Please look up some data and assumptions on google if needed.... just indicate t
ID: 3308123 • Letter: P
Question
Please look up some data and assumptions on google if needed.... just indicate the assumptions in the answer
Part D:
4. In this problem we will consider various aspects of how a shark maintains its depth in the water. Sharks have a problem in that, lacking swim bladders, they tend to be negatively buoyant. They have to keep swimming to generate lift to maintain their depth in the water. We will also in the process consider the process of "induced drag" (sometimes called "lift-induced drag"), which is drag caused by the generation of lift. (This question is worth twice as much as the previous questions.)Explanation / Answer
A:
Total density of shark d = 1076*0.86 + 964 *0.14 = 1060.32 kg/cu.m
86% made of density 1076 and 14% mad of 964
For the shark to keep floating at a const depth, the gravitaional force must be equal to the buoyant force, the density being more than the water it must apply additional lift force
Let Fl be the lift force,
volume of the shark V = 0.0138 cu.m
weight of the shark = dVg = 1060.32*0.0138 g
buoyant force is equal to the wieght of water dispalced by the shark body volume (0.0138 cu.m)
= 1024*0.0138g , with salt water
= 1000*0.0138g , with fresh water
shark weight = weight of water disaplced + Fl ( lift force)
Fls = 0.0138*9.8(1062.32 -1024) = 5.18 N -- in salt water
Flf = 0.0138*9.8(1062.32 -1000) = 8.43 N -- in fresh water
D :
In fresh water the lift force Ff = 8.43 N
In salt water the lift force Fs = 5.18 N
work done for moving a disatnce s = F*s
Power = energy /unit time = Fs/t , t is the time for the force applied.
ratio of power from fresh water to salt water = Ff /Fs = 8.43/5.18 = 1.63
The shark must spend 63 % more power in fresh water than in salt water.
F: let a be the volume of live and b that of rest of the body
density = 1076b + 964a
lift force in fresh water
= 0.0138*9.8(1076b + 964a -1000) = 5.13 ; lift force in salt water
964a +1076b = 1037.93
a +b = 1, total body
a = 0.34
The liver part of the shark must make 34% of the total body volume so that it compensates for the additonal drag in fresh water.