Question

Can someone help me on how to do this?

Answer 1

Explanation:

Recall that

$1\:\text{micron} = 1\:\mu\text{m} = 10^{-6}\:\text{m}$ so the speed of the bacterium is

$v = \dfrac{6.82×10^{-6}\:\text{m}}{3.5\:\text{s}} = 1.9×10^{-6}\:\text{m/s}$

Next, we convert this speed to km/hr. Recall that

$1\:\text{km} = 1000\:\text{m}$

$1\:\text{hr} = 60\:\text{minutes} = 3600\:\text{s}$

Therefore,

$1.9×10^{-6}\:\dfrac{\text{m}}{\text{s}}×\left(\dfrac{1\:\text{km}}{1000\:\text{m}}\right)×\left(\dfrac{3600\:\text{s}}{1\:\text{hr}}\right)$

$= 6.8×10^{-6}\:\text{km/hr}$