the magnitudes of the radii and take the logarithm base 10: \[ \frac{\text{Radius of Solar System}}{\text{Radius of Earth}} = \frac{10^{12}}{10^{7}} = 10^{12-7} = 10^{5} \] The order-of-magnitude difference for part (a) is 5.
03

Calculate the order-of-magnitude difference for part (b)

To find the order-of-magnitude difference for part (b), divide the magnitudes of the radii and take the logarithm base 10: \[ \frac{\text{Radius of Milky Way}}{\text{Radius of Proton}} = \frac{10^{21}}{10^{-15}} = 10^{21-(-15)} = 10^{21+15} = 10^{36} \] The order-of-magnitude difference for part (b) is 36.
04

Calculate the order-of-magnitude difference for part (c)

To find the order-of-magnitude difference for part (c), divide the magnitudes of the radii and take the logarithm base 10: \[ \frac{\text{Radius of Atom}}{\text{Radius of Neutron}} = \frac{10^{-10}}{10^{-15}} = 10^{-10-(-15)} = 10^{-10+15} = 10^{5} \] The order-of-magnitude difference for part (c) is 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Logarithms
Logarithms are a mathematical tool used to deal with very large or very small numbers. They help to transform multiplicative relationships into additive ones, making calculations more manageable. For example, the logarithm base 10 (common logarithm) of a number is the power to which 10 must be raised to produce that number.
A basic understanding of logarithms involves knowing that if \(a = 10^b\), then the logarithm of \(a\) to the base 10 is \(b\), written as \( \text{log}_{10}(a) = b \).
Understanding logarithms is crucial when comparing orders of magnitude because it allows us to convert multiplicative differences into simpler additive differences. For instance, comparing \(10^{12}\) meters (solar system radius) to \(10^7\) meters (Earth radius) gives us a logarithmic order-of-magnitude difference of \5\.
The Power of Scientific Notation
Scientific notation is a way to express very large or very small numbers in a concise form. It is written as the product of a number between 1 and 10 and a power of 10. For example, \(1,000,000\) can be written as \(1 \times 10^6\) in scientific notation.
This makes it easier to read, write, and compute with such numbers. Imagine dealing with the radius of a proton, written as \(0.000000000000001\) meters. In scientific notation, that cumbersome number becomes \(1 \times 10^{-15}\) meters, which is much simpler to handle.
When working with scientific notation, multiplying and dividing numbers is straightforward. You simply add or subtract the powers of 10: for instance, dividing \(10^{21}\) (Milky Way radius) by \(10^{-15}\) (proton radius) results in \(10^{21-(-15)} = 10^{36}\).
Performing Comparative Analysis
Comparative analysis involves comparing different entities to understand their relationships better. In the context of radii, we compare vastly different sizes to understand their scale differences using orders of magnitude.
To compare the sizes of different entities, we calculate the ratio of their sizes and express it in scientific notation to determine the order-of-magnitude difference. For example, comparing the solar system's radius of \(10^{12} \) meters to Earth's radius of \(10^7\) meters, we get a ratio of \( \frac{10^{12}}{10^7} = 10^5 \). This tells us that the solar system's radius is 5 orders of magnitude larger than Earth's.
By using comparative analysis, we can compare educational exercise objects like the radii of atoms (\( 10^{-10}\) meters) and neutrons (\( 10^{-15} \) meters). Calculating the order-of-magnitude difference (\(10^{-10+15} = 10^5\)) reveals that atoms are 5 orders of magnitude larger than neutrons. This makes it easier to grasp the relative size differences between these objects.

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Most popular questions from this chapter

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