I recall back in SFG, junior high dormitory, the senior prefect and bully who I had offended had asked me, "would you prefer to take a 'standard punishment' or perhaps a 'compound punishment'?" You really don't want to know what a 'standard' or 'compound' punishment is. However, hereafter, you would be learning about problems with compound units, their lists, as well as conversions between standard and compound units. Tighten your seat belt.
Problems on compound units
Problems of compound units are all tasks involving compound units that are not limited to but include the derivation, conversion, identification, and overall application of compound units in all fields.
If you don't get prepared for compound units, it may turn into a compound punishment, which may over time turn into a standard punishment.
Moving forward, it is indeed integral to define both compound and standard units.
What are standard units?
A standard unit is a simple and single unit of measurement generally used for a quantity.
For example, centimetres, seconds, kilograms, centigrade, etc.
Note that it is not in combination with any other unit. So, an area measured in \(m^2\) or even a volume in \(cm^3\) is still a standard unit since only one unit which is meters or centimetres is in operation.
What is a compound unit?
Compound units are units of measurement which comprise two, or more, different units.
They may be referred to in the unit as:
a combination of two, or more, standard units – such as speed measured in \(ms^{–1}\)
or as a new unit – such as force measured in newtons \((N)\).
Examples of compound units are \(ms^{–2}\), \(kgm^{–3}\), pascal \((Pa)\), joules, and watts.
Do not mistake standard units for SI units. SI units comprise both standard and compound units internationally (globally) used in measuring quantities.
Lists of compound Units
Apparently, there are several compound units. Herein, the basic standard units which are indeed needed in your routine calculations will be enlisted in the table below.
| Quantity | Compound unit |
| speed and velocity | \[ms^{-1}\] |
| acceleration and retardation | \[ms^{-2}\] |
| density | \[kgm^{-3}\] |
| force | \[N\] or \[kgms^{-2}\] |
| work | \[J\] or \[kgm^2s^{-2}\] |
| power | \[W\] or \[kgm^2s^{-3}\] |
| pressure | \[Nm^{-2}\] or \[Pa\] |
| molar concentration | \[moldm^{-3}\] |
Converting compound units
Having understood the definitions and examples of standard as well as compound units, it would become less challenging to convert from standard to compound units and vice versa.
How do we convert standard units to compound units?
This is the most common instance in unit conversion. It involves combining two or more standard units which are different. Such combinations involve simple to complex operations of multiplication and division. Note that some could involve exponentials as well as roots.
The two most important questions to ask when making these conversions or derivations are, "what standard units are involved?" Also, "what operations are involved?" This second question is usually answered when you know the formula used in calculating the compound quantity. A compound quantity is a quantity that is derived from the combination of two or more quantities.
Determine the unit of measuring a quantity \(H\) which is a quotient of distance and time.
Solution:
Here, the quantity \(H\) is derived from dividing distance by time.
Step 1: Find the unit of component quantities. Component quantities are quantities that are used in deriving a compound quantity. Here, the component quantities are distance and time. The unit of distance metres, \(m\), and the unit of time is seconds, \(s\).
Step 2: Apply the operation needed. In this case, it is the quotient of distance and time. Thus, we are dividing so that \[H=\frac{m}{s}\]
Hence, the unit of the quantity, \(H\) is \(ms^{-1}\).
Work is the product of force and distance. Find the unit of work.
Solution:
We have been told that work is the product of force and distance.
Step 1: Find the unit of component quantities. Here, the component quantities are force and distance. The unit of force is newtons, \(N\), and the unit of distance is meters, \(m\).
Step 2: Apply the operation needed. In this case, it is the product of force and distance. Thus, we are multiplying so that if work is \(W\), then, \[W=N\times m\]
Hence, the unit of the quantity, \(W\) is \(Nm\).
This is generally known as Joules, \(J\).
Converting compound units to standard units
Just as compound units are derived from a combination of standard units, standard units can be determined when compound units are broken by either standard or other compound units. Some examples below would elaborate better.
If the density of a material is \(5\, kgdm^{-3}\), find the mass when its volume is \(20\, dm^3\).
Solution:
Step 1: Look through the units given, and classify the units into compound and standard units. Note sometimes, both units may be compound units. In such cases, determine which compound unit is more complex. This is often easy to spot because the more elements in a unit, the more complex it is.
For instance, \(kgm^{-3}K^{-1}\) is more complex than \(dm^3\).
This question provides two units, a compound unit, \(kgdm^{-3}\) and a standard unit, \(dm^3\).
Step 2: Determine the relationship between both units by comparing both. Try to see differences. If the standard (or less complex compound) unit is seen to be part of the more complex unit, and it is not different in its exponentials, then, you divide. Otherwise, you should multiply both quantities. When doing this, do so with the units because you wish to determine the unit of the unknown too.
In this case, the unit \(dm^3\) which is found in \(kgdm^{-3}\) has a different exponential. Because the standard unit has an exponential of \(3\), while the compound unit has an exponential of \(-3\) for \(dm\). This suggests you are to multiply.
