Physics questions about density 10, physics questions about density

1. The amount of the substance contained in the property Shape State position Kilogram 2. Balance Water Platform Adjusting Balance Nut.

3. The mass per unit volume is kilograms per cubic meter.

4. Make it smaller.

5. Mass volume = m v

g error kg m g cm

Use a spring dynamometer to measure the gravity g of the stone in the air, submerge the stone in water, read the tensile force f1, and then submerge the stone in the juice to read the tensile force f2

(g-f2) water (g-f1).

The principle is that the V row is unchanged, (g-f1) water g = (g-f2) g

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

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.

Since the volume is the same, the mixed specific gravity is the average of the two:

Specific gravity of water: The specific gravity of brine is (.)

The density of the liquid after mixing is the cubic of kg m.

The cubic of kg m is cubic.

Let the volume be v

The quality of the water + the quality of the brine) 2v

1000v+1100v)/2v

1050 (kilograms, cubic meters).