A physics problem in the second year of junior high school requires a process

Pay attention to the several semicolons in the question, each telling you a condition, and the specific process is more verbose (for clear expression, P liquid table P is used here):

The gravity of the 1st metal block A = the buoyancy of the metal block A + the reading of the spring scale.

i.e. p gold * g * v gold = p liquid * g * v gold + f1 (1).

The second one is easy to understand, wooden block B is floating in liquid, and the buoyancy of wooden block B = the gravity of wooden block.

That is, p liquid * g * v row = p wood * g * v wood formula (2).

Since the title says that "the ratio of the volume of the exposed liquid surface to its total volume is 7:10", then V row = 3 10V wood, substituting equation (2) to obtain.

P wood = p liquid formula (3).

3rd only the wood block b is completely submerged, and the metal a is not subject to buoyancy, so the balance of force is:

Metal gravity + wood block gravity = buoyancy of wood block B + spring scale reading.

i.e. p gold * g * v gold + p wood * g * v wood = p liquid * g * v wood + f2 equation (4).

4th v wood = 20:13 v metal (5).

From these 4 explanations, we get 5 results, which are equations (1) and (5), and the following calculations are carried out step by step (to figure out which ones are known, only p liquid, f1, f2 are constants, and nothing else is the same).

Substituting Eq. (1) into Eq. (4) gives (parentheses are substitutions).

P liquid * g * v gold + f1 ) p wood * g * v wood = p liquid * g * v wood + f2 Eq. (6).

Substitute Eq. (3) into Eq. (6) to obtain (parentheses are substitutions).

P liquid * g * v gold + f1 + p liquid ) * g * v wood = p liquid * g * v wood + f2 Eq. (7).

P liquid * g * v gold + f1 + p liquid ) * g * (20:13 v gold) = p liquid * g * (20:13 v gold) + f2

Calculate the coefficient, i.e., p liquid * g * v gold = 13 (f1-f2) (think carefully that f1 is greater than f2).

Finally, V-gold = 13 (F1-F2) g p solution.

I don't know if there is a miscalculation, the process is for reference.

The density of water p=1000kg m3

Let the density of the object and the salt water be p, respectively'and p", the volume of the object is v in water:

Gravity mg = buoyancy f

mg=p'vg

f = gravity of the water drained = p * v row * g = p [(1-1 10)v] g = p'vg=

The density of the object p'=

After sprinkling the salt powder into the water and dissolving:

p'vg=p"[(1-1/4)v]g=(3/4)p"Density p of vg saline"=(4/3)p'=(4/3)*900=1200kg/m^3

An object floats on a certain liquid, and if the volume of the discharged water is n m of the total volume, then the density of the object is n m of the density of this liquid

So the density of the object The density of water = 9 10

The density of the object = the density of water * 9 10 =

Density of the object Density of salt water = 3 4

Density of salt water = density of the object (3 4) =

Answer: Let the object volume be v

When put in water, g=ffloat=p, water, g9v, 10, after adding salt, g=ffloat=p, brine g3v 4, the two equations are divided to get p, saltwater = water=dizziness, go out and have something to do, and come back to solve it later.

The finishing formula g=ffloat=pwaterg9v10=mg, you can get p=mv=water=

Let the density of the object be p1, the density of water is p0=1000kg m3, the density of salt water is p2, and the volume of the object is v

Then there is: the object is placed in water: p1*v*g=p0*(9 10*v)*g gravity is equal to buoyancy, and p1=900 is obtained

The object is placed in the eye water: p1*v*g=p2*(3 4)*g gravity is equal to buoyancy, so p2=1200

According to the formula, the buoyancy v (row) p (liquid) g, floating on the water surface, the force is balanced, that is, the buoyancy gravity, let the volume of the object be v and the mass is m, you can get two formulas, 9 10vp water g = mg (1).

3 4vp salt g = mg (2).

The density of water is known, the two formulas are different, the density of salt is obtained, choose any formula, divide V to the right, and the density of the object is obtained.

f float = g matter.

Dense water drainage g = dense object g

Dense water drainage body = dense object.

9 1 = dense salt =

Solution: (1) Perform force analysis on Figure A, which is: 1500N+G dynamic 2G person+40Nx 10Nkg

The force analysis of Figure B shows that 1500N+G dynamic 3 (G person-100N) solves this equation system, and obtains: G person 700N, G dynamic 300N(2) =w has w total.

gh)/(fs)

gh)/(f*3h)

g/（3f）

1500n / [ 3x ( 700n -100n）]＝

A: Xiao Ming's weight is 700N, and the mechanical efficiency is.

I don't know what to ask about.

The use of movable pulleys can save effort. Then it just takes.

a) The movable pulley passes through two ropes, then a pulling force greater than 750N (1500 2) is required to pull up the wooden box of 1500N.

b) If the movable pulley passes through three ropes, then a pulling force greater than 500n (1500 3) is required to pull up a wooden box of 1500n.

