Physical quantities and units

​​In a nutshell

In physics, it is important to be able to communicate measurements and experimental readings accurately and clearly. For example, solving a question in the metric system of units will yield a very different result to if you worked in the imperial system. In this lesson, you will learn about this system of units, apply and understand their prefixes, and be able to estimate approximate values to questions. 

Definitions

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​​Base units

Base units are quantities that can be measured directly. For example, length and time can be measured with one simple step - by means of a ruler, or a stopwatch. There are seven of these base units - also called SI units.

SI base units

​​Derived units

Derived units are quantities that cannot be measured directly and instead consist of multiple steps. Speed, for example, is measured by observing the distance an object has travelled in a specific time. Speed as a quantity is said therefore to be derived from both distance and time, which are two base units, and is measured in metres per second.

Some derived quantities have their own names for units. Force, for example, is derived from the acceleration of an object with a particular mass, and is measured in kilogram metres per second squared. This is a mouthful, and is instead given the name Newton. Here are a few examples of these named derived units.

Some derived units

Example

Pressure is defined to be force divided by the area the force acts on. What is the SI unit for pressure?

First, write down the equation for pressure:

$$pressure = \dfrac {force}{cross-sectional\, area}$$​

Next, rewrite the equation in terms of the SI units:

$$pressure = \dfrac{kg\,m\,s^{-2}}{m^2}$$

​Finally, simplify the equation:​

$$pressure = kg\,m^{-1}\,s^{-2}$$​​

Therefore, the SI unit for pressure is $$\underline{kg\,m^{-1}\,s^{-2}}$$. This unit has the name Pascal, with the symbol Pa.

Prefixes

Physicists use a system of prefixes that is added to the front of the unit, that abbreviates multiplying by a very large or small amount. 

Measuring in base units is helpful - that is, until you are asked to measure something incredibly large (like the radius of the Sun), or incredibly small (like the diameter of an atom). So, instead of saying that the radius of the Sun is $$696000000 \, m$$​, you instead would say that it is $$696 \space megametres$$​, or $$Mm$$​.

Important prefixes

Example

Write $$500\,nm$$​ in standard form.

First, write down the question and replace the prefix with the factor:

$$500\,nm = 500 \times 10^{-9}\,m$$

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Finally, simplify the equation:

​$$500\,nm = 5 \times 10^{-7}m$$

Therefore, $$500\,nm$$​ in standard form is written as $$\underline{5 \times 10^{-7}m}$$​.

​​Estimation

Estimations are rough calculations to find an approximate answer to a problem. Physicists aren't always interested in a perfectly accurate answer, and so will look for estimations instead. A perfectly accurate answer to a problem is only correct if the problem remains unchanged - but when conditions of the problem are tweaked slightly, then the answer will also change slightly.

Example

Estimate the top speed of an average car.

To get a specific answer, you need to know the type of car, the road it's driving on, and other such conditions that may change the outcome.

Assuming average conditions, it's easy to estimate that a car probably has a top speed of $$\underline{120\, mph}$$, or about $$\underline{50\,ms^{-1}}$$.

The order of magnitude of your estimation is the most important part to get right - it describes in which power of ten scale your answer lies. It helps to know what order of magnitude the answer you're looking for is, so that you can quickly notice if your answer is outrageously large or small.

Example

Estimate the top speed of an average car.

For your potential answer, analyse whether it makes sense in terms of its order of magnitude. 

An answer of $$5\,ms^{-1}$$ is too slow - it's about as fast as the average running speed. Similarly, $$500\,ms^{-1}$$ is obviously far too quick.

The average car probably travels in the $$10^1$$ order of magnitude, so $$\underline{50\,ms^{-1}}$$ is a good estimate.