Answer
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Hint :In order to solve this question, we are going to first know what the CGS and the SI systems of units are and how are they interrelated. Then, using that relational formula for the two systems, we are going to convert the CGS unit $ 1\,erg $ energy into the SI unit that is Joule.
The formula for the energy is given by
$ KE = \dfrac{1}{2}m{v^2} $
The dimensional formula for the energy is $ \left[ {M{L^2}{T^{ - 2}}} \right] $
Complete Step By Step Answer:
The CGS system of units means Centimeter-Gram-Second where centimeter unit is for the distance, gram is for the weight and the second is for time.
While the SI system of units is the standard one and consists of the meter for distance, kilogram for weight and second for time measurement.
Now, the formula for the energy is given by
$ KE = \dfrac{1}{2}m{v^2} $
The dimensional formula for the energy is $ \left[ {M{L^2}{T^{ - 2}}} \right] $
Taking the CGS system, we get the unit for one unit of energy as
$ gc{m^2}{s^{ - 2}} $ which is equal to $ 1\,erg $
Now converting this to SI system
$ 1erg = 1gc{m^2}{s^{ - 2}} = 1 \times {10^{ - 3}}Kg \times {\left( {{{10}^{ - 2}}} \right)^2}{m^2} \times 1{s^{ - 2}} \\
\Rightarrow 1erg = {10^{ - 3 - 4}}Kg{m^2}{s^{ - 2}} = {10^{ - 7}}Joule \\ $
Therefore, $ 1\,erg $ energy is equivalent to $ {10^{ - 7}}Joule $ of energy in the SI system.
Note :
Usually, we prefer to take the unit of the energy in Joules that is the units of the quantities in the SI system of units. If the energy value is so small that it is given in $ erg $ then, we convert it into joules before further processing as the SI system of units is the most widely used one and the easiest to approach.
The formula for the energy is given by
$ KE = \dfrac{1}{2}m{v^2} $
The dimensional formula for the energy is $ \left[ {M{L^2}{T^{ - 2}}} \right] $
Complete Step By Step Answer:
The CGS system of units means Centimeter-Gram-Second where centimeter unit is for the distance, gram is for the weight and the second is for time.
While the SI system of units is the standard one and consists of the meter for distance, kilogram for weight and second for time measurement.
Now, the formula for the energy is given by
$ KE = \dfrac{1}{2}m{v^2} $
The dimensional formula for the energy is $ \left[ {M{L^2}{T^{ - 2}}} \right] $
Taking the CGS system, we get the unit for one unit of energy as
$ gc{m^2}{s^{ - 2}} $ which is equal to $ 1\,erg $
Now converting this to SI system
$ 1erg = 1gc{m^2}{s^{ - 2}} = 1 \times {10^{ - 3}}Kg \times {\left( {{{10}^{ - 2}}} \right)^2}{m^2} \times 1{s^{ - 2}} \\
\Rightarrow 1erg = {10^{ - 3 - 4}}Kg{m^2}{s^{ - 2}} = {10^{ - 7}}Joule \\ $
Therefore, $ 1\,erg $ energy is equivalent to $ {10^{ - 7}}Joule $ of energy in the SI system.
Note :
Usually, we prefer to take the unit of the energy in Joules that is the units of the quantities in the SI system of units. If the energy value is so small that it is given in $ erg $ then, we convert it into joules before further processing as the SI system of units is the most widely used one and the easiest to approach.
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