A junior high school physics and mechanics question to find answers and detailed explanations

Because the pressure difference between the inside and outside of the boiler is to the fifth power of Pa, the support force of point A to the lever is f=to the fifth power of Pa square meter = 36N

According to the leverage equilibrium condition, f oa=g ob

i.e. g = f (oa ob) = 36n (oa ob) = 36n 1 4 = 9n

So the weight mass m=g g=

Analysis: This is a knowledge of leverage balance, then we must know the conditions of leverage balance: F1L1=F2L2, which requires the weight of point B, we must know the magnitude of the upward force at point A, and the length is set to OA=1, AB=3, then OB=4, and at the same time, the knowledge of pressure is involved in this problem.

Solution: It can be known from the meaning of the title:

p = 5 to the power of PA, s = 3 cm = 3 10 to the power of -4 m in the upward force at point a, according to f = ps.

f = ps = 5 to the power of pa * 3 10 to the power of -4 m = 36n the gravitational force of the object at point b, according to the conditions of the lever equilibrium: f1l1 = f2l2.

f*oa=g*ob.

g=f*oa/ob=36n*1/4=9n

Obtained according to g = mg.

m=g/g=9n/10n/kg=

You said that the answer to this question is, according to the meaning of the question and the data, whether it is the wrong answer or whether you made a mistake in the process of losing the question.

The pressure difference is divided by the valve area and then divided by 4, if you don't understand, the details are as follows:

1 The question is given is the pressure difference and area, the calculation is not to calculate the atmospheric pressure, if you want to understand the process to analyze the pressure received inside and outside the valve, internal pressure = internal pressure valve area External pressure = atmospheric pressure valve area + pressure on the valve at point A.

The pressure of point A on the valve is the gravity of the weight 0b: 0a = 4 times the gravity of the weight When the internal and external pressure is equal, find the weight of the weight.

The equation is Internal Pressure Valve Area = Atmospheric Pressure Valve Area + 4g can be derived (Internal Pressure - Atmospheric Pressure) Valve Area = 4g Internal Pressure minus Atmospheric Pressure is the difference between the internal and external pressure given in the title, and the result can be obtained g = 9n mass = g 10n kg =

If the result is kilograms, the problem should be a ratio of 1:3 to the length of OA or 1:2 to the length of OA

The fifth power of the pa 3 cm2 divided by 10,000 divided by 4 = 9 N = kilograms.

Pressure difference x valve area to get the pressure to be withstood by the valve The pressure xoa is equal to the weight of the weight xob can be found in the object.

The downward pressure at point A is equal to the pressure multiplied by the area = 36n, and according to the equilibrium principle of the lever, the mass of b multiplied by 4 equals the pressure at a multiplied by 1, and the physical gravitational force at b is 9n and the mass is .

Analysis: In this problem, the pressure formula p=f s is used to find the pressure of the steam in the pot on the A end, and then the mass of the object hanging on the B end is solved according to the lever balance condition.

The pressure of the steam in the pot on the A end: fa=ps=

According to the lever equilibrium condition: fa oa=g ob=mg (oa+ab) The mass of the object hanging on the B end is: m= fa oa g ob=36n 1 (4 10n kg)=

Explanation: To calculate the pressure problem, we need to use the formula p=f s, obviously, the pressure f and the force in the lever equilibrium condition are the same quantity, so the combination of the lever and the pressure problem must be realized through the relationship between these two forces.

(1) The car is in equilibrium, f traction force = f drag = 12000n *

2) The car is still in equilibrium, f traction = f drag = 12000N*

As long as it is driven at a constant speed, then the traction force is always equal to the resistance.

Knowledge preparation: It is necessary to know that the force of each section of the same rope is the same.

2. The total weight of the person and the hanging basket is equal to f + f + 2f = 4f

3. Bypass the rope on the pulley and connect it to the axle under the pulley.

First of all, the two pulleys are fixed pulleys, and the one above hangs the whole up, so each rope is subject to half of the force of the total weight, and then look at the pulley below, the resultant force of the two ropes below it is equal to the pulling force of the rope that pulls it upward, so the force on each rope below it is half of the tension of the rope above, and half of the half is a quarter, understand?

