Help, physics questions in the second year of junior high school! How does the pressure of the conta

Two situations: 1. The water in the original container is full, and if you put your hand into the water, the water will overflow, and at this time, the pressure and pressure of the container on the table will remain unchanged.

2. If the water in the original container is not enough, put it in your hands, the water will not overflow (or overflow a little), and the pressure and pressure of the container on the desktop will become larger.

Explanation: The pressure of the container on the desktop = the support force (action and reaction force) of the desktop on the container

Perform a force analysis on the bottom of the container, and the support force of the table top to the container = the pressure of the liquid on the bottom of the container (equilibrium force).

The pressure of the liquid on the container = density * depth * g, pressure f = pressure p * area s. In this case, neither the density nor the g, nor the bottom surface of the vessel change, so the pressure and pressure are only related to the depth of the liquid level.

In case 1, the liquid level height does not change, so the pressure does not change, and thus the pressure changes.

In case 2, the liquid level rises, the pressure increases, and the pressure also increases.

When the hand enters the water, the water will give the finger an upward buoyant force, according to Newton's third law, the hand will give the water a downward force equal to the buoyancy force, who has a greater pressure on the container, according to p=f s, it can be seen that p also becomes larger.

The pressure will increase, and so will the pressure.

Because the surface of the water rises when you dip your finger in the water, and that part of the water that rises is extra when you dip your finger into the water, the pressure and pressure increase.

If you don't understand, you can ask me again!

Dipping your finger in the water is equivalent to an increase in the mass of the water, and the gravity of the container increases, that is, the pressure on the table increases, according to P=F S=G S, S does not change, G increases, and P increases.

Pressure = Gravity + Buoyancy.

Therefore, the pressure increases.

Pressure = Pressure Contact Area.

The contact area remains the same, and the pressure increases, so the pressure increases.

The pressure of the water on the container will increase because the water surface will rise, but the pressure and pressure of the container on the table will not change. Unless your hand is still holding the container.

It depends on the conditions given by the question to determine the solution.

1) If the height h of the original water in the container is known, and the height h1 of the water surface rises after the object G is placed, then the bottom of the container is subjected to the pressure of water.

p (h+h1)p water g.

2) If the gravity g of the water in the container and the bottom area of the container s are known, and the buoyancy f experienced by the object after being placed in the water is also known, then the pressure of the water at the bottom of the container is known.

p＝（g+f）/s。

3) If you know the height h of the original water in the container and the area of the bottom of the container s, and know the gravity g of the object, put it into the water, and when it is stationary, it floats on the surface of the water, and the bottom of the container is under the pressure of the water.

p＝hpg+g/s。

p=ρgh’，h'is the height of the water after the g object is placed in the water.

p= gh ah The density of the object is less than the density of water If it is greater than the density of water The part of the object touching the bottom of the container is found separately.

How to find the pressure at the bottom of the container when the object is pulled by a spring and then put in water.

The pressure of the container on the table = the gravity of the container + the gravity of the water.

The pressure of water on the container = the bottom area of the container * the height of the water, which is not necessarily equal to the gravity of the water.

So the pressure of the container on the table is not equal to the pressure of the water on the container + the gravity of the container.

For example, when the mouth of the container is large and the bottom is small, or the bottom is small, the pressure of the water on the container and the gravity of the water are different) I hope it can help you

The pressure of the container on the table top is considered the pressure of the solid, which is equal to the total weight. In other words, no matter what shape the container is, the total weight of 10n water and 5n container is 15n, but if it is a non-cylindrical or cuboid container, the pressure of water on the container is not equal to the gravity of the water, that is, the pressure generated by 10n water is not equal to 10n. Because the pressure is distributed to the side walls of the vessel.

It can't be equal. Because the pressure of water on the container is related to the depth and density, and has nothing to do with the gravitational bottom area. The pressure of the water on the table is related to the force area and pressure. Therefore, these two quantities are not the same thing and cannot be equal.

Water and the container should be regarded as a whole, and the whole should be placed on the tabletop, so the pressure of the whole on the table is naturally the gravity of the whole. And I think the reason for your mistake is to ignore the fact that the "pressure of water on the container" is calculated by the water as the external environment, while the "pressure of the container on the tabletop" is calculated by the air for the external environment.

So, the "pressure of the container on the tabletop" should be equal to the "pressure of the container" plus the "gravity of the water".

There's something wrong with your habitual thinking. You think that the container is cylindrical, but in fact, if you look at the data, you can know that if it is 50*10cm 3 volume of water, the mass should be, and the weight should be 5N. In the question, 6n, it can be seen that the container has a large mouth and a small bottom. Herein lies the problem.

When an object floats in water, the pressure of the container on the table is made up of two parts: gravity and buoyancy.

The first part is due to the gravity of the object in the container. If Kai Zhao places a heavier object in a container filled with water, the container will be subjected to the gravity of the object, and then this weight will be transferred to the table through the bottom of the container. Therefore, if the heavier the object, the more pressure the container will put on the tabletop.

The second part is due to the buoyancy caused by the object immersed in the water. When an object is placed in water, its buoyancy as pushed upwards by the water increases. The magnitude of this buoyancy depends on the shape, size, and density of the object, as well as the density of the water and the depth at which it is located in the container, etc., and such buoyancy reduces the surface contact between the container and the tabletop.

This buoyancy is relatively greater for light objects, and some of the smaller items may even float upwards, so that the container has less pressure on the tabletop.

When an object is placed in water, the pressure of the container on the table depends on the weight and buoyancy of the object in the container, while the buoyancy depends on the size, shape, and density of the water. To reduce the pressure of the container on the tabletop, it is advisable to choose relatively light items to avoid overweight and fragile items that overwhelm the underweight container. <>

That is to say, the floating grinding force on the object is exactly 1N, which is about 102g, and at the same time produces an equal reverse force.

That is to say, the total pressure of the cylindrical container on the desktop is (500g+1000g+102g)*, assuming that the water temperature is 4, the mass of 1cm3 water at this time is just 1g), and then divide by the contact area, assuming that the blind height is the 100cm2 you mentioned, the pressure generated by the final pressure is.

That is, the buoyancy force on object A is exactly 1n, which is about 102g, and at the same time produces an equal reverse force.

That is to say, the total pressure of the cylindrical container on the table is (500g+1000g+102g)*, assuming that the water temperature is 4, the mass of 1cm3 of water at this time is just 1g), and then divide by the contact area, assuming that it is the 100cm2 you mentioned, the pressure generated by the final pressure is.

Answer: c is unchanged.

Solution: (1) After putting in the wooden block, the water overflows, but the height of the water surface remains unchanged. According to P=Gh, the density of water does not change, and the depth of water does not change, then the pressure of water on the bottom of the container does not change.

2) The container is filled with water, and the buoyancy of the wooden block is equal to the weight of the overflowing water, F float = G overflow. When the block floats, the amount of buoyancy experienced by the block is equal to the gravitational force of the block, f float = g wood. So g wood = g overflow.

According to f=g total, if the total gravity is constant, the pressure of the container on the table top is constant. According to p=f s, the contact area s between the container and the desktop is constant, and the pressure of the container on the desktop is constant, then the pressure of the container on the desktop remains constant.