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Chapter 4: Problem 19

The world population in 2005 was approximately 6.45 billion people. During that year the Coca-Cola company claimed that 15,000 of their beverages were consumed every second. What was the worldwide consumption of their beverages per year per person in \(2005 ?\)

Short Answer

Expert verified
73.34 beverages per person per year

Step by step solution

01

- Convert seconds to years

First, determine the number of seconds in one year. We know there are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, and approximately 365 days in a year. The total number of seconds in a year is calculated as: \[ 60 \times 60 \times 24 \times 365 = 31,536,000 \text{ seconds/year} \]
02

- Calculate total consumption per year

Given that 15,000 beverages are consumed every second, the total number of beverages consumed per year can be calculated by multiplying this rate by the number of seconds in a year: \[ 15,000 \text{ beverages/second} \times 31,536,000 \text{ seconds/year} = 473,040,000,000 \text{ beverages/year} \]
03

- Calculate consumption per person per year

To find the worldwide consumption per person, divide the total consumption per year by the world population in 2005 (6.45 billion people): \[ \frac{473,040,000,000 \text{ beverages/year}}{6,450,000,000 \text{ people}} \approx 73.34 \text{ beverages/person/year} \] Each person consumed approximately 73.34 Coca-Cola beverages per year in 2005.

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Key Concepts

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

Unit Conversion
In calculations involving large numbers and varying units of measurement, it's critical to convert all quantities to consistent units before proceeding. In this exercise, we first convert the given number of seconds in a year. There are
60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, and roughly 365 days in a year.
To find the total number of seconds in a year, we multiply these numbers together:
\[ 60 \times 60 \times 24 \times 365 = 31,536,000 \text{ seconds/year} \]
This conversion ensures that all calculations use consistent time units, which is essential for accurate results.
Annual Consumption
Annual consumption refers to the total amount of a product consumed in a year. For example, if 15,000 Coca-Cola beverages are enjoyed every second, we need to compute how many that would be over the entire year. Using the previously found number of seconds in a year, we calculate:
\[ 15,000 \text{ beverages/second} \times 31,536,000 \text{ seconds/year} = 473,040,000,000 \text{ beverages/year} \]
This enormous number represents Coca-Cola's annual consumption worldwide.
Per Capita Consumption
Per capita consumption calculates the average consumption per person, giving a clearer picture of individual consumption patterns. To determine this figure, divide the total annual consumption by the world's population in 2005:
\[ \frac{473,040,000,000 \text{ beverages/year}}{6,450,000,000 \text{ people}} \ \text{Per capita consumption } \ = \frac{473,040,000,000}{6,450,000,000} \ \text{鈮� 73.34 beverages/person/year} \]
Thus, each person consumed about 73.34 Coca-Cola beverages in 2005. This metric helps businesses understand individual consumer behavior across different regions.
Rate Multiplication
Rate multiplication involves multiplying a rate (in this case, beverages consumed per second) by a time period (seconds in a year) to get a total consumption figure. This is pivotal for understanding how small, consistent actions accumulate over time. Here, we knew that 15,000 beverages were consumed every second. By multiplying this rate with the total seconds per year:
\[ 15,000 \text{ beverages/second} \times 31,536,000 \text{ seconds/year} = 473,040,000,000 \text{ beverages/year} \]
Understanding this method is crucial for similar problems involving rates over time, helping predict yearly usage from shorter, more manageable time frames.

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Most popular questions from this chapter

The pH scale measures the hydrogen ion concentration in a liquid, which determines whether the substance is acidic or alkaline. A strong acid solution has a hydrogen ion concentration of \(10^{-1}\) M. One \(\mathrm{M}\) equals \(6.02 \cdot 10^{23}\) particles, such as atoms, ions, molecules, etc., per liter, or 1 mole per liter. \(^{6}\) A strong alkali solution has a hydrogen ion concentration of \(10^{-14} \mathrm{M}\). Pure water, with a concentration of \(10^{-7} \mathrm{M},\) is neutral. The \(\mathrm{pH}\) value is the power without the minus sign, so pure water has a \(\mathrm{pH}\) of \(7,\) acidic substances have a pH less than \(7,\) and alkaline substances have a \(\mathrm{pH}\) greater than 7 . a. Tap water has a pH of 5.8 . Before the industrial age, rain water commonly had a pH of about \(5 .\) With the spread of modern industry, rain in the northeastern United States and parts of Europe now has a \(\mathrm{pH}\) of about \(4,\) and in extreme cases the \(\mathrm{pH}\) is about \(2 .\) Lemon juice has a \(\mathrm{pH}\) of 2.1. If acid rain with a pH of 3 is discovered in an area, how much more acidic is it than preindustrial rain? b. Blood has a pH of 7.4 ; wine has a pH of about 3.4. By how many orders of magnitude is wine more acidic than blood?

Express each quantity in scientific notation. a. The mass of an electron is about \(\begin{array}{llllll}0.000 & 000 & 000 & 000 & 000 & 000 & 000 & 000 & 001 & 67 \text { gram. }\end{array}\) b. One cubic inch is approximately 0.000016 cubic meter. c. The radius of a virus is 0.00000005 meter.

Estimate the length of a side, \(s,\) of a cube with volume, \(V,\) of 6 \(\mathrm{cm}^{3}\) (where \(V=s^{3}\) ).

Radio waves, sent from a broadcast station and picked up by the antenna of your radio, are a form of electromagnetic (EM) radiation, as are microwaves, X-rays, and visible, infrared, and ultraviolet light. They all travel at the speed of light. Electromagnetic radiation can be thought of as oscillations like the vibrating strings of a violin or guitar or like ocean swells that have crests and troughs. The distance between the crest or peak of one wave and the next is called the wavelength. The number of times a wave crests per minute, or per second for fast-oscillating waves, is called its frequency. Wavelength and frequency are inversely proportional: the longer the wavelength, the lower the frequency, and vice versa-the faster the oscillation, the shorter the wavelength. For radio waves and other \(\mathrm{EM}\), the number of oscillations per second of a wave is measured in hertz, after the German scientist who first demonstrated that electrical waves could transmit information across space. One cycle or oscillation per second equals 1 hertz \((\mathrm{Hz})\). For the following exercise you may want to find an old radio or look on a stereo tuner at the AM and FM radio bands. You may see the notation \(\mathrm{kHz}\) beside the AM band and MHz beside the FM band. AM radio waves oscillate at frequencies measured in the kilohertz range, and FM radio waves oscillate at frequencies measured in the megahertz range. a. The Boston FM rock station WBCN transmits at \(104.1 \mathrm{MHz}\). Write its frequency in hertz using scientific notation. b. The Boston AM radio news station WBZ broadcasts at 1030 \(\mathrm{kHz}\). Write its frequency in hertz using scientific notation. The wavelength \(\lambda\) (Greek lambda) in meters and frequency \(\mu\) (Greek mu) in oscillations per second are related by the formula \(\lambda=\frac{c}{\mu}\) where \(c\) is the speed of light in meters per second. c. Estimate the wavelength of the WBCN FM radio transmission. d. Estimate the wavelength of the WBZ AM radio transmission. e. Compare your answers in parts (c) and (d), using orders of magnitude, with the length of a football field (approximately 100 meters).

For each equation determine the value of \(x\) that makes it true. a. \(6.3 \cdot 10^{x}=0.00063\) b. \(10^{-3}=x\) c. \(5^{x}=\frac{1}{125}\) d. \(x^{3}=\frac{1}{1000}\) e. \(4^{-3} \cdot 2^{-5}=2^{x}\) f. \(9^{-1} \cdot 27^{-2}=3^{x}\)
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