Grade 9 Physics Pressure Calculation Question 10

Known: G1470N

Meter: p1=?

Solution: p1=f s=g s=1470 Pa.

Answer: The pressure of water on the bottom of the drum p1 is 5880 Pa.

Known: g bucket + water.

2000ns

1 m g-wood.

100n seek: p2=?

Solution: g total = g wood.

g bucket + water = 2000 + 100 = 2100n

P2F S=G S=2100 1=2100Pa.

Answer: The pressure p2 of the plank against the ground is 2100 Pa.

Hope it helps.

A 300 cm3 bottle contains 200 cm3 of water, and a thirsty crow throws a pebble with a mass of 12 grams into the bottle at a time, and when the crow throws 25 identical pebbles, the water rises to the mouth of the bottle. (g takes 10 N kg).

Find: (1) What is the total volume of the stones in the bottle?

2) What is the density of the stones?

3) How buoyant is each stone in the water?

4) If the contact area between the bottom of the bottle and the ground is 50cm2, how much pressure does the bottle have on the ground when the water surface rises to the mouth of the bottle? (Bottle mass is not counted).

It is equal to the pressure divided by the area of action.

The pressure is equal to the gravitational force: f g mg = 3kg x 10n kg = 30n

Force area: s 100cm2=

Pressure: P=F S=30N

p=f/s f=mg=3*10=30n

s = 100 * 10 minus 4 power = square meters.

This gives 3000pa

Large-scale equipment with a mass of 200 tons is transported on a horizontal road with a large flatbed truck with a mass of 13 tons. It is known that the contact surface between each wheel of the flatbed truck and the ground is 140 square centimeters, and the maximum pressure that can withstand is to the 6th power of PA, so how many wheels does the flatbed truck have at least? (g takes 10 n kg).

Answer: 200t = 200000kg 13t = 13000kg 140cm2 = multiplied by 10 to the power of -2.

f=g=mg=(200000kg, 13000kg)*10nkg=sixth power n

Let there be at least x wheels, according to the title.

6 to the sixth power divided by (multiply 10 by -2 x x) less than or equal to 6 to the power of pa

x is greater than or equal to 102 (to use the introductory method).

An object is a cube, 10 cm wide and 1 cm thick, and it is full of water. Throw it into 100m deep water during a voyage to find the pressure on each surface of the cube.

Pressure: There is a smooth inclined plane with a slope of 30 degrees that is 1m high, and there is a small block of 10m at a speed of less than a second from the bottom of the inclined plane to the inclined plane, and the pressure of the object on the inclined plane when reaching the center height is asked.

Physical pressure calculation questions in the second year of junior high school.

Hang an iron block with a mass of 158g under the spring dynamometer, if the iron block is completely immersed in the water of a cylindrical container with a cross-sectional area of 200, find:

1. > buoyancy of the iron.

2> indication of the spring dynamometer.

3> The increase in the pressure of water on the bottom of the container due to the insertion of the iron block. ( Iron = .)

I looked at it - I didn't write it carefully, I just knew that the mass was kg, that is.

The following calculations, alcohol is taken , g is taken.

Because "the pressure at points A and B at the bottom meter of the two containers is 980 Pa", it can be seen that the depth of water = and the depth of alcohol = in the above. So what are the conditions for starting the depth of water and alcohol in the container? The answer is the depth of water = and the depth of alcohol =.

If water and alcohol are pumped at the same depth, so that the pressure of water and alcohol on the bottom of the two containers A and B is equal, of course, the depth of each should be extracted, so the same volume of extraction = cubic meters, mass: water 3kg, alcohol.

Start the depth of the water as.

Start with a depth of alcohol.

Pumping cubic meters of liquid in the same volume.

Suppose the depth of the extraction is h water g (wine g ( H=

This should be based on the pressure formula water multiplied by g times the height of A = alcohol multiplied by g times the height of B, because the height is equal, so as long as the density of alcohol is equal to the density of water!

The magnitude of buoyancy is equal to the weight of the overflowing water f buoyancy = g row = mg

The block is floating in the water, the lower part of the block is immersed in the water, and the upper part is exposed to the water. According to the buoyancy force equal to the difference between the upper and lower pressure of the liquid on the object, the buoyancy force is the pressure of the water on the lower surface of the wooden socks or socks, and the pressure is f=6n.

The pressure of water on the lower surface of the block p=f s=6n (

The pressure difference crack grinding type is the weight of the beam pressure limiting valve divided by the area of the outlet = 100g*10 (-3)*pie*(

PA15258PA atmospheric pressure.

At least how much force can be withstood at the interface between the lid and the pot?

Pressure Guess Difference * Pot Area = 15258 * Pie * (

Cattle. Note: Indicates to the power of the power.

There are many ways to do this, but you said it was a pressure topic in the third year of junior high school.

That is, to increase the pressure, and the higher the pressure, the higher the boiling point.

Answer: 1. The pressure handle is made wide to reduce the pressure.

2. Two erasers are padded at the bottom to increase the friction between the stapler and the desktop.

3. The staple (needle edge) is very sharp and thin, which is conducive to increasing the pressure.

4. The base is made wide to reduce the pressure.

According to the analysis, the same depth of footprints indicates the same pressure.

p1=p2g1=p1*s1

g2=p2*s2

So the mass of a dinosaur is 400 times that of a fawn, and the mass of a dinosaur is 200kg

PS: The measured area of dinosaur footprints mentioned in the title is 4dm, which is understood as the total area of two feet).

The footprints are the same depth, so the pressure on the ground is the same.

Pressure = Force Force Area under force.

The deer has four legs, that is, its ground force area is s1 = 4 * 10 cm 2 = dragon feet, the same way s2 = 8dm 2=

So f1 s1 = f2 s2

That is, 20*g

Dragon mass x = 400kg