Answer:
8.67807 N
34.7123 N
Explanation:
m = Mass of shark = 92 kg
= Density of seawater = 1030 kg/m³
= Density of freshwater = 1000 kg/m³
= Density of shark = 1040 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
Net force on the fin is (seawater)

The lift force required in seawater is 8.67807 N
Net force on the fin is (freshwater)

The lift force required in a river is 34.7123 N
from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"
here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?
Answer:
(A) 26 m/s
(B) 32.4 m
(C) v = 15.4 m/s
Explanation:
initial speed (u) = 6.4 m/s
acceleration due to gravity (a) = 9.9 m/s^[2}
time (t) = 2 s
(A) What is its speed after falling for 2.00s?
from the equation of motion v = u + at we can get the speed
v = 6.4 + (9.8 x 2) = 26 m/s
(B) How far does it fall in 2.00s?
from the equation of motion
we can get the distance covered
s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)
s = 12.8 + 19.6 = 32.4 m
c) What is the magnitude of its velocity after falling 10.0m?
from the equation of motion below we can get the velocity

v = 15.4 m/s
Please show picture of diagrams
The ones that interact would be Atmosphere, biosphere, and cryosphere. :)