Simple physical pressure problem, physical pressure problem

Not necessarily. It depends on the shape of the glass.

If it is straight up and straight down, the pressure of the glass on the table is the pressure of the water on the bottom of the cup plus the pressure of the empty cup on the tabletop.

If the cup is small at the top and large at the bottom, then the pressure of water on the bottom of the cup and the pressure of the empty cup on the table is greater than the pressure of the cup and water on the tabletop.

If the cup is large at the top and small at the bottom, then the pressure of the water on the bottom of the cup and the pressure of the empty cup on the table is less than the pressure of the cup and water on the tabletop.

The specific reason is that the pressure of water is not simply F s, but is calculated by gh, so the pressure of water is not directly related to the weight of water and the area of the bottom of the cup.

The empty glass has no pressure on the water. However, what kind of glass are you?

Yes, water glass can be a whole.

The pressure of the whole glass on the table is stronger than the pressure of water on the bottom of the glass + the pressure of the empty glass on the water.

There is an equivalent to this case.

When the thickness of the cup wall is not timed, and the bottom of the cup is flat, it is okay ("The pressure of an empty glass against water"Is it wrong, it should be the pressure of the empty cup on the table).

Not equal. The pressure of the water on the bottom of the glass = the pressure of the empty glass on the water.

The pressure of the entire glass on the tabletop.

Gravity on the glass + gravity of water) The contact area between the glass and the tabletop.

Unless the thickness of the cup wall is not counted.

No. The pressure of the entire glass on the table top is equal to the weight of the entire glass divided by the area of the bottom of the glass.

It should be smaller: the triangle and rectangle are now cut apart but still close together, and the overall pressure remains the same. According to p=f s, the pressure of the triangular part must be smaller than that of the rectangle, that is to say, p(triangle) p(average) field touch p (rectangle), then the rectangle is cut off, and the p(rectangle) larger than the mean will have less influence on p (average) because the weight of the whole figure is reduced by the number of songs, so p (average will decrease).

Therefore, choose B.

If it is an object with an upper and lower cross-sectional area such as a cube or cuboid, select C

The root width and space are p= gh, and when the difference is cut in the vertical direction, the density of the object does not change, and the height does not change, then the pressure of the object on the ground does not change.

If it is an object with an upper and lower cross-sectional area such as a positive difference and a cuboid, then select C

According to p= gh, when the vertical direction is cut, the width and space density of the object does not change, and the height does not change, then the pressure of the object on the ground does not change.

The pressure of the inner liquid on the bottom of the container should be calculated as p= gh.

That is, the early pressure at the bottom of the container is related to the density and depth of the liquid.

Since Zhengxian Hall A< B If you want to satisfy pA=pB.

then it needs to meet the requirements of ha" hb.

That is, what method is to be used to make hA" hB.

Again, since sA, sB, the same volume of liquid v is withdrawn

Then δH=V S then δH A < δH B (i.e., the liquid level of A drops slower than B's), so pumping out the same volume of liquid can satisfy H A" H B that is, BP A = F A S A. f A = f B, S A is greater than S B, which can only mean that P A changes more slowly than P B.

If at the beginning pA" = pB draws the same mass, then it is impossible to have the same pressure at the bottom of the container.

Not necessarily if you start with a p-a-form.

Pressure = gh

The home mass is smaller than the B mass, i.e. A rrh< B rrh

Since r>r, Ah< Bh are multiplied by g on both sides- A pressure "B pressure.

If you want to make A and B have the same pressure at the bottom, when reducing the pressure, you need to reduce A less. When the pressure is increased, the nail needs to be increased more.

a.The same mass of wisdom is extracted separately, and the pressure decreases: Methyl GδH= Methyl G[M ( Methyl Rr)]=mg ( Rr).

B pressure decreases: B g δh = B g [m ( B rr)] = mg ( rr).

Apparently mg (rr)b.The same volume v is extracted separately, and the pressure decreases: AgδH= Ag[v (rr)].

B pressure decreases: B GδH = B G[v ( rr)].

If the diagram is drawn in proportion, it is obvious that the density of A is small, and then the pressure of A is reduced, and the pressure at the bottom of container A and B can be the same.

If the diagram is not drawn to scale, it is difficult to determine because the relationship between the density of the two is unknown.

c.Pour the same mass M separately to increase the pressure of the forma: formagδh= formag[m(a-rr)]=mg (rr).

B pressure increases: B g δH = B g [m ( B rr)] = mg ( rr).

Apparently mg (rr)dPour the same volume of V separately, and the pressure of the Forma increases: Forma GδH= Forma G[v (rr)].

B pressure increases: B g δH = B g [v ( rr)].

If the diagram is drawn in proportion, it is obvious that the density of A is small, and the pressure of the former mountain and then A is small, so the pressure at the bottom of the container A and B cannot be the same.

If the diagram is not drawn to scale, it is difficult to determine because the relationship between the density of the two is unknown.

Conclusion: If the diagram is drawn to scale, a b is correct.

Choice D, this question should be regarded as the density of the fence is uniform, widened, thickened twice, then the volume is four times the original, that is, the gravity becomes four times the original, the pressure is equal to the gravity divided by the following surface area, the gravity becomes four times the original, because the widening is doubled, the lower surface area becomes twice the original, so the pressure p=4g (2s)=2*g s, that is, the pressure is twice the original. Therefore, choose D

Select D, set the original pressure to A, and the bottom area to S. From the question: Later, the pressure is 4a, and the base area is 2s, so the pressure is divided by the pressure by the base area, so it is 2 times the original pressure.

b Thicken and raise the original wall twice as much, that is, 2 times the mass of S unchanged, it is B.

d,, the pressure is multiplied by a constant by gravity... The pressure is divided by the ground area, so only the heightened part will increase the pressure.

The mass of d has been quadrupled and the contact area has been doubled.

1 The pressure of water on the object in it p=density*g*h=1g cm3*10n kg*25cm= pressure f=p*s=

2.Acceptance analysis of plastic sheets Water pressure (upward) = gravity + external force So f external force = 3. The fall just happens to indicate that the resultant force at this time is 0 instantly, which is the gravity of the added liquid = external force density * (10 * 20) * g = 2n, so the density is 1g cm 3

Bernoulli's equation p+ gh + (1 2) * v 2 = c

For the same height h, the velocity increases in the flow, and the pressure decreases; As the velocity decreases, the pressure increases;

With the same velocity, the smaller the height, the greater the pressure.

Bernoulli's equation p+ gh + (1 2) * v 2 = c

The water flows downward, the height decreases, the velocity increases, the gravitational potential energy is converted into kinetic energy, and the pressure does not necessarily increase or decrease.

However, the landlord said p=f a, and the object of study should be the stressed object, not water, so it cannot be said whether the pressure of water becomes larger or smaller.

This process has the conversion of potential energy to kinetic energy, and the formula is established to be based on the condition of energy conservation, and the reason why water becomes thinner is also due to external forces doing work.

Calculate the Hypertherm mass: m= ·v

Hypertherm's pressure on the ground f=mg

Hypertherm pressure on the ground p=f s

F pressure = g = mg = vg = number with their own.

p=f=s=number your own brought.