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Chapter 4: Problem 5

How many droplets of water are in a river that is \(100 \mathrm{~km}\) long, \(250 \mathrm{~m}\) wide, and \(25 \mathrm{~m}\) deep? Assume a droplet is 1 milliliter. ( Note: one liter \(=\) one cubic decimeter and 10 decimeters \(=\) I meter.)

Short Answer

Expert verified
625 trillion droplets

Step by step solution

01

- Convert River Dimensions to Meters

First, convert all the given dimensions of the river to meters to ensure consistency in units. The river's length is already in meters (100 km is 100,000 meters), width is 250 meters, and depth is 25 meters.
02

- Calculate the Volume of the River

Find the volume of the river using the formula for the volume of a rectangular prism: \[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Depth} \]Plug in the values:\[ \text{Volume} = 100,000 \text{ m} \times 250 \text{ m} \times 25 \text{ m} \]\[ \text{Volume} = 625,000,000 \text{ m}^3 \]
03

- Convert Cubic Meters to Liters

1 cubic meter is equal to 1,000 liters. To convert the volume of the river from cubic meters to liters, multiply by 1,000:\[ 625,000,000 \text{ m}^3 \times 1,000 \text{ liters/m}^3 = 625,000,000,000 \text{ liters} \]
04

- Convert Liters to Milliliters

Since 1 liter equals 1,000 milliliters, convert the volume in liters to milliliters by multiplying by 1,000:\[ 625,000,000,000 \text{ liters} \times 1,000 \text{ milliliters/liter} = 625,000,000,000,000 \text{ milliliters} \]
05

- Interpret the Result

Since each droplet of water is defined as 1 milliliter, the number of droplets of water in the river is equal to the volume in milliliters:\[ 625,000,000,000,000 \text{ droplets} \]

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Key Concepts

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

unit conversion
Unit conversion is an essential skill in many fields of science and engineering.
To solve problems accurately, all measurements must be in the same unit.
For this exercise, we converted different units of measurement to meters.
This step is crucial for consistent calculations.

To recap the conversions from the exercise:
  • 1 kilometer (km) = 1,000 meters (m).
  • The river length of 100 km is converted to 100,000 meters.
  • The width and depth are already in meters (250 m and 25 m, respectively).
This method ensures all dimensions are in the format needed for volume calculation.
Always double-check your conversions to avoid errors.
rectangular prism volume
The volume of a rectangular prism can be calculated using a simple formula.
The formula is
\[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Depth} \]
This formula is straightforward and allows you to find the volume using three measurements.

In the given exercise, we applied the following steps:
  • Length: 100,000 meters
  • Width: 250 meters
  • Depth: 25 meters
Plug these values into the formula:
\[ \text{Volume} = 100,000 \text{ m} \times 250 \text{ m} \times 25 \text{ m} = 625,000,000 \text{ m}^3 \]
Easy right?
This volume signifies the total space the river occupies.
metric system
The metric system is a measurement system used by most countries worldwide.
It's based on powers of ten, which makes conversions straightforward.

In this exercise, we utilized the metric system to convert volumes:
  • 1 cubic meter (m^3) = 1,000 liters (L)
  • 1 liter = 1,000 milliliters (mL)
To convert the volume from cubic meters to liters:
\[ 625,000,000 \text{ m}^3 \times 1,000 \text{ L/m}^3 = 625,000,000,000 \text{ L} \]
Further conversion to milliliters:
\[ 625,000,000,000 \text{ L} \times 1,000 \text{ mL/L} = 625,000,000,000,000 \text{ mL} \]
The result (625 trillion milliliters) represents the number of water droplets in the river.
The metric system's simplicity is evident here.
It's a consistent and widely adopted method for scientific measurements.

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Most popular questions from this chapter

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).

(Requires the use of a calculator that can evaluate powers.) If a rope is wound around a wooden pole, the number of pounds of frictional force, \(F,\) between the pole and the rope is a function of the number of turns, \(N\), according to the equation \(F=14 \cdot 10^{0.70 \mathrm{~N}}\). What is the frictional force when the number of turns is: a. 0.5 b. 1 c. 3

An electron weighs about \(10^{-27}\) gram, and a raindrop weighs about \(10^{-3}\) gram. How many times heavier is a raindrop than an electron? How many times lighter is an electron than a raindrop? What is the order-of- magnitude difference?

Evaluate without a calculator: a. \(\sqrt{10,000}\) b. \(\sqrt{-25}\) c. \(625^{1 / 2}\) d. \(100^{1 / 2}\) e. \(\left(\frac{1}{9}\right)^{1 / 2}\) f. \(\left(\frac{625}{100}\right)^{1 / 2}\)

Evaluate the following without a calculator. a. Find the following values: i. \(\log 100\) ii. log 1000 iii. \(\log 10,000,000\) What is happening to the values of \(\log x\) as \(x\) gets larger? b. Find the following values: i. \(\log 0.1\) ii. \(\log 0.001\) iii. \(\log 0.00001\) What is happening to the values of \(\log x\) as \(x\) gets closer to \(0 ?\) c. What is \(\log 0 ?\) d. What is \(\log (-10) ?\) What do you know about \(\log x\) when \(x\) is any negative number?
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