Two balls with the same mass, their density ratio is 3 2

Suppose the density of A is 3, and the body is 1; Then the density of B pants is 1, and the volume is 3

So it can be calculated that the mass of A is 3*1=3, and the mass of B is 1*3=3

So the mass ratio is 3:3=1:1

Assuming that the mass of A is 3a and the volume is b, then the quality of B is 2a and the volume is 3bAccording to the density = mass volume, the density ratio of X0d A and B = 3a b: 2a 3b = 9:2

Root potato with hail = m v can be obtained, v1=m1 1, v2=m2 2v1 number sail v2=m1 2 m2 1=(m1 m2)( 2 1)=(2 hidden respect 1)(2 3)=4 3

That is, v1:v2=4:3

Answer: Slightly.

According to =m v, v1=m1 1, v2=m2 2v1 v2=m1 2 m2 1=(m1 m2)( 2 1)=(2 1)(2 3)=4 3

That is, v1:v2=4:3

Answer: Slightly. ,1,4:3,2,p=m divided by v, and the example is 4:3,1,2 3:1 2,1,

The ratio of the masses of the two objects A and B is 3:2, and the ratio of the volume of the collision is 1:3;

From the above conditions, it can be seen that if the volume ratio is 1:1, the mass ratio is 9:2, and when the volume is the same, the density ratio and the quality ratio are the same.

It can also be assumed that the mass of object A is 1800 kg, the mass of object B is 1200 kg, and the volume is 1 cubic meter and 3 cubic meters, respectively.

Then the densities are kilograms per cubic meter and kilograms per cubic meter, and the ratio is 9:2

The ratio of densities is 2:3, and the masses are equal.

The volume ratio is 3:2

When submerged, the buoyancy ratio is 3:2

The buoyancy of water against the two balls fA:fB=6:5

Explain that A is floating and B is submerged.

That is, V A: V A row = 15:12 = 5:4

then p A * v A = p water * v A row.

P A: P water = 4:5

p A = 800kg cubic meters.

p B = 1200kg cubic meters.

The ratio of the volume of balls a and b.

v1/v2=(m/ρ1)/(m/ρ2)=ρ2/ρ1=3/2，v1/v2=3/2

If both balls are floating, the buoyancy ratio of balls A and B is 1:1. Not on topic.

If both balls are submerged, the buoyancy ratio of balls A and B is 3:2. Not on topic.

Let's set: A and B balls are buoyant to f1 and f2 respectively.

According to the title, f1 f2=6 5, a ball floats, b ball sinks, f1 f2=(1v1g) (water v2g) = (1v1) (water v2) substitute data.

A ball density 1=800kg m3

b-ball density 2=1200kg m3

First judge Hu Xuan to determine the state of A and B:

First of all, it is impossible for both A and B to do the flotation (basis: the mass is equal, buoyancy = gravity, but the buoyancy ratio is 3:2).

Secondly, it is impossible to sink all of them (basis: if they all sink, the volume ratio is 2:1, and the buoyancy ratio should be: 1).

Therefore, A floats, and B sinks or levitates.

Let the buoyancy of A be 3f and B 2f

Then the volume of B is v B = 2f g * p water, and the volume of A v A = 4f g * p water.

The gravitational force on A is g = f float = 3f

V A = 3 F g * P A = 4 F g * P water.

Get P A = water.

P B = water = 1500 kg m 3

Because it is immersed in the liquid, the F float density * object volume * g has F float A liquid density * methyl volume * g ......(1) F float B liquid density * B volume * g ......(2) Volume mass density, i.e., volume is inversely proportional to density.

1) (2) Get: F Float A: F Float B 5 4*2 3=10:12=5:6