A physical problem about density, and a physical problem about density

1. Fill the sink with water, submerge the bracelet in water, and measure the amount of liquid level drop l1.

2. Fill the sink with water, immerse the wooden blocks in the water freely, and measure the amount of liquid level drop l2.

3. Fill the sink with water, and then freely soak the wooden block and bracelet in the water to measure the amount of liquid level drop l3.

4. The density of the bracelet is P water*(L3-L2) L1

1.Use a scale to measure the side length a of the cube block, so as to calculate the base area a;

2.Place the cube wooden block into the water and measure the height h of the outcropping water;

3.Place the hand casting on the wooden block and measure the height of the outcropping water again h1;

4.Tie the hand-cast to the wooden block with a thin wire, put the hand-cast together with the wooden block in the water (the hand-cast is under the wooden block), and measure the height of the wooden block above the water surface h2

Density p=m v

m=pwater*a (h-h1);

v=a²（h2-h1）；

p = p water (h-h1) (h-h1).

1. Place the wooden block in the water and measure the size of the wooden block above the water surface.

2. Place the bracelet on the floating wooden block and measure the size of the wooden block on the water surface again.

3. Hang the bracelet under the wooden block with a thread, place it in the water, measure the size of the wooden block on the water surface three times 4, sort out and calculate, and get the density of the bracelet.

Fill the sink with water, put the wooden block in the water, take out the wooden block after a part of the water overflows, and measure the depth of the overflowing water x1 with a scale. In the same way, tie the jade bracelet with a thin wire and put it in the water, and then measure the depth of the overflow water x2. Measure the depth and side length of the cube wooden block submerged in the water, you can know the volume of (x1-x2)s=v jade bracelet, and then measure the quality of the jade bracelet, you can measure the density of the jade bracelet.

According to the problem, the volume v of the part does not change, and the density decreases and the mass decreases. Therefore, the original mass of the part [steel]*v is changed, and the mass becomes [aluminum]*v, and the difference is m=104kg:

Steel]*v- [aluminum]*v=m==>v=104 (7900-2700)m3

The required mass of aluminum metal m = [aluminum]*v=2700*104 5200=54kg

The first one is very simple, first measure the quality of the mineral water bottle, and then measure the weight of the bottle filled with water and milk separately to calculate the net weight of water and milk, the volume is equal, the mass is proportional to the density, and the milk density can be obtained.

The second, 1Weigh the beakers separately.

2.A beaker is filled with water and weighed.

3.Place the metal block into a beaker filled with water, and the overflow is caught by the second beaker. Weigh separately.

4.Calculate the volume of the metal block using the water remaining in the second cup.

5.The weight of the first cup is used to calculate the mass of the metal block.

6.Mass divided by volume gives the density.

The third, 1Place the beaker on both ends of the scale, one with a metal block and the other with a measuring cup to weigh the water until the balance is balanced. The volume of water is calculated, from which the mass of the metal block is derived.

2.Repeat step 1, but fill both beakers with water and calculate the volume of water separately until the difference between the two sides is the volume of the metal block.

3.Mass divided by volume, OK!

Question 1: 1The bottle is filled with water, and the mass m water is measured with a balance. 2.The bottle is filled with milk, and the quality of the milk is measured with a balance. 3.Density of milk = m milk multiplied by the density of water in the mass divided by water. (Rolled out according to the same volume of water and milk).

(1): Fill a certain amount of water in the mineral water bottle, mark it, measure the weight with a balance, then pour out the water and drain it. The milk is then filled to the mark and the quality is measured. Divide the mass of milk by the mass of water and multiply the ratio by the density of water, which is the density of milk.

2):1: Use the balance to measure the mass m1 of the metal block

2: Fill the beaker with appropriate water and measure the mass m2

3: Put the metal block into the water and measure the mass m3

4: The buoyancy of the metal block with m1 + m2-m3 = m4, that is, the mass of the water discharged, 5: m4 divided by the density of the water is the v of the metal block

6: m1 divided by v, which is the density.

