The United States possesses the ten largest warships in the world, aircraft carriers of the Nimitz class. Suppose one of the ships bobs up to float 11.0 cm higher in the ocean water when 50 fighters take off from it in a time interval of 25 min, at a location where the free-fall acceleration is 9.78 m/s2. The planes have an average laden mass of 29 000 kg. Find the horizontal area enclosed by the waterline of the ship.

Question

anonymous · Answer

I can't figure out why it has a time frame ....

anonymous · Answer

*****I don't know if this is right*****  but it's at least a start :P

The time frame is there to imply the ship is in equilibrium at both times methinks.  Anyway

the volume of the water the boat displaces is given by

$$m=\rho V_{disp}$$
so the change in volume would just be 
$$\Delta M=\rho \Delta V_{disp}$$
$$ \rho = \ \text{density of water} = 1000 \ kg/m^3$$
the change in mass is 
$$\Delta M =  (\cancel{M_{ship}} - 50*29000 \ kg)-\cancel{M_{ship}} \ =-m= -1450000 \ kg$$

$$|\Delta V_{disp}|=\frac{|m|}{\rho} = 1450 \ m^3$$

The height of the displacement is given in the problem as 

$$h=.11m$$

So, in theory dividing the volume by its height should give the area

$$A_{waterline} = \frac{V_{disp}}{h} = \frac{1450 \ m^3}{.11 \ m} \approx 13182 \ m^2$$

And according to Wikipedia it's about 317 meters long, which would make it 41 meters wide, so that's not too far off maybe? . 

Again, no idea, just thought I'd throw my 2 cents in.

anonymous · Answer

@ahmedmido dunno if this helps :P