A cylindrical container is filled with water, and if a piece of wood is placed in the water, the pre

cp=pgh,p,g,h three quantities have not changed!!

Oh is the same, choose c

There are many ways to analyze this.

First, look at the pressure formula PGH.

p refers to the density of the liquid, which has nothing to do with the wood block.

Wood is put in, and the h is still the same, because the excess water will overflow.

So all three quantities remain the same. So according to the formula p=pgh is constant.

Second, the pressure is equal to the pressure divided by the area p=f s.

In the beginning, the pressure was the gravitational force of all water.

Later put in the wooden block, and the pressure becomes.

Later gravity of the water (= gravity of the original water - gravity of the overflowing water) + gravity of the wooden block.

But according to Archimedes' buoyancy theorem, the buoyancy experienced by the object = the gravitational force of the water being discharged.

And because it's floating, so.

Buoyancy force experienced by the object = gravitational force exerted by the object = gravitational force expelled by the water.

So the total pressure does not change to:

It turns out that the gravity of the water - the weight of the water + the gravitational force of the object.

It turns out that the gravity of the water - the buoyancy of the object + the gravity of the object.

It turns out that the gravity of the water - the gravity of the object + the gravity of the object.

It turns out that the gravity of water.

It's still the same as before. So there is no change.

Of course, it has become bigger, one is that the overall mass has increased, and the other is that the water level has risen.

So the pressure is high.

Choose C, it's definitely right, how many times did you do it in junior high school.

Oh is the same, choose c

There are many ways to analyze this.

First, look at the pressure formula PGH.

p refers to the density of the liquid, which has nothing to do with the wood block.

Wood is put in, and the h is still the same, because the excess water will overflow.

So all three quantities remain the same. So according to the formula p=pgh is constant.

Second, the pressure is equal to the pressure divided by the area p=f s.

In the beginning, the pressure was the gravitational force of all water.

Later put in the wooden block, and the pressure becomes.

Later gravity of the water (= gravity of the original water - gravity of the overflowing water) + gravity of the wooden block.

But by. Archimedes.

Buoyancy theorem. Buoyancy experienced by the object = gravitational force discharging the water.

And because it's floating, so.

Buoyancy force experienced by the object = gravitational force exerted by the object = gravitational force expelled by the water.

So the total pressure does not change to:

It turns out that the gravity of the water - the weight of the water + the gravitational force of the object.

It turns out that the gravity of the water - the buoyancy of the object + the gravity of the object.

It turns out that the gravity of the water - the gravity of the object + the gravity of the object.

It turns out that the gravity of water.

It's still the same as before. So there is no change.

Of course, the same height.

First of all, let's take a look at the pressure formula of the liquid p= gh, we can see that the pressure is only related to the height of the liquid column, and the water in the container is full at the beginning, and the water is obviously full after putting it into the wooden block, so the pressure remains unchanged.

The round container is filled with water, and when a piece of wood is placed in the water, the water will overflow and the depth of the water will remain the same, according to p= water gh, then the pressure of the water on the bottom of the cup will not change

So the answer is: no change

(1) Left picture, object floating on water, there are 2

The volume of 5 is exposed to the water, V row = 35V, F float = water V row G = water 35

vg=g wood, let the bottom area of the container be s, then.

V row = 35v = s h, from the inscription, after putting in the wooden block, the pressure of water on the bottom of the container increases by 300pa, that is, p= water hg=300pa,--2) in the figure, hang the wooden block a at the left end of the lightweight lever at point b, the pressure of water on the bottom of the container is reduced by 100pa compared with the wooden block a when floating, so the pressure of water on the bottom of the container is increased by 200pa, that is: p = water h g=200pa,-- by can be obtained:

h = 23 h, while s h = 35 v and v row = s2

3△h=23s△h=23×3

5v = 25v at this time:

f float = water v drain g = water 25

VG, the tensile force experienced at the B-end:

fb = g wood - f float = water 35

vg- water 25

vg=15 water vg, leverage balanced, fb ob=g oc, i.e.: 15

Water vg ob=g oc---3) on the right, change the water in the container to another liquid, so that the exposed liquid surface of the wooden block A is the same as that of Figure B, subject to the buoyancy of the liquid:

ffloat = liquid-25

VG, the tensile force experienced at the B-end:

fb = g wood - f float = water 35

VG-Liquid 25

vg, lever balance, fb ob=g od, i.e.: ( water 35

VG-Liquid 25

vg) ob=g od--- ob: water3

1 2 water is 120 ml, which means that the volume of this cylindrical container is 240 ml. The volume of a conical container of equal height is 1 3 of the volume of this cylinder, i.e. 80 ml. Pouring 120 ml of water into an 80 ml container will spill 40 ml of water.

The volume of the cylinder = 90 3 5 = 150 ml.

Then the cone volume = 150 * 1 3 = 50 ml.

90-50 = 40 ml.

The water will overflow, spilling 40 ml.

90x5 3 = 150 cylindrical total volume.

The volume of the cone is 1 3 of the cylinder

So cone volume = 150 3 = 50 ml.

3/5-1/3=4/15

The water will overflow, spilling 40 ml.

The volume of a cone of equal height to the bottom of a cylinder is 1/3/3 of the volume of the cylinder, 3/5 > 1/3 of the volume of the cylinder

So it will overflow.

Nothing to do with 90 ml.

The base area s=丌r 2=

The volume of the iron block v = the volume of water between the old and new levels of the container = sh=

Cubic centimetre.