In the second year of junior high school, science and physics topics are understood

Solution: The submarine moves at a constant speed at a certain depth, indicating that the buoyancy f (floating) = the weight of the submarine g (diving) at this time

i.e. (water) v (boat) g = m (diving) g

1030kg m 3 * v (boat) = 8000 * 10 3kg can be obtained The appearance volume of the submarine v (boat) =

It can be seen that the buoyancy force of the submarine f (floating) = m (diving) g = 8 * 10 7n The submarine rises at a constant speed, indicating that the buoyancy f (floating) = the mass of the submarine g (rising) + the resistance of the seawater f (resistance).

i.e. f(float) = g (rise) + 2*10 7n

g(rise)=6*10 7n

Then the mass of the submarine when it rises at a constant speed m (rise) = g (rise) g = 6000t The submarine is stationary on the sea surface, indicating that the buoyancy f (buoyancy on the sea surface) = g (rise), that is, v (volume under the sea surface) g = g (rise).

Then v(volume under the sea surface)=

So v (volume outcropping) v (boat).

v (boat) - v (volume under the sea surface)] v (boat) so when the submarine is stationary at the sea surface, the volume exposed to the sea surface is 1 4 of its total volume.

8000t=8*10^6kg

1.According to the title, when the submarine descends, its gravity = buoyancy + seawater resistance of the submarine, 8 * 10 6kg * g = 2 times 10 to the 7th power n + density * volume * g

Substituting the data.

v = cubic meter.

2.According to the title, when the submarine rises, its gravity + seawater resistance = buoyancy on the submarine, the column formula is obtained, m*g + 2 times 10 to the 7th power n = density * volume * g

Substituting the data.

m=4000t

3.When the submarine floats on the water, its mass m=4000t, at this point gravity = buoyancy.

mg = density * j volume immersed in water * g

Volume immersed in water v = cubic meter.

When the submarine is stationary at the surface, the volume of the exposed sea surface is a fraction of its total volume = 1 - the volume immersed in water v total volume = 1 3

Hope it helps.

The exterior volume of the submarine is cubic meters

When a submarine is stationary at sea, the volume of the sea surface is its total volume cubic meter.

The volume of the outcropping of the sea is its total volume 1 4

ROU seawater: multiply 10 by 3 kilograms m, g takes 10n per kilogram, is this the weight of the water discharged? If not, then I don't know how.

The mass of the object lying on the liquid is equal to the volumetric mass of the liquid discharged, so the depth is directly proportional to the mass of the bottle sand and inversely proportional to the density of the liquid. I need to explain the specific calculation.

Gravitational buoyancy.

Volume = buoyancy 10 density of water =

Density = g cm 3

The graph reading given does not show the number.

Solution: It is known that the cross-sectional area of the hollow column hole s = 10 square millimeters = minus 5 square meters of 10, and the mass of the pressure limiting valve m = 100g = the atmospheric pressure at that time p0 = 1 10 to the 5th power of the square, 1If the maximum pressure in the pressure cooker is p, then p=p0+mg s=2 10 to the 5th power of Pa.

2..It can't be used, because the maximum pressure set by the original pressure limiting valve exceeds the maximum allowable pressure, so it is necessary to replace a lighter pressure limiting valve, and let its mass be a, then there are 5 power pascals = p0 + ag sa =

1,20 N

2,20n/

Let's do the calculations yourself This question is mainly to test the understanding of the concept of pressure.

In 1 ml of concentrated sulfuric acid (the mass fraction of the solute is 98%, the density is grams of cubic centimeters), the mass of H2SO4 is grams; If 1ml of concentrated sulfuric acid is mixed with 4ml of water to prepare 1:4 dilute sulfuric acid, the mass fraction of the solute in the dilute sulfuric acid solution is

Answer: The first step is to find the mass of 1ml of concentrated sulfuric acid = 1ml*. Then it is possible to calculate the mass of sulfuric acid = . Then calculate the mass of this 5ml of dilute sulfuric acid =. The mass of H2SO4 in dilute sulfuric acid remains unchanged, so the mass fraction of dilute sulfuric acid = (.