40kg is equal to 400n, and it is not mentioned in Figure A, so at this time, Xiao Ming only mentioned 1500n-400n=1100n. And because Figure A is one-half of the force, so Xiao Ming's tensile force is 550N, and Fig. B is one-third of the force of the province, so Xiao Ming can lift it, and because the force is mutual, Xiao Ming's tensile force at this time is 500N, and the tensile force he is subjected to is also 500N, and with Xiao Ming's pressure on the floor 100N, Xiao Ming's gravity is 600N.

If the wooden box is lifted by one meter, the useful work is 1500n times 1m, and the total work is 500n times 3m, because one-third of the force in Figure B will take three times the distance, so the mechanical efficiency is 1500 divided by 1500 is equal to 100%. Because you ignore friction and don't do extra work, the mechanical efficiency is 100%.

Hope it can help you and adopt my answer, thank you.

1. Let Xiao Ming's weight be g, when Xiao Ming uses the device shown in Figure A to lift heavy objects, the following equation holds:

1500N-40kg 10N kg+G) 2=G..1)

When Xiao Ming lifts a heavy object with the device shown in Figure B, the following equation holds:

1500N+G) 3=G-100N.2)

Lianli above two formulas, solution g = 300N, substitution (1), solution g = 700N2, Xiao Ming lifting the wooden box when the mechanical efficiency of the pulley block is:

n=g nf=1500n (700n-100n) 3=I hope you can understand, and I hope you can adopt and agree!

(1) W useful = m1gh = 600kg 10n kg 12m = (2) according to = w useful w total.

1=m1gh/3f1h

2=m2gh/3f2h

It can be obtained f1 f2 = m1 2: m2 1 = 6: 5 (3) p elevator = 75% p output = 6000w

v rope = p f1 =

v elevator = v rope 3=

The first question is whether it is a normal operation and a day's power consumption.

If so, the electricity consumption in one day = 1000 * 24 1000 = 24 kWh = 24 kWh.

Operating current = p u = 1000 220 =

The second question is the daily energy saving.

30 days = degrees = 75kw h

1.Normal operation indicates that it is the working state at the rated power, which is 1000W·h

Solution: (1) If the ball is solid, its volume should be: v iron = m iron = = because v iron v, so the ball is hollow

A: This ball is hollow

2) v null = v - v iron =

m water = water v empty =

Answer: The hollow part can be filled with 300g of water

1. Calculate the density of this ball, if the density is equal to iron, it is solid, and if it is less than iron, it is hollow.

2. If it is solid, directly multiply the density of iron by the volume of the ball is 3, divide the mass of the iron ball by the density of iron, and calculate the volume of iron in the iron ball, and subtract the volume of iron from the total volume is the volume of water. The density of the water is multiplied by the volume of the water to calculate the mass of the water, and the two add up.

The arithmetic is your own calculation, and the physics idea is the most important. Because many of the questions are just changed numbers.

2. If it is solid, its mass m= iron v iron =

1. The actual mass of the iron ball is m ball = , so the iron ball is hollow.

3. If the hollow part is filled with iron, its mass is m iron = m - m ball = v empty = v iron = m iron iron = =

m water = water v empty =

Suspension needs to maintain a balance of forces, and the suction force in the minimum direction upward (i.e., the pressure difference between up and down) should be balanced by gravity. Pressure calculation force area, disc area r 2 = square meter, then the pressure difference between up and down is.

Standard atmospheric pressure 100kPa

For the analysis of the force of the glass, except for the part of the disc sucked, the atmospheric pressure of the other parts are balanced, only the disc part of the glass is considered, there is gravity downward, the size is, atmospheric pressure, upward, the air pressure in the suction cup, downward, the elastic force of the suction cup deformation, downward, to ensure that the glass is sucked, the elastic force must be greater than 0. Otherwise, it's dropped. If the air pressure in the suction cup is x, then there is an equation:

s (suction cup area) x + g (glass gravity) + f (suction cup elastic force) = s p (standard atmospheric pressure), known amount is brought in, and f must be greater than 0, x<96000 can be obtained. That is, the maximum air pressure can only be 96kpa.

I'm in my first year of high school, and the content of the pressure is a bit rusty, forgive me.

Thought Process Pressure * Force Area = Force Area There are 2 Yes is a circle and the other is a square with a side length of 1 m, but because it is a suction cup that absorbs the glass and the area is smaller than the glass (diameter. 04m > 1m) So the stressed area is the area of the suction cup.

Calculate the pressure = force force area ( OK (I don't know if it's right or wrong, the unit also forgot.) ）

p=fs

f= s=atmospheric pressure-p=the answer you want.

Isn't the standard atmospheric pressure 101kpa?