Hurry!! Small problems of physical pressure in the second year of junior high school!!

Whether you can talk face-to-face is simple and troublesome.

Big brother: You can find someone in person to help you with your homework

You don't want to use your brain to say directly, if you don't study hard, what will you eat in the future?

S Large: S Small = 10:7

Let the container be subjected to gravity x, (is m (unit n)

The system of equations is as follows).

m+f3)/s=(m+xn)/(7/10)s(m+f4)/(7/10)s=m/s

f3+f4=

3m=7f3=-10f4

f3+f4=

f4=3m=7f3=10f3-9

f3 = 3n (down), f4 = or upward).

m=7,x=

The mass is g g = .

Because the water has 650g;

The second question in the question is simple, inverted proportion, H1 is 15cm;

It's been a night's work, hope! Thank you.

I feel that this question does not have much practical significance, and it feels that it is no longer interesting to come up with such a question.

Pressure of water on the bottom of the barrel: p=h g=20cm*1000kg m 3*pressure of water on the bottom of the barrel: f=ps=1960pa*3dm 2=pressure of the bucket filled with water on the horizontal ground:

f = 80n + 10n = 90n pressure on the horizontal ground of the bucket filled with water: p = f s = (80n + 10n) 3dm 2 = 3000pa

1) Increase the pressure.

Straighten out the diagram.

Where's the picture? (1) The contact area of the same size can be increased by increasing the pressure, what does this mean, is it missing?

The height difference between the mercury column above the CD column in the left tube is smaller than that of the original two tubes, and the gas pressure in the tube is P=P0- P, so it can be seen that the gas pressure in the left tube increases and the volume decreases. Option D

5. Note that the pressure inside the confined gas is equal everywhere. Although the mercury column of the right tube remains the same, the gas pressure remains the same, and the height difference between the two tubes remains the same. Therefore, choose C

1) The height of the column in the A tube is equal to 0.

2) Gas p1 = 75 + (35-10) = 100, v1 = 30p2 = p0 - p p0, you can set the length of pipe b gas to 40. Find p2Namely.

100*30=p2*40 P2=75, which is exactly equal to atmospheric pressure.

Therefore, all the mercury in tube B is returned to the horizontal tube, and the length of gas in tube B is 40cm

Pressure = PGH

Pressure Qi Liang strong difference cave transport = pg * distance difference from the heart of the macro . (

Answer (a) The indoor atmospheric pressure is 100,000 Pa

The atmospheric pressure in the center of the tornado can reach 90,000 pap indoor》p in the center of the tornado.

The equivalent pressure is 100,000 pa-90,000 pa=10,000 pa

According to the pressure formula p=f s f=ps=10000x200=2000000n

So the roof is subjected to an upward force of 10,000n.

Choose A. The reason: the gas pressure in the room acts on the roof in an upward direction.

Outdoor atmospheric pressure acts on the roof in a downward direction. The gas pressure inside is greater than that outdoors. So up.

The magnitude of the pressure is the difference between indoor and outdoor pressure. That is, the roof of the factory building will be likely to be lifted upwards! Instead of being pressed down and collapsing.

At this time, the gas in the room is equivalent to a closed gas, and the original pressure is maintained.

then the force f=p'*s = cattle.

The room is a pa, and the roof is a pa, and the pressure on the roof is less than that in the room, so the pressure is upward. a

The question is whether your 90,000 pa is absolute or relative.

From the inscription, it can be seen that the internal pressure is stronger than the external pressure, so the pressure of the atmosphere on the roof is in the upward direction. The magnitude of the pressure is the pressure difference multiplied by the force area, which is 2000000N. Therefore, I choose A.

Choose C, the tornado force is upward, the indoor pressure and the outdoor force on the roof are balanced, that is, the force is upward, f=ps=9000x200=1800000

a Indoor pressure minus tornado centre pressure multiplied by area.

a, because the indoor pressure is strong with the outdoor, so the direction is upward. Then (100000-90000)*200

Answer: c is obtained by p=f s=g s, g A s A = g B s B; In this question, S A is less than S B, so G A is less than G B.

Solution 1: Limit method thinking, remove the same mass, remove the mass of G A, then A has no remaining, the pressure on the ground is zero, and B is still left, so the pressure on the ground is strong, but this method is not very good.

Solution 2: In the conventional solution, if the part with the same gravity is g, then for A, the bottom area of the remaining part is (GA-G)s AA GA.

In terms of B, the base area of the remaining part is (gB-g)s B g B.

Since the object of the same mass is intercepted and placed on the same mass, the total gravity of A and B is the same as the original one for the second time, which is g A and g B, so:

p'A=gA, <(GA-G)S, GA, GA"; So: 1 p'A = (s A g A) x (G A-g) g A(1).

p'B = g B < (g B - g) s B g B "So: 1 p'B = (s B g B) x (g B - g) g B (2).

1), (2) Divided: p'B p'A = "(G A-G) G A" i.e.: p'B p'A = (1-g g A) (1-g g B) is greater than 1, so p'B is greater than p'Armor.

The bottom surface of A A is smaller than that of B, and the same mass is shared on A to increase the pressure.

Since they have the same pressure on the ground, and the title shows that cube A is smaller than cube B (the height of A is less than the height of B). Therefore, this question can be analyzed by the method of "extreme thinking", which is relatively simple.

Because the same height is cut off respectively, the height of the cut off can be set to be just equal to the height of A, then the pressure of A on the ground will be zero, and B still has the remaining part, and the pressure on the ground is not zero, so it can be known that the relationship between the remaining part of the ground pressure after the interception is:

p'A is less than p'Second.

Answer: c