High school physical pressure problem, please write a specific solution, thank you

The air quality in the pipe is certain, and the extra pressure comes from air gravity.

Let the air quality be m, the cross-sectional area of the tube be s, the gravitational acceleration g, and all pressures will be expressed in cmhg (meaning that only the values are considered and the units are not considered).

Since the liquid level on the mercury column no longer moves, the pressure is balanced, and there is:

15mg/s + 70=75

The solution yields mg s = 1 3

When the reading is 67 there is.

70+15-67)mg/s + 67=18mg/s + 67=6+67=73

So the actual atmospheric pressure is 73 cmHg

I don't know, but I think my answer should be fine, and I usually don't have a lot of fragmentary questions, most of them are relatively neat.

What grade are you from? Different grade levels can use different methods to help inspire ideas.

The answer is correct, when the reading is 70cmHg, the pressure mixed with the air = 75-70 = 5cmHg, and the volume is 15cm*s (cross-sectional area).When this barometer reading is 67cmHg, the volume of mixed air is 18cm*s, because the gas pressure is inversely proportional to the volume, the pressure of mixed air = 5*15 18=, so the actual atmospheric pressure = the pressure of mixed air (barometer reading (67cmHg) =

The answer is correct.

See me come up with the answer...

When the standard barometer reads 75 cmHg, it only reads 70 cmHg, and the upper air column is 15 cm long, so the 15 cm air column generates a pressure of 5 cmGh.

When the reading is 67 cmgh, the length of the air column is 15 + 70-67 = 18 cm, which is given by the formula p1*v1=p2*v2

At this time, the air column pressure is 5*(15 18)=5*(5 6)=25 6 cmgh

Add the height of the mercury column to 67cmgh, and the actual atmospheric pressure at this time is 67+ Answer: What is the actual atmospheric pressure?

Make up the answer and you're done. Give points!

When the temperature is constant, the product of the gas volume and the air pressure is unchanged, so when the standard air pressure is 75, the air pressure is compatible, so 5 15 is equal to the fixed value, so when the air pressure is 67, the volume at this time is 85-67 = 18 times the air pressure = 5 15 = 75, and the air pressure in the barometer is plus 67 is equal to for reference only.

<> usually use the hypothetical approach.

Assuming that the height difference of the mercury column remains unchanged, due to the tilt, the air column becomes longer and the air pressure becomes smaller, and the mercury column can be balanced by the rise of the side.

However, once the mercury column rises, the pressure difference between the inside and outside becomes larger, so the air column cannot return to its original length.

So AC is correct.

When the reading is 70 cmHg, the pressure mixed with the air = 75-70 = 5 cmHg, and the volume is 15 cm*s (cross-sectional area).When the barometer reading is 67cmHg, the volume of the mixed air front book is 18cm*s, because the gas pressure is inversely proportional to the volume of the body, so the pressure of the mixed air = 5*15 18=, so the actual atmospheric pressure mask at this time = the pressure of the mixed air (barometer reading (67cmHg) =

When the plug returns to equilibrium, the two states before and after heating are an isobaric transformation. You can also think of it this way, the outside atmospheric pressure is unchanged, and the piston is kept in balance before and after heating, and it is easy to conclude that the pressure of the piston by the gas in the cylinder is equal to the external pressure, so the pressure in the cylinder of the piston returns to equilibrium.

However, it should be noted that after the piston is balanced, if the piston is in a state of outward movement during the heating process, the air pressure of the punching surface in the cylinder is different from the initial state.

When the plug returns to equilibrium, the two states before and after heating are an isobaric transformation. You can also think of it this way, the outside atmospheric pressure remains unchanged, and the piston is balanced before and after heating, and it is easy to come to the conclusion that the pressure of the piston by the gas in the cylinder is equal to the pressure outside, so the pressure in the cylinder remains unchanged when the piston quietly returns to equilibrium.

However, it should be noted that after the piston is restored to balance, if the piston is in a state of outward movement during the heating process, the air pressure in the cylinder is not the same as the initial state.

When the plug returns to equilibrium, the two states before and after heating are an isobaric transformation. You can also think of it this way, the outside atmospheric pressure is unchanged, and the piston is kept in balance before and after heating, and it is easy to conclude that the pressure of the piston by the gas in the cylinder is equal to the external pressure, so the pressure in the cylinder of the piston returns to equilibrium.

However, it should be noted that after the piston is restored to balance, if the piston is in a state of outward movement during the heating process, the air pressure in the cylinder is not the same as the initial state.

When the reading is 70cmhg, the pressure mixed with the air = 75-70 = 5cmhg, and the volume is 15cm*s (cross-sectional area) When the barometer reading is 67cmhg, the volume of mixed air is 18cm*s, because the gas pressure is inversely proportional to the volume, so the pressure mixed with the air seepage pants = 5*15 18=, so the actual atmospheric pressure when this is flame resistant = the pressure of the mixed air (barometer reading (67 coal production chemical industry) =

The liquid level in the tube is stationary, so the pressure is balanced. The downward pressure is inside P, and the upward pressure is outside P + PH, so P inside = P outside + PH = P0 + Gh.