Step 3: Carryout the operation explicitly. Thus, \[5\, kgdm^{-3}\times 20\, dm^3=5\times20\times kg\times dm^{-3}\times dm^{3}\]
Note that \[dm^{-3}\times dm^{3}=1\]
Hence, the mass \(m\) of the material is \[m=5\times 20\times kg\times1\] This gives \[m=100\, kg\]
Notice that our answer has a unit, therefore, the unit for mass is \(kg\).
Converting from one compound unit to another
Sometimes you may need to convert from one compound unit to another. This may take place within the same unit, or it may involve an entirely different unit used to measure the same quantity.
Conversion between similar compound units
This takes place when you wish to convert the same units which vary based on the exponential of \(10\). Exponential of \(10\) means, \(10^a\) or \(10^{-a}\), where \(a\) is \(1\), \(2\), \(3\)...\(n\).
The power produced by a machine is \(15\, kW\). Find the power of the same machine in \(MW\).
Solution:
Step 1: Identify the units involved. Your value has been given in kilowatts, \(kW\), while your answer is in megawatts, \(MW\).
Step 2: Define the relationship between units. If \[1\, kW=10^3\, W\] and \[1\, MW=10^6\, W\] with \[10^6=10^3\times10^3\] we can say \[1\, MW=10^3\times10^3\, W\]
Since \[1\, kW=10^3\, W\] it surely means that
\[1\, MW=10^3\times1\, kW\]
Hence,
\[1\, MW=10^3\, kW\]
or
\[1\, MW=1000\, kW\]
Step 3: So we would have to convert \(15\, kW\) to \(MW\) which is
\[15\, kW=\frac{15}{1000}\]
So our answer is \(0.015\, MW\)
Converting between compound units for the same quantity
Sometimes, when a quantity has more than one unit you may be required to present your answer in another compound unit.
If \[1\, Pa=7.5\times10^{-3}\, mmHg\]
and the pressure exerted on a body is \(4\, Pa\), express the pressure in \(mmHg\).
Solution:
Since
\[1\, Pa=7.5\times10^3\, mmHg\]
then,
\[4\, Pa=4\times7.5\times10^{-3}\, mmHg\]
So we have
\[4\, Pa=3\times10^{-2}\, mmHg\]
Compound units and rates
Essentially, rates tend to compare quantities, during such comparisons, compound units are formed. Most times, rates are ideal when it involves money or time.
Wage and salary in compound units
Wage and salary tell us how much someone is paid over a period. This relationship is also known as wage rate or salary rate. It provides a compound unit which is usually a certain amount per time.
A man is paid \(£120\) for \(6\) hours of labour. What is his hourly wage?
Solution:
We intend on knowing how much he is paid for 1 hr. So, \[£120=6\, hrs\] Divide both sides by \(6\) to arrive at \[£20=1\, hr\]
Now to derive our compound unit, divide, by \(1\, hr\) to get \[£20\, hr^{-1}=1\]
The \(1\) signifies \(1\) unit of labour which is interpreted as \(1\) unit of labor at the rate of \(£20\, hr^{-1}\).
Note that the compound unit here is \(£hr^{-1}\).
Price in compound units
The price of goods is also very relevant as compound units can be derived from it. For instance, \(£6barrel^{-1}\) of fuel, \(£2sweet^{-1}\) and so on.
Examples of compound units
More practice would increase your ability to solve questions on compound units.
Classify the following quantities in a table based on the compound and standard units; displacement, velocity, volume, area, force, pressure, electromagnetic induction, and electric current.
Solution:
The compound and standard units can be arranged as
| Standard Units | Compound Units |
| electric current | velocity |
| displacement | force |
| volume | pressure |
| area | electromagnetic induction |
Imisi covers a distance of \(500\, m\) in \(160\, s\), find the Imisi's speed in miles per hour.
Solution:
We know speed is usually measured in \(ms^{-1}\), but our answer this time is to be in another compound unit that measures speed.
Step 1: Convert the related components. Here, we have two components, distance and time. So, convert the distance in meters to miles. If \[1\, m=6.2\times10^{-4}\, mi\]
Then,
\[500\, m=500\times6.2\times10^{-4}\, mi\]
\[500\, m=3.1\times10^{-1}\, mi\]
Likewise, convert seconds to hours. If
\[1\, s=2.78\times10^{-4}\, hr\]
Then,
\[160\, s=160\times2.78\times10^{-4}\, hr\]
\[160\, s=4.448\times10^{-2}\, hr\]
Step 2: Calculate the speed with new values.
\[\frac{3.1\times10^{-1}\, mi}{4.448\times10^{-2}\, hr}\]
Hence the speed in \(mihr^{-1}\) is \(7\, mihr^{-1}\).
Compound Units - Key takeaways
- Problems of compound units are all tasks involving compound units that are not limited to but include the derivation, conversion, identification and overall application of compound units in all fields.
- Compound units are units of measurement which comprise two, or more, different units.
- Just as compound units are derived from a combination of standard units, standard units can be determined when compound units are broken by either standard or other compound units.
- Sometimes you may need to convert from one compound unit to another.
- Rates tend to compare quantities, during such comparisons, compound units are formed
Explanations 
Exams 
Magazine 
