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Suppose your hair grows at the rate \(1 / 32\) in. per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of \(0.1 \mathrm{nm},\) your answer suggests how rapidly layers of atoms are assembled in this protein synthesis.

Short Answer

Expert verified
The hair grows approximately \(1.01 \mathrm{nm}/\text{second}\). This indicates that roughly 10 layers of atoms are added per second.

Step by step solution

01

Convert inches to nanometers

First, convert the rate from inches to nanometers. We know that 1 inch is approximately equal to \(2.54 \times 10^{7}\) nanometers. Therefore, we can multiply the given rate \(1 / 32\) in. per day by \(2.54 \times 10^{7}\) to get the rate in nanometers per day.
02

Convert days to seconds

Now, convert the time from days to seconds. We know that 1 day is equal to 24 hours, each hour contains 60 minutes, and every minute has 60 seconds. So, we multiply the rate obtained in step 1 by the reciprocal of the number of seconds in a day which is \(1 / (24 \times 60 \times 60)\).
03

Calculation

Finally, by performing the multiplication from step 1 and step 2, we will get the rate at which hair grows in nanometers per second.

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Key Concepts

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

Nanometers
Nanometers are incredibly small units of measurement used to measure things as tiny as molecules and atoms. One nanometer (nm) is one billionth of a meter or 0.000000001 meters. Because of their small size, nanometers allow us to describe dimensions that are otherwise too tiny to comprehend with larger units like millimeters or inches.

To put this in perspective, a human hair is about 80,000 nanometers wide. When you're working with anything at the molecular level, such as protein synthesis or atomic distances, using nanometers gives a precise understanding of minute differences in size. Nanometers are commonly used in scientific fields like biology, chemistry, and materials science due to their precision.
Rate of Growth
The rate of growth refers to how quickly something increases in size over a specific period. In this context, we're discussing the growth rate of hair, which was originally given as \(\frac{1}{32}\) inches per day. Calculating this rate in different units can help us better understand and apply these measurements in different contexts.

To find the growth rate of hair in nanometers per second, the growth rate needs to be converted from inches to nanometers first, and then from days to seconds. Understanding rates of growth is important in various scientific and mathematical applications, including biology and physics, where such calculations can illustrate real-life dynamics like how quickly a virus spreads or how fast plants grow.
  • Inches to nanometers: 1 inch = \2.54 \times 10^{7}\ nm
  • Days to seconds: 1 day = 86400 seconds (the product of 24 hours, 60 minutes, and 60 seconds)
Mastering unit conversions for rates of growth can enhance your problem-solving abilities, especially in high-precision situations.
Protein Synthesis
Protein synthesis is a fundamental biological process that involves creating proteins, which are essential for the structure and function of all living cells. This process occurs in different stages, including transcription and translation.

During protein synthesis, cells assemble layers of atoms rapidly, akin to the calculation of hair growth. Understanding how quickly cells can construct proteins gives insight into the efficiency and speed of biological processes.

Conversion of units, like from inches per day to nanometers per second, can help relate macroscopic observations to the microscopic assembly of molecules within cells. This analogy is useful for appreciating the rapid and nuanced behaviors within biological systems, as even small errors in these rates and conversions can significantly affect outcomes in research and practical applications.

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Most popular questions from this chapter

The nearest stars to the Sun are in the Alpha Centauri multiple-star system, about \(4.0 \times 10^{13} \mathrm{km}\) away. If the Sun, with a diameter of \(1.4 \times 10^{9} \mathrm{m},\) and Alpha Centauri A are both represented by cherry pits \(7.0 \mathrm{mm}\) in diameter, how far apart should the pits be placed to represent the Sun and its neighbor to scale?

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The distance from the Sun to the nearest star is about \(4 \times 10^{16} \mathrm{m} .\) The Milky Way galaxy is roughly a disk of diameter \(\sim 10^{21} \mathrm{m}\) and thickness \(\sim 10^{19} \mathrm{m} .\) Find the order of magnitude of the number of stars in the Milky Way. Assume the distance between the Sun and our nearest neighbor is typical.

Which of the following equations are dimensionally correct? (a) \(v_{f}=v_{i}+a x\) (b) \(y=(2 \mathrm{m}) \cos (k x),\) where \(k=2 \mathrm{m}^{-1}\).

The paragraph preceding Example 1.1 in the text mentions that the atomic mass of aluminum is \(27.0 \mathrm{u}=27.0 \times 1.66 \times 10^{-27} \mathrm{kg} .\) Example 1.1 says that \(27.0 \mathrm{g}\) of aluminum contains \(6.02 \times 10^{23}\) atoms. (a) Prove that each one of these two statements implies the other. (b) What If? What if it's not aluminum? Let \(M\) represent the numerical value of the mass of one atom of any chemical element in atomic mass units. Prove that \(M\) grams of the substance contains a particular number of atoms, the same number for all elements. Calculate this number precisely from the value for u quoted in the text. The number of atoms in \(M\) grams of an element is called Avogadro's number \(N_{\mathrm{A}} .\) The idea can be extended: Avogadro's number of molecules of a chemical compound has a mass of \(M\) grams, where \(M\) atomic mass units is the mass of one molecule. Avogadro's number of atoms or molecules is called one mole, symbolized as 1 mol. A periodic table of the elements, as in Appendix \(\mathrm{C},\) and the chemical formula for a compound contain enough information to find the molar mass of the compound. (c) Calculate the mass of one mole of water, \(\mathrm{H}_{2} \mathrm{O} .\) (d) Find the molar mass of \(\mathrm{CO}_{2}\).

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