## Unit Conversions

We often need to convert units from one standard to another for example, mile to meter, hour to second, meter to inch, feet to meter, kilogram to gram etc. You need to know the length equivalent of meter to inch or kilometer to mile etc. such as $1\text{m}=39.37\text{in}$ or $1\text{km}=0.6214\text{mi}$.

There are various standards of units and we are considering only the SI units and you need to know the equivalent of SI units to other units as above to convert the units to the SI system. For example you are currently converting $2.5\text{hr}$ to second. The easy and memorable way to do this is multiplying $2.5\text{hr}$ by $\frac{3600\text{s}}{1\text{hr}}$ since you already know $1\text{hr}=3600\text{s}$. Notice that $\frac{3600\text{s}}{1\text{hr}}$ is $1$ and multiplying by $1$ doesn't change the value. Another thing is you can cancel or multiply the units which also helps to determine whether the units are consistent or not on both sides of an equation.

Another example is to convert $45\text{km}$ to mile. You may know $1\text{mi}=1.609\text{km}$. Now we multiply $45\text{km}$ by $\frac{1\text{mi}}{1.609\text{km}}$ in order to cancel $\text{km}$. Note that multiplying by $\frac{1\text{mi}}{1.609\text{km}}$ is the same thing as multiplying by $1$ as you already know $1\text{mi}=1.609\text{km}$.

$$45.00\overline{)\text{km}}\left(\frac{1\text{mi}}{1.609\overline{)\text{km}}}\right)=27.97\text{mi}$$Another thing in unit conversions is that you can carry the units throughout the calculation in problems to see whether the units are consistent or not. It means that that if you are finding the speed of a car, the unit of speed should come as $\text{m/s}$ on both side of the equation $v=\text{distance/time}$. We always carry the units through the calculation as it is a good way to know the units are consistent or not in our calculation.

Example 1| Convert one cubic inches to cubic meter.

You should know $1\text{m=39}\text{.37in}$ and $1{{\text{m}}^{3}}={{(9.37\text{in)}}^{3}}$ so we multiply $\text{1i}{{\text{n}}^{3}}$ by the ratio $\frac{1{{\text{m}}^{3}}}{{{(39.37)}^{3}}\text{i}{{\text{n}}^{3}}}$ and multiplying by $1$ does not change the value.

$${\text{1in}}^{3}=\overline{)1{\text{in}}^{3}}\left\{\frac{1{\text{m}}^{3}}{{(39.37)}^{3}\overline{){\text{in}}^{3}}}\right\}=1.64\times {10}^{-5}{\text{m}}^{3}$$## Rule and Theory

Physics tries to understand the phenomena which are occurring in nature as well as how the physical universe behaves. Simply you know physics as a collection of facts based on phenomena happening in the nature and also as rule which associates certain phenomena in a more general idea. There is something called *range of validity* which specifies the range in which a particular rule is valid.

You can also understand a rule as theory but there is a little difference between them. The difference is that theory is not just a random statement; it’s a more general rule which gives an explanation of phenomena based on facts and experiment. No theory can be taken as an ultimate explanation. Every theory has a limit which is valid only in certain and special condition also known as range of validity.

## Dimensional Analysis

Dimensions are another way to express the physical quantities (or the standards of physical quantities) in a *straight way*. We can check whether the dimensions are consistent or not in both side of an equation. We denote the dimensions of length, mass and time to be the upper case letters L, M and T respectively. Now the dimension of velocity is $\frac{\text{L}}{\text{T}}=\text{L}{{\text{T}}^{-1}}$ whose SI unit is ${\text{m}}/{\text{s}}\;$. Similarly the dimension of acceleration is $\frac{\text{L}}{{{\text{T}}^{2}}}=\text{L}{{\text{T}}^{-2}}$. Now we check whether the dimensions are correct or not on both side of the following equation.

\[{{v}_{x}}={{a}_{x}}t\]

Now we equate dimensions on both side of the equation.

