Why is ( kg ⋅ m 2 ) / s written as "Joule second" when there is no joule present, and time is divided and not multiplied?

Question

Why is   (  kg  ⋅      m    2    )      /    s written as &quot;Joule second&quot; when there is no joule present, and time is divided and not multiplied?

Jeremy Mayo · Accepted Answer

One Joule is defined as  1   J  =  1            kg      ⋅              m        2                    s      2        .Hence,  1   J  ⋅  s  =  1            kg      ⋅              m        2              s    .

hexaedru8p · Accepted Answer

The Joule-second is important because it is the unit of   ℏAs far as the Joule is concerned, we know work is force (  F  =  m  a) times distance, so energy has units:      k    g    ⋅    m          /              s      2        ×    m    =    k    g                  m        2                    s        2              =    k    g                  (                    m      s                                )                    2      which agrees with kinetic energy being       1    2    m      v    2  Perhaps   ℏ&#039;s most famous role is relating frequency and energy:  E  =  ℏ  ωor  ℏ  =      E    ω  So that unit is Joules per Hertz (J/Hz), which is a more natural way to view the relation than Joules times seconds, though they are equivalent.

Why is (\text{kg} \cdot \text{m}^2)/\text{s} written as"Joule second when there is no joule present?

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