1.Pull at a constant speed, so g = f pull + f float, f float = density volume g = 1030kg m 3, because f pull =, so g matter =

2.Because people put their hands on the air raft, the weight of the person and the air raft and the metal box can be regarded as a whole, so g object = g person + g raft + g box =, because it is floating, so f float = g object = density v row g = 1030kg m 3 v row 10n kg, so v row =, the volume of man is m person human density = because the 5 6 of the body of the person is exposed to the water surface when the person is stationary, so 1 6 is immersed in water, so the volume of the air raft immersed in water is, so the volume of the air raft immersed in water is.

The first question is better to understand the uniform speed is the force balance The force analysis of the metal box is only affected by the upward pull of the person The gravity of the box itself The pressure of the sea water does not tell the depth F=g g = The second question is still calculating The hand is slow.

The second question is the analysis of the overall force.

f float = g person + g metal box f float = sea density * g v drain v drain = v person 5 6 v metal box v gas valve Just untie the v gas valve.

Since it is the air valve, gravity is ignored.

(1) The buoyancy of the metal box is the volume of the box multiplied by the density of the seawater, and the buoyancy plus the pulling force of man is gravity.

2) You can calculate what the volume of the person is, and then calculate what the buoyancy of 1 6 volumes is, and add the rest of the weight to the raft, plus the weight of the metal box, and divide it by the density of the seawater.

When the bus brakes suddenly, the passenger reverses in the direction of the vehicle (fill in "the same" or "opposite"), depending on the passenger's possession. (Answer: Same; Inertia) is moving a football in the air, in addition to the action of air resistance, it is also affected by the force, and the object of this force is

Answer: Heavy; Earth).

The iron box, weighing 500 Newtons, is subjected to a horizontal thrust of 200 Newtons, and moves in a straight line at a uniform speed on the horizontal ground. At this time, the friction between the iron box and the ground is friction, and the magnitude of the frictional force is oxen. (Answer: Swipe; 200

At a depth of 1 meter below the surface of the water, the pressure generated by the water is Pascal. (Answer: 9800).

The crane wire rope hangs an object weighing 5 10 N under the wire rope, and when the object descends at a constant speed, the tension of the wire rope.

5 10 Ox. (Fill in "greater than", "equal to" or "less than"). Answer: equal)

A metal block weighs 80 N, and the contact area with the ground when placed on a level ground is meters, and its pressure on the ground is Pascal. (Answer: 4000).

Xiao Hao uses 400 N vertical upward force to pull the object with a mass of 50 kg and placed on the horizontal ground at rest, and the resultant force of the object is Then Xiao Hao uses 100 N horizontal to the right force to go to the object, and the object moves in a uniform straight line on the horizontal ground at a speed of meters and seconds, and the size of the frictional force on the ground is the direction of the resultant force on the object Answer:

0；100 Niu; horizontally to the left; 0）

If someone lifts an object with a mass of 10kg placed on a horizontal table with a vertical upward pull force of 50N, the net force of the object is (0)n, the net force of the pulling force and the supporting force is (98N), and the direction (vertically up or perpendicular to the table top).

Rally force 50n upward.

Gravitational acceleration g=

The gravity of the object is 10kg*

The object is balanced by force!

and objects on the table not moving!

The support force is gravity-pull force = 48n, and the direction is opposite to gravity, vertically upwards or perpendicular to the tabletop upwards!

0 100 direction straight up! The object is balanced by the force at rest, and the net force is 0! The resultant force of the supporting force and the pulling force is equal to the gravitational force and is in opposite directions.

If g is taken, then the net force on the object is multiplied by 10n kg and subtracted by 50n equals 48nThe sum of the resultant force of the pulling force and the supporting force is that the gravitational force of the object is 98n and the direction is upward.

0；98 (g, taken; Vertically up.

Because the object weighs 98 N, the pulling force does not move, so the object is at rest, then the net force is 0; The direction of the pulling force and the supporting force are both vertically upward, and the resultant force of the other two is in great opposition to the gravitational force.

It is the acceleration of gravity The knowledge point of the first year is the acceleration corresponding to the object when it is only affected by gravity.

The resultant force is 0n, the resultant force of larry and the supporting force is 100 N, downward.