\[\begin{align*} \left[ \text{L}{{\text{T}}^{-1}} \right] &=\left[ \text{L}{{\text{T}}^{-2}} \right]\left[ \text{T} \right] \\ \text{or,}\quad \left[ \text{L}{{\text{T}}^{-1}} \right] &=\left[ \text{L}{{\text{T}}^{-2+1}} \right] \\ \text{or,}\quad \left[ \text{L}{{\text{T}}^{-1}} \right]& =\left[ \text{L}{{\text{T}}^{-1}} \right] \end{align*}\]

Now we consider ${{a}_{x}}\propto {{v}_{x}}^{m}{{t}^{n}}$ and find out the actual exponents by dimensional analysis. Now we equate the dimensions on both side of ${a_x} = k{\kern 1pt} {v_x}^m{t^m}$ where $k$ is the proportionality constant.

\[\begin{align*} \left[ \text{L}{{\text{T}}^{-2}} \right]=&k{{\left[ \text{L}{{\text{T}}^{-1}} \right]}^{m}}{{\left[ \text{T} \right]}^{n}} \\ \text{or,}\quad \left[ {{\text{L}}^{1}}{{\text{T}}^{-2}} \right]=&k\left[ {{\text{L}}^{m}}{{\text{T}}^{-m+n}} \right] \end{align*}\]

The equation is correct if the dimensions on both sides are correct. Now we solve for $m$ and $n$, and we get $-m+n=-2$ and $m=1$. So $m=1$ and $n=-1$. Now putting the value of $m$ and $n$, we get, ${a_x} = k{\kern 1pt} {v_x}^1{t^{ - 1}}$ and therefore,

$${a_x} = k{{{v_x}} \over t}$$

And in this case $k=1$ and ${{a}_{x}}=\frac{{{v}_{x}}}{t}$.

## Estimating Problems

Estimating problems is a kind of problem solving where we estimate or guess the situation and find out the solution. The solution may not be the exact value but a rough estimation. We make a rough estimation and make a general conclusion about the situation imposed in the problem and our target is to get the value nearest as possible to the exact value. Such kind of calculation is also called the **order-of-magnitude estimates**. For example estimate how many times we blink our eyes in our life time. Firstly we estimate that we blink 10 times per minute. And let the average life time of a person is $\text{85}$ years. You should be a little bit closer to the average life time of a person which may not be $\text{85}$ years. Now we calculate how many minutes there are in 85 years which is

The total number of blinks is,

$44676000\overline{)\mathrm{min}}\left(\frac{10\text{\hspace{0.17em}}\text{times}}{1\overline{)\mathrm{min}}}\right)=446760000\text{\hspace{0.05em}}\text{\hspace{0.17em}}\text{times}$which is about $\text{4}\text{.47}\times \text{1}{{\text{0}}^{8}}\,\text{}$ times.

Example 2| Estimate the total number of breaths a person takes in the life time.

First we estimate the average life time of a person to be $85$ years. Suppose that a person takes 10 breaths per minute. Now we calculate the total number of minutes in $85$ years and multiply it by 10 to get the total number of breaths that a person takes in all life time.

$$\text{85}\overline{)\text{yr}}\left(\frac{365\overline{)\text{day}}}{1\overline{)\text{yr}}}\right)\left(\frac{24\overline{)\text{hr}}}{1\overline{)\text{day}}}\right)\left(\frac{60\mathrm{min}}{1\overline{)\text{hr}}}\right)=44676000\mathrm{min}$$So the total number of breadths that a person takes in the life time is

$44676000\overline{)\mathrm{min}}\left(\frac{10\text{\hspace{0.17em}}\text{breaths}}{1\overline{)\mathrm{min}}}\right)=446760000\text{\hspace{0.17em}}\text{breaths}$which is about $\text{4}\text{.47}\times \text{1}{{\text{0}}^{8}}\,\text{breaths}$. Again if you are estimating, be closer to the data as much as you can.