Answer:
Principle Archimedes is applied in building a ship and submarine using the manipulating that buoyancy, is controlled the ballast tank system.
Explanation:
Submarine is rather had they focused on main parts of the submarine,he is complex and long process implementation,the most submarine design like submarine stability.
Submarine stability is complete and the fundamental Archimedes principle to arrive the weight of submarine is equal to buoyancy force.
Submarine into the parts and components of ballast tank the sequence in diving and surfacing,there two vital parts:- flood parts and air vents
flood parts:- at the bottom position and allow water to enter or leave that tank.
air vents:- air vents at the top of the pressure hall,and that they submarine dive.
this time submarine is most modern system is depth is 300 to 450 meters,high pressure air is 15 bar is tank air valve.
submarine is basic of the effective volume of all the submarine surfaced condition,submarine minus to the free water flood is equal to the fully pressure hull,submarine is the surfaced condition.
Answer:
the answer should be the third statement
You used density, because water/ice has a density of 1, and ice will sink in anything with a lesser density
Answer:
The height is 
A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )
Explanation:
From the question we are told that
The height is 
The angle of the slope is 
According to the law of conservation of energy
The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

Where I is the moment of inertia which is mathematically represented as this for a sphere

The angular velocity
is mathematically represented as

So the equation for conservation of energy becomes
![mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5B%5Cfrac%7B2%7D%7B5%7D%20mr%5E2%20%5D%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B2%7D%7B5%7D%20%2B1%20%5D)
![mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]](https://tex.z-dn.net/?f=mgh_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%5B%5Cfrac%7B7%7D%7B5%7D%20%5D)
![gh_s =[\frac{7}{10} ] v^2](https://tex.z-dn.net/?f=gh_s%20%3D%5B%5Cfrac%7B7%7D%7B10%7D%20%5D%20v%5E2)

Considering a circular hoop
The moment of inertial is different for circle and it is mathematically represented as

Substituting this into the conservation equation above
![mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2](https://tex.z-dn.net/?f=mgh_c%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28mr%5E2%29%5B%5Cfrac%7Bv%7D%7Br%7D%20%5D%20%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Where
is the height where the circular hoop would be released to equal the speed of the sphere at the bottom



Recall that 


Substituting values

Answer:
Explanation:
Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3