Pressure vs. buoyancy, pressure vs. buoyancy issues

1. In the process of the scientific research submarine floating from 10 meters below the surface of the water to 5 meters below the surface of the water, the water is on its ( ).

a Buoyancy becomes less and pressure remains the same.

b Both buoyancy and pressure are reduced.

c Both buoyancy and pressure remain constant.

d The buoyancy remains the same, and the pressure becomes smaller.

Answer: D2, the submarine that has been submerged in the seawater, in the process of continuing to dive into the deep sea, it is affected by (

a The pressure increases, and the buoyancy increases.

b Both pressure and buoyancy are constant.

c The pressure increases, but the buoyancy remains unchanged.

d The pressure remains unchanged and the buoyancy increases.

Answer: C3, about the sinking of the submarine "Curs", this is because of the ** after ( ).

a The wreckage of the submarine is no longer subject to buoyancy.

b The buoyancy of the wreck of the submarine is less than the force of gravity.

c The wreckage of the submarine drained a larger volume of water.

d The density of seawater has become less.

Answer: B4, in the process of the submarine surfacing from the water, the pressure of the water on the wall of the submarine ( ) between the non-surface, the submarine is subject to the buoyancy of the water ( ) when it surfaces, the buoyancy of the water on the submarine ( ) is filled in "larger", "smaller" or "unchanged"). When the submarine is floating on the water and moored at the dock, the buoyancy of the water against the boat ( ) The self-weight of the boat.

Fill in "equal", "greater than" or "less than").

Answer: Smaller, unchanged, smaller, equal.

5. Submarines are an important piece of equipment of the navy. The underwater displacement of a certain submarine is 1500 tons. When the submarine dives 100 meters below the surface of the sea, the pressure of the submarine is ( ) Pa and the buoyancy of the submarine is ( ) Ox.

The density of sea water is 3 kilograms, and g is 10 N kilograms).

6. A submarine with an aquatic displacement of 990t, an underwater displacement of 1200t, and a maximum diving depth of 250m cruising 10m below the sea surface in Bermuda, from the known conditions or adding one more condition, to find two physical quantities of pressure and buoyancy. (Open-ended questions).

A cube object with a density of 2 cm on the side is placed in the water (the bottom side is parallel to the surface of the water).

Find the pressure on the bottom surface.

If you don't know how to do it, you will be able to solve it, and you should be able to do it.

When floating at 100pa, the object: liquid v immersion: v total v immersion = 4cm3

h=v submerged s=

p=ρgh=100pa

This is easy to understand, for example, the ice cube is an iron block, you can consider the ice and the iron block separately, because the ice will not change even if the liquid level melts, so you only need to consider the iron block, because the iron block is floating, that is to say, the buoyancy force on the iron block is equal to the gravity of the iron block, and the density of iron is greater than that of water, so the volume of water to float the iron block must be greater than the volume of the iron block itself.

In fact, it is two cups of water of the same height, one with 1 mass of iron, and the other with 1 mass of water, so the water added is significantly higher than the iron added.

According to the relationship between density, volume, mass, and the equilibrium of forces on the two spheres (combined with the law of buoyancy), the equation can be obtained:

m vm g=gm= water g vm-t

n vn g=gn= oil g vn-t

t is a fine wire pull).

Thus there is: t=( water - m) g vm = ( oil - n) g vn and is known by: water - m> oil - n

So VM so choose A

A because of the equal restraint of the tensile force in the line, fm-gm = fn-gn, that is, (p water-pm)*vm = (p oil-pn)*vn, and (p water-pm) > (p oil-pn), so vm

Question 1: At the beginning, the contact area between the wine bottle and the desktop is S1, the height of the water is H1, the pressure is P1, the contact area between the bottle and the desktop after inversion is S2, the height of the water is H2, the pressure at this time is P2, it is known that P=G S, G is the total gravity is unchanged, it can be known from the shape of the wine bottle, S2 S1, H2>H1, the pressure on the table is > P2P1, if your problem is that the bottle mouth is the bottom after the handstand, then by P= Gh, you can get the pressure on the bottom of the bottle, if you still refer to the bottom of the original bottle, then it is P1 at the beginning, after the handstand, because the water is not full, then the pressure becomes zero.

Question 2: We know that the density of steel is not the same as that of lead, and their buoyancy is equal by the following two points: first, their mass is equal; Second, they are equal in buoyancy; You can judge that they should all be floating, that is, the density of this liquid is greater than steel, and it is also larger than lead, the density of steel I remember is about 7000 kg m, and the density of water, alcohol, and salt water are less than this value, only the density of mercury is larger, it seems to be more than 13000!

Therefore, it can be judged that this liquid is mercury.

The pressure of the bottle on the tabletop and the pressure of the wine on the bottom of the bottle increases.

The pressure of the wine bottle on the table top is the same pressure, but the force area becomes smaller.

The pressure of the wine on the bottom of the bottle is due to the fact that the bottle is small at the top and large at the bottom, and the depth of the liquid becomes greater when it is inverted.

When a solid steel ball of the same mass and a solid lead ball are put into the same liquid, the buoyancy is equal in magnitude, and this liquid must be: (mercury).

Mercury is denser than both.

1. The pressure of the wine bottle on the table will become larger, because p=f s, f does not change, and s becomes larger. The pressure of the wine on the bottom of the bottle becomes greater because p = ghG doesn't change, but H gets bigger, so the pressure gets bigger!

2.Mercury. Because both belong to floating, buoyancy is equal to gravity, and their masses are equal, so gravity is equal.

To make the buoyancy equal, then only the gravitational force of the object is equal to its buoyancy, so only in the floating state to meet this condition, so the mercury with the highest liquid density is selected.