# State the units for intensity using only fundamental units? (Hint: work from W/(m^2))

Aug 6, 2018

$\frac{k g}{{s}^{3}}$

#### Explanation:

Steps shown below:

$\frac{W}{m} ^ 2 = \frac{\frac{J}{s}}{m} ^ 2 = \frac{\frac{N m}{s}}{m} ^ 2 = \frac{\frac{\frac{k g m}{s} ^ 2 \cdot m}{s}}{m} ^ 2 = \frac{\frac{\frac{k g {m}^{2}}{s} ^ 2}{s}}{{m}^{2}} = \frac{\frac{k g {m}^{2}}{s} ^ 3}{{m}^{2}} = \frac{k g}{{s}^{3}}$

Aug 6, 2018

$\frac{k g}{s} ^ 3$
kilograms over seconds cubed

#### Explanation:

${m}^{2}$ is already in fundamental units, so let's work on $W$ (which stands for power)

$W = \frac{F d}{t}$

The fundamental unit for distance is $m$ and for the time it's $s$

$F = m a$

The fundamental unit for mass is $k g$ and the unit for acceleration is $\frac{m}{s} ^ 2$

So, we know that $\frac{\frac{k g \cdot \left(\frac{m}{s} ^ 2\right) \cdot m}{s}}{m} ^ 2$

Simplifying we get $\frac{k g}{s} ^ 3$