Calculate how long it would take for diatoms to settle onto the titanic... depth

Question

Calculate how long it would take for diatoms to settle onto the titanic...

depth to the titanic = 12,500 feet or 3810 meters

2) From Wikipedia: In the deep ocean, marine snow occurs a continuous shower of mostly organic detritus falling from the upper layers of the water column. It is a significant means of exporting energy from the light-rich photic zone to the aphotic zone below. The term was first coined by the explorer William Beebe as he observed it from his bathysphere. As the origin of marine snow lies in activities within the productive photic zone, the prevalence of marine snow changes with seasonal fluctuations in photosynthetic activity and ocean currents. Marine snow can be an important food source for organisms living in the aphotic zone, particularly for organisms which live very deep in the water column. one component of marine snow, diatoms, range in size from about 10-150 m. An article in the Journal of Plankton Research (van lerland and Peperzak 1983) reported diatom mass density to be around 1.1 g cm3. a) Calculate how long it would take for diatoms to settle onto the Titanic. That is, about how old was the marine snow that James Cameron saw when he looked around down there?

Explanation / Answer

As the diatoms are falling from the uppermost layer of the ocean, they travel the entire length of 3810 meters before reaching Titanic.

The diatoms experience buoyant force in the water which leads to sinking of the diatoms is given by

Fb=density* Volume of diatom*g

Assuming the diatoms to be perfectly spherical, the Volume of the diatom for the largest diatom= (4/3)(3.14)(d/2)^3=1.766*10^(-12)m^3

Fb=1100kg/m^3*1.766*10^-12m^3*9.8m/s^2

Fb=1.9*10^-8N

Thus we know that acceleration a= 9.8 m/s^2

Assuming that the diatom falls linearly, we use the equation of motion

x=x0+v0del(t)+(at^2)/2

x is the distance traveled by the diatom = 3810 m

x0 is the initial displacement= 0

v0 is the initial velocity=0

Rearranging the equation,

time needed to travel the distance t=sqrt(2x/a)=sqrt(2*3810/9.8)=27.88 seconds

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