Elementary 3 physics questions, only for details, detailed and detailed

The density of the prepared brine is required to be p= x 10 kg m, and now there are some brines with a volume of v= and a mass of m=, so the density of this brine is.

p=m v=600g 500cm=, its density is higher than the required brine density.

So it doesn't meet the requirements. The existing brine is dense, so it is diluted with water.

Suppose you need to add m grams of water, the density of water is p= x 10 kg m 1g cm After adding m grams of water, the density of the brine meets the requirements, and the volume of m grams of water added is v water, and its mass is m water.

At this time, the total mass of the brine is: the mass of the original brine The mass of the water added m grams, i.e.

m+m water. The total volume of the brine at this time is: the volume of the original brine The volume of m grams of water added, i.e.

v+v water. At this time, the brine meets the requirements, and its density is p p ( m + m water ) (v + v water) = (600 + m water) (500 + v water) =

Replace m as m=pwatervwater, and substitute it into the above formula, i.e., there is.

600 1 V water water).

Water 50v water 500cm, that is, you need to add 500 cubic centimeters of water.

Calculated in terms of v=.

Density * Volume = Mass.

m water = 500g

The quality of the prepared brine is 500*

When 50g of salt is added, it just meets the standard.

There are 60g of salt and 10g left

With a ratio of 10 500 = 10 100, add 100g of water

Solution: The use of movable pulley is labor-saving and not work-saving, although the force used by the movable pulley is half of that of the fixed pulley, but the length of the portable rope is 2 times that of the fixed pulley, w = FH and w = 1 2FX2H 2 FH w = W and both t are the same P = P moving.

Solution: (1) When the water of the tank is just immersed in the pontoon A, the water depth is: H=H1+H2=At this time, the pressure at B is: P= GH=1000 10 The area of the cover piece B: S sheet = 60cm =

The pressure of water on cover sheet B is: f = ps sheet = 5000 (2) s cylinder = 400cm =

Because, when the water flowing into the tank just submerges the pontoon A, the cover piece B is opened, and g+f=f floats.

G= vg-f= S cylinder H1G-F=1000 (3) Because, when the water flowing into the tank just submerges the pontoon A, the cover plate B is opened, and at this time, the water depth is H=

Therefore, when the water depth in the tank is lower, the cover piece B is automatically closed again.

The gravitational force of pontoon A needs to be the cause of the pressure on cover sheet B by subtracting the buoyancy force on it.

Answer: Taking "pontoon A + cover sheet B" as a whole, it is subjected to the vertical downward gravity g (the weight of A), the water pressure f on the B sheet, and the vertical upward buoyancy force "F float", when the resultant force of these three forces is 0, it is the critical point of the switch cover.

Since he is a pontoon, how can he be submerged?

Because there is a link under the pontoon A, the cover piece B can pull the pontoon.

What does it mean that the coverslip is set aside? What is set aside?

It should be a typographical error, it should be "open".

A is subjected to three forces, gravity, buoyancy and the pull of the rod. And the tensile force of the rod is equal to the pressure on b.

Although the buoyancy of the pontoon is greater than the gravitational force, A is also affected by the tensile force of the rod; Just like a hydrogen balloon, although the buoyancy force is greater than the gravitational force, it can still sink when it is pulled by hand.

Qiqi with dark circles under his eyes, it's so silly to ask, hehe.

The role of the first step is to use the edge of the table as a fulcrum, the left of the fulcrum is the gravity acting on the center of gravity, and the right is the hook yard gravity acting on the right end.

The second step is to measure the strength arms of both.

The third step is moment balancing.

Your question: Because g2 is the gravity of the wooden ruler and G1 is the gravity of the hook yard, take a closer look at the final gravity of the wooden ruler

You are given a custom resistor labeled "10 ohm amp" and a sliding rheostat marked "30 ohm amp", on the premise of ensuring that all circuit components are complete, if connected to the circuit in series, the maximum current allowed through the circuit is amperes, and the maximum voltage allowed to be added at both ends is 12 volts. Detailed analysis is required.

In order not to burn out the appliance, the maximum passing current should be lower than the current of the electrical applicator with the minimum allowable passage in the electrical appliance, i.e. amperes.

The total resistance of the circuit is 40 ohms, and the maximum current is amps, so the maximum voltage is u=ir=ohms=12V

Hope, thank you!

A and B are two identical rectangular wooden blocks, long L, stacked together, placed on the horizontal tabletop, the end face is parallel to the tabletop, A wooden block is placed on the B wooden block, and the right end has 4 L stretched out, in order to ensure that the two wooden blocks are not turned over, the length of the wooden block B extending out of the table can not exceed? If there is no limit to the distance between the right end of block A, what is the maximum distance between the right end of block A and the edge of the table in order to ensure that the two blocks do not fall over? The answers are 3 8l and 3 4l

When fully loaded, the boat is floating on the water, F float = g = mg = 7 * 10 to the 6th power kg * 10n kg = 7 * 10 to the 7th power n

(1) P gh = 1 103kg m3 10n kg f ps = 5000ps 60 10-4m2 = 30n (2) f full floating v full discharge g = 1 103kg m3 400 10-4m2

GA F Full Float F 80N 30N 50N

3) When the depth of the cylindrical pontoon A immersed in water is H1, the cover sheet B is automatically closed then the F float Ga

Water v row g = ga 1 103kg m3 400 10-4m2 h1 10n kg = 50n

h1= depth of water in the tank h2=