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Chapter 2: Problem 179

Is reporting a measured mass as \(580 . \mathrm{g}\) the same as reporting it as \(580 \mathrm{~g}\) ? Explain.

Short Answer

Expert verified
Both expressions "580 . g" and "580 ~g" can be better represented as "\(580 \mathrm{g}\)". Although these two expressions are essentially equivalent in representing the mass, it is recommended to remove unnecessary symbols (decimal point or approximation) when reporting the measurement.

Step by step solution

01

Understand the given notations

We have two expressions for the mass: 1. "580 . g" 2. "580 ~g" The "." in the first case is an isolated symbol or is a decimal point without any further decimal values. In the second case, the "~" symbol can be understood as an approximation or as no relevance at all.
02

Analyze the expressions for equivalence

In the first expression, the decimal point without any further decimal values does not add any useful information to the mass. In fact, it may create confusion about the measured value. Therefore, it is more appropriate to write this measurement as "\(580 \mathrm{g}\)" without the decimal point ".". In the second expression, if we consider that "~" has no relevance at all, then the mass can be simply reported as "\(580 \mathrm{g}\)".
03

Conclude the comparison

In conclusion, as both measurements can be better represented as "\(580 \mathrm{g}\)", reporting the mass as "580 . g" or "580 ~g" is essentially equivalent, although it is recommended to remove unnecessary symbols (decimal point or approximation) when reporting the measurement.

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Most popular questions from this chapter

The students measure another ball with the ruler in Problem \(2.93\) and determine that its diameter is \(2.55 \mathrm{~cm}\). What is the radius of the ball to the correct number of significant figures?

A metal sphere has a radius \(r\) of \(4.00 \mathrm{~cm}\). What is the volume \(V\) of this sphere in cubic centimeters? The formula for the volume of a sphere is \(V=(4 / 3) \pi r^{3}\), where \(\pi=3.14159 .\)

A train traveling at \(45.0\) miles \(/ \mathrm{h}\) has to make a trip of \(100.0\) miles. How many minutes will the trip take? Use unit analysis to calculate your answer, and show your work.

Write two conversion factors that express the relationship between: (a) Grams and kilograms, using 1 and 1000 (b) Kilograms and grams, using 1 and \(0.001\) (c) Yards and feet (d) Meters and centimeters, using 1 and 100 (e) Meters and centimeters, using 1 and \(0.01\)

Write each number in standard notation: (a) \(1.79 \times 10^{-2}\) (b) \(8.76 \times 10^{-9}\) (c) \(4.88 \times 10^{10}\) (d) \(7.52 \times 10^{1}\) (e) \(8.37 \times 10^{\circ}\) (f) \(4.184 \times 10^{4}\)
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