3):1: Put a measuring cylinder and an appropriate amount of beaker and water on the right and left of the balance respectively, and add water to the graduated cylinder to balance the balance and mark the water level of the graduated cylinder A

2: Put the object on the left side of the scale, add water to the right plate continuously to balance the scale, and note that this is the water level B of the graduated cylinder

3: b-a=m, where m is the mass of the metal block.

4: Use a graduated cylinder to measure the V of the metal block

5: m v is density.

The water from the large cup has not yet reached the overflow level (or the large cup of water is not just filled).

You know. The overflowing water was not exactly the volume of the stone, as there was still a portion of the large cup that was not counted.

You'd better bring the unit with you, which is 67g of water (m2-m1), because water, m1, m2 actually contain units.

The volume of the ball is the volume of the spilled alcohol, i.e. v ball = v wine = m wine =

The buoyancy received by the ball in the water = the gravity of the ball, i.e. m ball g = f float = m water discharge g, so the mass of the ball is the mass of the water discharged =

Ball = M Ball V Ball =

If the pellet sinks in the alcohol, the volume of the pellet is equal to the volume of the discarded alcohol.

So. v ball = v row = m row alcohol alcohol =

The ball floats on the surface of the water, then.

m-ball = g ball g=g drain boiling water g=(m drain boiling water * g) g=m drain boiling water = so. Ball = M Ball V Ball =

F float = g row = liquid v row g

Solution: Let the liquid that needs to be A is M A kg, then.

M liquid = M A + M water.

m liquid liquid = m A A + m water water.

Replace m water = 1 kg; Water = 1000kg m; liquid = 900kg m; A = 800kg m, substituting the above equation to obtain:

M liquid = M A + 1

M liquid 900 = M A 800 + 1 1000

Solve the above binary linear equation system to obtain: m liquid =, m A =

A: The maximum kilogram of solution can be prepared.

One less condition: the volume of liquid A does not change after fusion with water.

Liquid = (M A+1) (V A+1) =

v A = M A

Liquid = (M A+1) (M A

Solution: M A = M liquid = M A + 1 =

What is written upstairs is all wrong. Binary linear equations cannot be solved!!

Remember to adopt and recommend Ha

It is impossible to formulate, and the density of liquid A is not as large as liquid B!

A density, B density, evaporation solvent.

Is a system of equations sufficient?

60%×v1+30%×v2）÷（v1+v2）=60%

v1+v2=1000

Solution: If 60 degrees of liquor x ml is required, then 30 degrees of liquor (1000-x) ml 60% x + 30% * (1000-x) = 42% * 1000 solution is obtained: x = 400 (ml).

So, 1000-x = 600 (ml).

Answer: Slightly. Description: Because after blending, the volume reduction after mixing is not considered. Therefore, the volume of pure alcohol after blending should be equal to the volume of pure alcohol in 60 degrees + the volume of pure alcohol in 30 degrees.

There is no need to involve density and quality in this topic, otherwise it will become quite troublesome and waste unnecessary time.

The density of the prepared brine is p=mv=>, and water should be added.

Let the addition of water xkg that is xdm3, then , the solution is x=

The more salt, the denser the brine.

The actual density is calculated from the volume and mass given, and compared with the required density, if it is small, salt should be added, and water should be added if it is large.

If you want to add salt, the volume of salt is negligible, and the volume of salt water remains the same, the topic requires the density to meet the requirements, which means that the quality must meet the requirements, and the actual quality and required quality (how much weight should this bottle of salt water) can be reduced.

If you want to add water, set the water x ml, calculate the weight of y grams with density, and require density = weight after adding water divided by the volume after adding water, a one-dimensional equation, and solve it.

Now the density of brine is kilogram cubic meter = 1200 kilogram cubic meter = gram cubic centimeter.

Because gram cubic centimeter "gram cubic centimeter so to add water set to add water x gram, then.

600 g + x gram.

500 cubic centimeters + x grams 1

600+ x=550+

x=500 so add 500 grams of water.

I don't know if it's right.

Density = m v

600g/5*10^2cm^2

So add water.

Let the added water be x

The steps to solve the equations are omitted, and the math class should be taught.

x=500g

So add 500g of water.