Chapter 15: Problem 12
Write each of the following using base units only, and without the use of a prefix: a) \(5643 \mathrm{kN} / \mathrm{t}\).d b) \(453000 \mu \mathrm{H} /(\mathrm{kg} \cdot \mathrm{h})\) c) \(654.0004 \mathrm{nS.m} /(\mathrm{h} . \mathrm{kg})\)
Short Answer
Expert verified
Question: Convert the given values to their base units and eliminate the prefixes.
a) 5643 kN/t.d
b) 453000 μH/(kg·h)
c) 654.0004 nS·m/(h·kg)
Answer:
a) 0.0654 N·s/kg
b) 0.00012583 H·s/kg
c) 1.81 × 10⁻¹⁰ S·m/(s·kg)
Step by step solution
01
Identify prefixes and their corresponding exponential values
First, we need to identify the prefixes and find their corresponding exponential values. In this case, we have the following:
- k (kilo) in kilonewtons (kN): \(10^3\)
- t (tons): \(1000 \mathrm{kg}\)
- d (day): \(86400 \mathrm{s}\)
02
Replace the prefixes with their exponential values
Now, we replace the prefixes with the exponential values and perform the necessary calculations:
\(5643 \mathrm{kN} / \mathrm{t}\).d = \(\frac{5643 \times 10^3 \mathrm{N}}{(1000 \mathrm{kg})(86400 \mathrm{s})}\)
03
Simplify the expression
Finally, simplify the expression:
\(\frac{5643 \times 10^3 \mathrm{N}}{(1000 \mathrm{kg})(86400 \mathrm{s})} = \frac{5643}{(1000)(86400)} \mathrm{N.s/kg} = 0.0654 \mathrm{N.s/kg}\)
b) \(453000 \mu \mathrm{H} /(\mathrm{kg} \cdot \mathrm{h})\)
04
Identify prefixes and their corresponding exponential values
First, we need to identify the prefixes and find their corresponding exponential values. In this case, we have the following:
- \(\mu\) (micro) in microhenries (\(\mu\)H): \(10^{-6}\)
- h (hour): \(3600 \mathrm{s}\)
05
Replace the prefixes with their exponential values
Now, we replace the prefixes with the exponential values and perform the necessary calculations:
\(453000 \mu \mathrm{H} /(\mathrm{kg} \cdot \mathrm{h}) = \frac{453000 \times 10^{-6} \mathrm{H}}{(\mathrm{kg})(3600 \mathrm{s})}\)
06
Simplify the expression
Finally, simplify the expression:
\(\frac{453000 \times 10^{-6} \mathrm{H}}{(\mathrm{kg})(3600 \mathrm{s})} = \frac{453000}{(10^{6})(3600)} \mathrm{H.s/kg} = 0.00012583 \mathrm{H.s/kg}\)
c) \(654.0004 \mathrm{nS.m} /(\mathrm{h} . \mathrm{kg})\)
07
Identify prefixes and their corresponding exponential values
First, we need to identify the prefixes and find their corresponding exponential values. In this case, we have the following:
- n (nano) in nanosiemens.m (nS.m): \(10^{-9}\)
- h (hour): \(3600 \mathrm{s}\)
08
Replace the prefixes with their exponential values
Now, we replace the prefixes with the exponential values and perform the necessary calculations:
\(654.0004 \mathrm{nS.m} /(\mathrm{h} . \mathrm{kg}) = \frac{654.0004 \times 10^{-9} \mathrm{S.m}}{(\mathrm{h})(\mathrm{kg})}\)
09
Simplify the expression
Finally, simplify the expression:
\(\frac{654.0004 \times 10^{-9} \mathrm{S.m}}{(\mathrm{h})(\mathrm{kg})} = \frac{654.0004}{(10^{9})(3600)} \mathrm{S.m/(\mathrm{s.kg})} = 1.81 \times 10^{-10} \mathrm{S.m/(\mathrm{s.kg})}\)
So, the expressions simplified to their base units are:
a) \(0.0654 \mathrm{N.s/kg}\)
b) \(0.00012583 \mathrm{H.s/kg}\)
c) \(1.81 \times 10^{-10} \mathrm{S.m/(\mathrm{s.kg})}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Prefixes in Measurement Units
The use of prefixes in measurement is a convenient way to express large or small quantities in physics and various scientific disciplines. Prefixes represent powers of ten, and when attached to a unit, they denote multiplication by that power. This method of unit conversion simplifies understanding and conceptualizing quantities that are otherwise too large or too small to be conveniently written in standard decimal form.
