Middle school physics questions about air pressure and buoyancy

1.Quickly pull the leather bowl upwards, so that the total volume of gas between the leather bowl and the sundries becomes larger, the pressure becomes smaller, the blockage is reduced by the downward pressure, less than the upward pressure, the debris will move upward, thus becoming loose, (for example: put a piece of paper on the table, cover it tightly with your hands, quickly lift the palm, and the paper will move upwards.)

Once loosened, it can be carried away by the current.

2.(1) p = water density * g * h

2) f buoyancy = water density * g * v row.

3) F buoyancy = g ice = m ice g = ice density * v ice * g

It's time to start school, study hard.

When sinking to the bottom, for the object, it is subjected to gravity, buoyancy, support force, equilibrium state gravity = buoyancy + support force, so support force = gravity buoyancy, the support force is that the pressure of the object on the container is equal, that is, pressure = gravity buoyancy, plus the pressure of the liquid on the container, that is, the total pressure = liquid gravity + object gravity object buoyancy.

Rope that's right.

The last one, levitation, floating, and sinking to the bottom, all belong to the equilibrium state, and the pressure exerted by the object on the bottom of the container is actually equal to the gravitational force of the liquid part of the liquid occupied by the object, so it is equivalent. Think of it this way: in the case of suspension, the density of the object and the liquid is the same, so the pressure at the bottom of the container is the same.

In the floating state, buoyancy = object gravity = gravity of discharging water, so it is equivalent to pulling out that part of the water and putting in water of equal gravity, so the pressure on the bottom of the container is still unchanged, which is equal to the rest of the bottom of the container. In the state of sinking to the bottom, buoyancy = gravity of draining water = gravity of the object Support force, then the gravity of the object = the gravity of the drained water + the support force (the pressure of the object on the container), that is to say, part of the gravity of the object directly acts on the bottom of the container as pressure (or pressure), and the other part replaces the gravity of the water in the original position, this part is equivalent to the water discharged, so the pressure (or pressure) of the water on the bottom of the container is also the same.

Dizzy, I'll add to that.

1.Enlarge. Can't use p=pgh, because the depth of the liquid changes! Do your own research. The wooden block should rise, and the liquid level will fall. However, considering the whole, the total g increases, and the bottom area remains the same, so the pressure increases.

2.No change. Floating, buoyancy equals gravity, and the gravity of the wooden block does not change, so the buoyancy does not change!

Hope it works for you.

Regarding the bottom of the cup (the change in the pressure of the internal water on the bottom is determined by the change in the rise of the water level, I think you know this, the situation analysis is the same for both times).

About cup-pair tabletops.

The first question is simple:

Since the object is already in the cup, it is clear that the overall increase in pressure on the table equals the gravitational force of the object, so you are right.

The second one is different:

Since the object does not fall into the cup, it is also subject to a pull force of the rope, so the increase in pressure is less than the size of this pulling force, so it should be removed after the weight of the object is reduced by the rope tension.

Do you understand? Friend.

, the increased pressure at the bottom of the cup is equal to the pressure generated by increasing the depth of water after the object is placed, and the increased pressure on the table is equal to the pressure generated by the increased gravity of the object.

2. Hang an object with a string and put it into the cup, the pressure increased at the bottom of the cup is equal to the pressure generated by increasing the depth of water after putting into the object, and the increased pressure on the table is equal to the increased gravity of the object minus the pulling force of the string

This is the original question on the ninth grade physics Qidong topic 4, and the answer is:

Let the density of the liquid be p, the physical density be p1, the bottom area of the container be s, and the volume of the object be v.

The first time: p*g*v row = p1*g*v

Because the water surface rises by 7cm, V row =

2nd time: v row = v = (

Substitute v=.

--p1:p=7:5

Because p=1*10 3kg m3

So p1=

1.The volume of the aluminum ball.

V-ball = m aluminium p aluminium = 540g

Aluminum balls are submerged. The buoyancy experienced while in the water.

f float = p water.

V-ball g = 1000 kg m *

2.The height at which the water surface rises.

h=v-ball s=

After the aluminum ball is inserted, the pressure of the water on the bottom of the container increases.

p = p water gh = 1000 kg m * 10n kg * 3The gravity of an aluminum ball.

G-ball = M-ball G=

When the rope is just broken, f pull = 4n.

f float = g ball - f pull =

v row = f float p water g =

v dew = v ball - v row =

1. From the force analysis, it can be seen that the buoyancy force on the rectangular block is equal to the gravitational force 10n, and the buoyancy = the density of the liquid multiplied by the volume of water discharged by the object and then multiplied by g

That is, 10 = 1000xvx10

p=1*10^3*10*

2. The large cardboard will be "lifted" by the plastic bag first.

The pressure of the large cardboard on the plastic bag is small.

p=f sf=g coin.

p large cardboard = g coin s large.

p small cardboard = g coin s small.

p. Large cardboard p. Small cardboard.

The large cardboard will be "lifted" by the plastic bag first.

The first question is problematic, the wooden block may be sitting diagonally.

Even if it is flat into the water, there are two different ways, lying down, vertically to solve the problem: calculate the depth of the wooden block immersed in the water, the bottom area * depth = 10 steak volume of boiling water, the pressure is equal to 10 ox divided by the bottom area. Count it yourself.

There is a problem with the second question. It must be the one who is close to the straw to get up first, and if you blow hard, you will jump.

Problem solving: The pressure problem, more in-depth is the dimensional problem. The large cardboard weighs twice as much as the small cardboard, but the area is 2*2 times, so the pressure is small and it is lifted first.

ps。What kind of question is this, the mentally handicapped still wants to be a teacher, you can go and give him a big mouth.

1.The 10 Ox cuboid is 10 cm, 10 cm, and 20 cm long on each side. When it floats on the surface of the water, the block is immersed in water to a depth of 5 cm, and the pressure of the water on the bottom of the block is 500 Pa.

I think that if it is a basically uniform wooden block, the 20 cm side should be parallel to the water, not standing in the water.

2. The large cardboard gets up first.

The pressure inside the plastic bag is the same, the area of the large cardboard is 4 times the area of the small cardboard, the upward force is 4 times that of the small cardboard, and the downward pressure (the gravity of the coin) is only 2 times that of the small cardboard, so the large cardboard rises first.

Physical wolf pack.

When the hot air balloon is suspended in the air...

f float = kilogram cubic meter * 10 N kilogram * 1000 cubic meter = 12900n and because buoyancy is equal to gravity when suspended

So the dead weight is also 12900N

Levitation definition: ffloat=g, levitation.