Based on the horizontal line on the way, the gas pressure p is inside the tube, and the atmospheric pressure + the pressure gh above the horizontal line outside the tube

So p=p0+ gh

pa=65cmhg

pb=70cmhg

Analysis of the force The upper surface of the liquid in the tube is subjected to the pressure of the air column in the tube, the atmospheric pressure p0s and h ps, the direction of p0s and h ps is downward, and the pressure of the gas in the test tube is p0+ gh

1.At a certain temperature, the product of the pressure of the confined gas and the volume (in this case, the length of the gas column can be substituted) does not change. Let the air pressure be p at this time, then (76-70)*30=(p-68)*32 gives p=

2. Under a certain volume, the ratio of the pressure of the closed gas to the absolute temperature remains unchanged. Let the air pressure be p at this time, then.

76-70) 300=(p-70) 270 to get p=

When the air temperature is 27 degrees Celsius, the standard barometer shows 76cm Hg, and the barometer shows 70cmHg, that is, the air pressure of 30cm is equal to the pressure of 6cm of mercury, 1), PV T = constant value, 32P = 30 * 6, P = 45 8cmHg, and the actual air pressure is 68 + 45 8, the unit is cmHg

2), PV t=constant value, P 270=6 300, P =, the actual air pressure is 70+

Stepping on a balloon is caused by excessive pressure on the balloon.

The pressure of the person on the balloon:

p1 = f s = 80 (8 * 10 -4) = 1 * 10 5 Pa atmospheric pressure p0 is also 1 * 10 5 Pa

So the air pressure in the balloon p2 = p1 + p0 = 2 * 10 5 Pa, is two atmospheres.

The maximum pressure on the surface f2 = p2 * s2 = 2 * 10 5 * (20 2) = 2 * 10 6 n

The maximum pressure on the surface of the balloon is 2*10 6 N, and the air pressure inside the balloon is twice the atmospheric pressure.

**, the relative pressure is 80 8x10 -4=10 5Pa;

In this way, the absolute pressure inside the balloon is 2x10 5pa;

That's twice the atmospheric pressure.

2x10 5x20x10 -4=400N, i.e. the maximum relative pressure at the balloon** is 400N;

The maximum pressure is the limit that all surfaces can withstand, so it is 80*20 8=200N

Atmospheric pressure p0 is taken as 100kpa, and the internal and external pressure is always balanced, so the air pressure in the balloon = atmospheric pressure p0 + (80 8) * 10000 = 200kpa

It is about 2 times the atmospheric pressure.

The surface of the balloon is subjected to a maximum pressure of 80 8 * 20 = 200N

**The air pressure inside the balloon is 1atm+80 (8 10000)=1atm+100000pa=2atm

Therefore, the air pressure inside the balloon is 2 times the atmospheric pressure.

First, find the pressure p1 given to the ball when stepping on it, and the maximum pressure on the surface is multiplied by the atmospheric pressure + the force of the balloon **.

Intra-ball pressure at detoxification: Intra-ball air pressure multiplied by surface area equals maximum pressure.

Pressure Equation:

1atm+80n 8cm2=p.

Pressure equation: 80n+12cm2*1atm=p*20cm2 in the last life 20cm2 is the area of the inner surface of the sphere.

Two equations dissolve the answer.

The maximum pressure on the surface of this balloon is: 80N divided by 8cm2

**The air pressure in the balloon is: 80N divided by 8cm2 and divided by the standard atmospheric pressure, times the atmospheric pressure.

The conditions are not enough.

In balloon blowing is not aware of the air pressure inside the balloon at this time.

If you know, you can solve it.

The surface area you give doesn't match the volume, but anyway, find the radius first. Since the balloon is not deformed, the force on the balloon is uniform. The maximum pressure on the balloon in this way is:

Atmospheric pressure +80 area. The area here is calculated using the radius you just found. Then divide this pressure by atmospheric pressure to get how many times.

Rationale: If the object (the mercury column in the diagram is also an object) is balanced, the resultant external force is zero; If the acceleration state, then the resultant external force f=ma

1) Since the mercury column is in equilibrium, the resultant external force is zero.

Let the cross-sectional area of the mercury column be s, then: ps=mg+p0s= sδhg+p0s then: p=p0+ δhg

The latter two questions are the same, and the equations are not listed.

2) Taking the mercury column in the δh section as the research object, its lower edge pressure is the same as that of the outside world is p0, and the equation is listed according to the above principle.

3) The same is true.

Gas pressure: 7Fluid flow velocity as a function of pressure.

For chamber A systems: (1 3ls) 400 = (1 3L-x) S 300 For chamber B systems: (2 3ls) 400 = (1 3L-Y+x) S 300

Solution: x=l 12

y=l/4