For example, the prefix 'kilo-' (k) means a factor of one thousand, or \(10^3\), so 1 kilogram (kg) is equivalent to one thousand grams. Similarly, 'milli-' (m) means a factor of one-thousandth, or \(10^{-3}\), which means that 1 millimeter (mm) is one-thousandth of a meter. Understanding these prefixes is crucial because it permits a standardized way to express quantities and perform conversions between different scales of measurement.
When working with these prefixes, it is essential to recognize and convert them into their exponential forms. This allows for the accurate performance of calculations, especially when different units are involved. For instance, translating kilonewtons (kN) to newtons (N) multiplies the amount by one thousand, due to the 'kilo-' prefix, which is the same as multiplying by \(10^3\).
For example, the prefix 'kilo-' (k) means a factor of one thousand, or \(10^3\), so 1 kilogram (kg) is equivalent to one thousand grams. Similarly, 'milli-' (m) means a factor of one-thousandth, or \(10^{-3}\), which means that 1 millimeter (mm) is one-thousandth of a meter. Understanding these prefixes is crucial because it permits a standardized way to express quantities and perform conversions between different scales of measurement.
When working with these prefixes, it is essential to recognize and convert them into their exponential forms. This allows for the accurate performance of calculations, especially when different units are involved. For instance, translating kilonewtons (kN) to newtons (N) multiplies the amount by one thousand, due to the 'kilo-' prefix, which is the same as multiplying by \(10^3\).
Base SI Units
The International System of Units (SI) provides a foundation for all measurements in science and technology. Base SI units are the standard units of measurement for fundamental physical quantities, and all other units are derived from them. There are seven base SI units for length (meter, m), mass (kilogram, kg), time (second, s), electric current (ampere, A), temperature (kelvin, K), amount of substance (mole, mol), and luminous intensity (candela, Cd).
For instance, force is measured in newtons (N), which is derived from base units as \( \text{kg} \times \text{m/s}^2 \). Converting complex units back to these base units simplifies calculations and standardizes results across various fields of study. In the provided exercise solutions, tons (t) and days (d) were converted into kilograms (kg) and seconds (s), respectively, aligning the units with the base SI units for mass and time. This process is integral to making sure measurements are accurately comparable and understood in a global context.
For instance, force is measured in newtons (N), which is derived from base units as \( \text{kg} \times \text{m/s}^2 \). Converting complex units back to these base units simplifies calculations and standardizes results across various fields of study. In the provided exercise solutions, tons (t) and days (d) were converted into kilograms (kg) and seconds (s), respectively, aligning the units with the base SI units for mass and time. This process is integral to making sure measurements are accurately comparable and understood in a global context.
Scientific Notation
Scientific notation is a method used to express very large or very small numbers in a concise form. It consists of a decimal part, which is a number greater than or equal to 1 but less than 10, and an exponential part, which shows the power of ten that the decimal part must be multiplied by. This notation is particularly useful in physics and other sciences where extreme ranges of values are common.
For example, the number 453,000 is written in scientific notation as \(4.53 \times 10^5\). In the context of unit conversion, scientific notation helps to clearly represent the magnitude of quantities, especially after converting prefixes into their corresponding exponential values. When simplifying expressions, the final result can often be most conveniently expressed in scientific notation, as seen in the conversion resulting in \(1.81 \times 10^{-10} \text{S.m/s.kg}\). This approach alleviates confusion and makes arithmetic with large or small numbers more manageable.
For example, the number 453,000 is written in scientific notation as \(4.53 \times 10^5\). In the context of unit conversion, scientific notation helps to clearly represent the magnitude of quantities, especially after converting prefixes into their corresponding exponential values. When simplifying expressions, the final result can often be most conveniently expressed in scientific notation, as seen in the conversion resulting in \(1.81 \times 10^{-10} \text{S.m/s.kg}\). This approach alleviates confusion and makes arithmetic with large or small numbers more manageable.