The problem of physical pressure in the second year of high school, the problem of physical pressure

Let the gas pressure in the middle section of the right side be p, and the liquid height of the upper section of the right pipe is h', then p=p0+pgh'

For the liquid level of the lower section of the right tube: p0+pgh=p, it can be seen that h=h'

Therefore, add mercury dropwise to the left tube, due to the h of the liquid at the upper end of the right tube'If it does not change, the left tube must also maintain this height.

Add mercury dropwise to the right tube, H of the liquid in the upper part of the right tube'As it increases, the left tube h will inevitably increase as well.

The gas heating and cooling will not change the h=h above'So changing the gas temperature can not change h option b

First of all, why can't C or D.

When the temperature changes, the weight of the mercury on the right plus the weight of the gas does not change. The left and right sides should be balanced, so the amount of mercury on the left side does not change. It will only push the mercury up or down at the upper right end.

When you add mercury to the left, how much the mercury column on the left increases, and how much the level height of the mercury below the air column on the right still increases. So h is constant.

Only when mercury is added to the right, the mercury in the upper part of the air increases, and the air is compressed, so h changes.

After the increase of c t, pv=nrt and the small section of hg column on the upper right are balanced by force, and the pressure of the air column remains unchanged no matter how the state of the air column below changes. Balanced on the left, h should be unchanged.

If H increases, the gas pressure increases, and mercury can only be injected from the right side.

Heating and cooling are not possible, due to isobaric changes, h should be kept constant.

Isn't the temperature rise also isobaric, h is only related to pressure, p=p0+h right hg

The liquid pressure formula p=pgh, cmhg corresponds to the pressure generated by the mercury column at the corresponding height. CMHG can be used to express pressure, just as we can express j (energy) in kilowatt-hours.

According to the ideal gas equation pv=nrt

After the air is pressed into the tank, it just becomes a two-part system, and it is an isothermal process, and the front and back of the right part are the same underwater pressure of 200m is 20 atmospheres.

pv0=p1v1+p2v2

p*2=95*2+20*10

p = 105 is 105 atmospheres, approximately.

The title is not clear, you haven't played it all; Also, this is not a topic for the first year of high school.

I don't think I remember the pressure in the first year of physics...... it's not strong

Your problem is that you're ignoring an important property of friction – friction has nothing to do with the contact area.

The formula for calculating sliding friction (including maximum static friction) is:

f ＝ n；

Where: is the sliding friction factor: it is related to the roughness of the contact surface between the object and the supporting surface, but not to the contact area;

n is the positive pressure: it refers to "the component of the interaction force between the object and the supporting surface, in the direction perpendicular to their contact surface". Since the "interaction force" is actually a pair of action and reaction forces, the positive pressure mentioned above can be either of the two.

Remember: positive pressure only requires perpendicular to the contact surface, but does not require pointing from "whom".

The two objects involved in friction in this question are: suction cups and walls; So, the "positive pressure" that is sought is:

the pressure of the suction cup on the wall; Or: the support force of the wall to the suction cup;

According to the balance of the force of the suction cup in this direction, it can be seen that:

The support force of the wall on the suction cup The total atmospheric pressure on the suction cup;

So, the area used to find the atmospheric pressure is all and not 4 5.

Because there is no air between the untouched parts of the middle 1 5, the vacuum, like the 4 5 attached to the wall, is subject to atmospheric pressure.

When calculating friction, the contact surface is only 4 5 circles.

p0v0 = p1v1+p2v2

v0 = 2 m^3

p1 =v1 = 2 m^3

p2 = 1 atm + gh

is the density of seawater, h is the depth, g is the acceleration due to gravity, take 10 m s 2p2 = 10 5 pa + 1000 10 200 pa = 10 6pa = 21 x10 6pa so v2 = 10 m 3

p0 =( p1v1+p2v2)/ v0

95* 2+ 21 *10) / 2= 200 atm

The pressure of the original compressed air in the air reservoir:

b moving? If not, then it will not be moved.

For a, there is: pol1s t=p1(l1+l)s tx tx to be found.

For b, we have: pol2s=p2(l2-l)s and p2=p1 substituting the above equation to obtain the tx value.

f=p1s If b moves, the above equation is not true!

Derive the rest for yourself!

1.Question 3. Since the liquid is relatively stationary, i.e., the pressure is balanced. Air pressure in the tube + liquid pressure = atmospheric pressure.

Liquid height h=lsin30°

2.The pressure inside and outside the tube is equal, and the pressure of the gas on the piston in the hall is f1 = s in p, the gravity of the piston mg is mg, and the pressure of the atmosphere on the piston is f2 = p0s

More force direction analysis. Draw the diagram yourself.

f1+mg=f2+f

then f1 = f2 + f-mg

f = ps i.e. p1=p0+(mg-f)/s

Question 1.

Second, I won't.

The original air pressure of the container P, your pumping will inevitably lead to a decrease in air pressure, how can it be possible to pump out the air with pressure P, the pumped V0 is in a state of connection with the original V, imagine the vacuum pump as a pump; When inflating, the gas is pumped from the outside, which can be different from the air pressure in the container, p0v0=n1rt, pv=n2rt, if the temperature remains the same, the amount of gaseous substances after inflation is (n1+n2), and the pressure p3=(n1+n2)rt v=(p0v0+pv) v, I don't know if I said it clearly.

1. There is a pressure to the 5th power of the overall force balance.

2. After calculation, the isobaric change is excluded, and the pressure is reduced to the 5th power of multiplying 10 at most, at this time, A moves five-sixths L to the left, and B pushes to the left to the head. The AB rope tension is zero.

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