/*! 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 22 The average lead pencil, new and... [FREE SOLUTION] | 91Ó°ÊÓ

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The average lead pencil, new and unused, is \(19 \mathrm{cm}\) long. What is its length in millimeters? In meters?

Short Answer

Expert verified
The pencil is 190 mm or 0.19 m long.

Step by step solution

01

Understanding the Units

To understand the problem, we first need to know the conversion factors between centimeters, millimeters, and meters.
02

Conversion from Centimeters to Millimeters

Since there are 10 millimeters in 1 centimeter, we multiply the length of the pencil in centimeters by 10 to get its length in millimeters.
03

Calculating Length in Millimeters

Using the conversion factor, we can calculate: \[ 19 \text{ cm} \times 10 = 190 \text{ mm} \] So, the pencil is 190 millimeters long.
04

Conversion from Centimeters to Meters

We need to know how many meters there are in a centimeter. Since there are 100 centimeters in 1 meter, we divide the length of the pencil by 100 to convert it into meters.
05

Calculating Length in Meters

Using the conversion factor, we calculate: \[ 19 \text{ cm} \div 100 = 0.19 \text{ m} \] So, the pencil is 0.19 meters long.

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Key Concepts

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

Centimeters to Millimeters
Converting centimeters to millimeters is quite straightforward. In the metric system, where both centimeters and millimeters are standard units of length, the transition between these two units is based on powers of ten. Specifically, there is a simple conversion factor to keep in mind:
  • 1 centimeter (cm) equals 10 millimeters (mm).
To convert from centimeters to millimeters, you multiply the number of centimeters by 10. This multiplication stems from the definition of these units within the metric system. For example, if you have an item that is 19 centimeters long, you would calculate its length in millimeters as follows:\[ 19 \text{ cm} \times 10 = 190 \text{ mm} \]With this conversion, your pencil, originally 19 cm, is 190 mm long. This simple multiplication rule applies universally, making it easy to convert any length in centimeters to millimeters.
Centimeters to Meters
The conversion from centimeters to meters is another effortless task in the metric system. Unlike the centimeter to millimeter conversion, moving to meters requires understanding the relationship between centimeters and meters:
  • 1 meter (m) is equal to 100 centimeters (cm).
Thus, to convert a length from centimeters to meters, you divide the number of centimeters by 100. This process involves scaling down by a factor of 100, reflecting the larger measurement unit that meters represent as compared to centimeters. For instance, if a length measures 19 centimeters, its equivalent in meters would be calculated as follows:\[ 19 \text{ cm} \div 100 = 0.19 \text{ m} \]Therefore, the pencil, which is initially 19 cm, measures 0.19 meters in length. This division rule is consistent and reliable for converting any length from centimeters to meters, emphasizing the consistent nature of conversions in the metric system.
Metric System Conversions
The metric system is arguably the most straightforward system for unit conversions, as it is based predominantly on powers of ten. This characteristic makes conversions within the metric system easier than those in systems with arbitrary conversion factors. Here’s why the metric system is user-friendly:
  • It uses a decimal (base-10) structure, making calculations simpler.
  • Conversions involve simple multiplication or division.
  • Prefixes used in the system reflect the power of ten involved (e.g., "centi-" means one hundredth, "milli-" means one thousandth, etc.).
When working with metric conversions, you generally deal with simple arithmetic. Whether you’re going from centimeters to millimeters or centimeters to meters, the process involves understanding these multiplication or division steps. As such, the metric system enhances clarity and reduces errors during calculations, making it the go-to choice for scientific and international applications. By practicing these conversions, you not only understand the length of objects more precisely but also enhance your overall mathematical fluency.

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Most popular questions from this chapter

A In July \(1983,\) an Air Canada Boeing 767 ran out of fuel over central Canada on a trip from Montreal to Edmonton. (The plane glided safely to a landing at an abandoned airstrip.) The pilots knew that 22,300 kg of fuel were required for the trip, and they knew that 7682 L of fuel were already in the tank. The ground crew added 4916 L of fuel, which was only about one fifth of what was required. The crew members used a factor of 1.77 for the fuel density-the problem is that 1.77 has units of pounds per liter and not kilograms per liter! What is the fuel density in units of kg/L? What mass of fuel should have been loaded? \((1 \mathrm{lb}=453.6 \mathrm{g} .)\)

A 26 -meter tall statue of Buddha in Tibet is covered with 279 kg of gold. If the gold was applied to a thickness of \(0.0015 \mathrm{mm},\) what surface area is covered (in square meters)? (Gold density \(=19.3 \mathrm{g} / \mathrm{cm}^{3} .\) )

The solder once used by plumbers to fasten copper pipes together consists of \(67 \%\) lead and \(33 \%\) tin. What is the mass of lead in a \(250-\mathrm{g}\) block of solder?

Give the symbol for each of the following elements: (a) silver (b) aluminum (c) plutonium (d) tin (e) technetium (f) krypton

You can identify a metal by carefully determining its density (d). An unknown piece of metal, with a mass of \(2.361 \mathrm{g},\) is \(2.35 \mathrm{cm}\) long, \(1.34 \mathrm{cm}\) wide, and \(1.05 \mathrm{mm}\) thick. Which of the following is this element? (a) Nickel, \(d=8.90 \mathrm{g} / \mathrm{cm}^{3}\) (b) Titanium, \(d=4.50 \mathrm{g} / \mathrm{cm}^{3}\) (c) Zinc, \(d=7.13 \mathrm{g} / \mathrm{cm}^{3}\) (d) \(\operatorname{tin}, d=7.23 \mathrm{g} / \mathrm{cm}^{3}\)

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