/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 The human body contains about \(... [FREE SOLUTION] | 91Ó°ÊÓ

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The human body contains about \(10^{14}\) cells, and the diameter of a typical cell is about \(10 \mu \mathrm{m}\) Like all ordinary matter, cells are made of atoms; a typical atomic diameter is \(0.1 \mathrm{nm}\). The volume of a cell is about \(10^{-10} \mathrm{m}^{3}\) b. \(10^{-15} \mathrm{m}^{3}\) c. \(10^{-20} \mathrm{m}^{3}\) d. \(10^{-30} \mathrm{m}^{3}\)

Short Answer

Expert verified
The volume of a typical cell in the human body is approximately \(10^{-15} m^{3}\).

Step by step solution

01

Convert the diameter of the cell from micrometers to meters

Given that \(10 \mu m = 10 \times 10^{-6} m = 10^{-5} m\), the diameter of the cell in meters is \(10^{-5} m\). Next, we calculate the radius of the cell which is half of the diameter, i.e. \((10^{-5} m)/2 = 5 \times 10^{-6} m\).
02

Apply the formula for the volume of a sphere

The volume \(v\) of a sphere is given by the formula \(v = (4/3) \pi r^{3}\). Substituting the radius of the cell (\(5 \times 10^{-6} m\)) into the equation, we find that the volume of the cell is \(v = (4/3) \pi (5 \times 10^{-6})^{3}\).
03

Calculate the volume of the cell

By simplifying the equation, we find that the volume of the cell is approximately \(10^{-15} m^{3}\). Therefore, the correct choice is (b) \(10^{-15} m^{3}\).

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Key Concepts

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

Atomic Diameter
Understanding atomic diameter is essential when discussing the microscopic world. An atom is the smallest unit of a chemical element, and its size is incredibly tiny, usually on the order of nanometers (nm). One nanometer is one billionth of a meter, or written as a decimal, it is 0.000000001 meters. To put this into perspective, a typical atomic diameter is about 0.1 nm, which is 0.0000000001 meters. While this may seem inconsequential, it plays a significant role in determining properties of substances, such as their density and how closely they can pack together. Considering that a cell is composed of countless atoms, knowing the atomic diameter gives us a basis for understanding the scale at which cell structures operate.

Atoms are the building blocks of molecules, which in turn make up cells. By comparing the diameter of an atom to the diameter of a cell, which is often measured in micrometers, we begin to comprehend the complexity of cellular structures. The vast difference in scale between atoms and cells also highlights the remarkable organizational structures of biological systems, where each cell functionally organizes a mind-boggling number of atoms.
Convert Micrometers to Meters
The process of converting micrometers to meters is a fundamental skill in scientific measurement, particularly in the study of biology where cells are often measured in micrometers. One micrometer (also written as 'micron' and symbolized as \(\mu m\)) is one-millionth of a meter, or \(1 \mu m = 1 \times 10^{-6} m\). When handling numbers this small, using scientific notation helps to simplify calculations and reduces the chances of making errors.

To convert micrometers to meters, you simply multiply the number of micrometers by \(10^{-6}\). For example, a cell with a diameter of \(10 \mu m\) would be \(10 \times 10^{-6} m\), or \(10^{-5} m\) when expressed in meters. It's also important to remember that when converting measurements, all parts of the process must maintain the same unit of measure to ensure accuracy and consistency. This conversion is critical when you're calculating something like cell volume, as we did in the original problem, since formulas often require meter measurements.
Formula for the Volume of a Sphere
The volume of a sphere can be determined using a mathematical formula, which is instrumental when trying to understand the volume of spherical objects like cells. This formula is expressed as \((4/3)\pi r^3\), where \(r\) is the radius of the sphere. It's important to note that the radius is half of the diameter, an easy conversion that is sometimes overlooked but crucial for correct calculations.

To use this formula, you first need to ensure that the sphere's radius is measured in meters, as the volume will then be in cubic meters. For a cell with a diameter of \(10^{-5} m\), the radius would be half of that, which is \(5 \times 10^{-6} m\). By inputting this value into the volume formula, you obtain the cell's volume in cubic meters. These calculations help scientists and students alike in understanding not just the physical size of cells but also in applications like determining how much substance a cell can hold or how it might interact with other cells and materials at a microscopic level.

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Most popular questions from this chapter

You're writing a biography of the physicist Enrico Fermi, who was fond of estimation problems. Here's one problem Fermi posed: What's the number of piano tuners in Chicago? Give your estimate, and explain to your readers how you got it.

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In the 1908 London Olympics, the intended 26 -mile marathon was extended 385 yards to put the end in front of the royal reviewing stand. This distance subsequently became standard. What's the marathon distance in kilometers, to the nearest meter?

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Express the following with appropriate units and significant figures: (a) \(1.0 \mathrm{m}\) plus \(1 \mathrm{mm},\) (b) \(1.0 \mathrm{m}\) times \(1 \mathrm{mm},\) (c) \(1.0 \mathrm{m}\) minus \(999 \mathrm{mm},\) and (d) \(1.0 \mathrm{m}\) divided by \(999 \mathrm{mm}\)

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