Solve 1 junior high school physics problem and solve 1 junior high school physics problem

Solution: (1) Because p=w t w=fs

So p=fv

So 20kw = 50000n v

So v=20000w 50000n=

2) From p=fv, f=p=16000w, ffloat=g-fpull=50000n-40000n=10000nffloat 1 2=5000n

3) F float = vg, i.e. v = f float g = 10000n 1000kg m3 10n kg = 1m3

Because v=sh is s=v h=1m3 2m=

3.Fill the glass bottle as shown in Figure 1 with water, and then plug the cork with a thin glass tube into the glass bottle.

Since there is no diagram, then I can only roughly say the solution.

Look at the height of the liquid level of the glass tube in the middle of the glass bottle.

Holes higher than the liquid level in the glass tube will not allow water to flow out.

Holes that are lower than the liquid level in the glass tube will flow water.

This is because the air pressure at the liquid level in the glass tube is equal to the water pressure here, and in the area above it, the water pressure is lower than the air pressure, so the air is drawn in, and the water pressure is higher than the air pressure in the lower area, and the water shoots out.

Answer: Because there is no force with B, spring A is deformed. When the force acts on spring B until spring A returns to its original state (without force), the elongation of spring A rises by object m, and the connecting line between object M and spring B remains unchanged, so when the force acts on spring B, spring B will move.

Therefore, the elongation of spring a should be added.

The distance traveled at point c is equal to the deformation of a + the deformation of b.

The initial state is a compression -mg k1. (minus sign indicates compression); The end state is a long time. At the end of the minus at the beginning, the mg k1 was elongated

Initial state b original length; In the terminal state, the B deformation is mg K2. At the end of the minus the beginning, elongation mg k2

CD distance = mg k1 + mg k2 = mg (1 k1 + 1 k2) = mg (k1 + k2) k1k2

The distance between CDs should be the elongation of spring B from the absence of force to the amount of MG.

i think so too

When it is located at point C, the upper spring has no deformation, that is, the pulling force given to m is zero, and the gravity of M acts completely on the spring below, and the deformation is mg k2, when it is located at point d, the spring below returns to its original length, that is, m will rise mg k2, and at the same time, the gravity of the object m is offset by the tensile force given by the spring above, so the spring above will elongate mg k1, so the distance of c and d is the answer you said.

When the spring is at point C, mg=k1*a is deformed, and when the spring is at point d, mg=k2*b is deformed.

Solution: (1) The game code on the balance should be moved to the zero line at the left end of the scale, so choose a to measure the same quality of edible oil and water

So the answer is: balance

M oil = m water, while M oil = oil v oil = oil H2S, M water = water v water = water H1S, thus deriving oil = H1H2 water.

So the answer is: h1h2 water.

2) The reasons for the experimental error are many inaccurate height measurements of water and oil, inaccurate quality measurement of oil, different cross-sectional areas of glasses, etc

So the answer is: the cross-sectional area of the two glasses is not exactly equal

Place the glass with the appropriate amount of cooking oil and water in the left and right plates of the balance, and use the rubber tip dropper to increase and decrease the amount of water in the cup to balance the scale.

Take off the two glasses, and use a scale to measure the height of water and oil in the two cups h1 and h2 respectively; then the density of the edible oil =

h1 h2 * water.

2) Xiao Ming thinks that the quality of two cylindrical glasses is not exactly the same, which will cause experimental error Please also mention a possible reason for experimental error: the cross-sectional area of the two glasses is not exactly equal.

List them according to the equal mass, and then reduce the same amount. I'm a Form 3 Physics teacher, so you can ask if you have something.

God, this question is too difficult, how do you answer it if you don't have a question?

Distance from the virtual image to the eye (5m) = distance from the eye to the plane mirror (the distance from the actual eye chart to the plane mirror;

So the distance from the eye chart to the plane mirror = ?

Distance from the eye chart to the plane mirror (eye to plane mirror (distance from the eye to the actual eye chart (;

The first is what is the formula because of symmetry: the distance from the actual eye chart to the plane mirror = the distance from the virtual image to the plane mirror.

The principle of reflection can be known: the distance from the eye to the virtual image is 5m, and the distance from the eye to the mirror is the distance from the eye chart to the mirror.

Learn that the eye chart should be meters behind the eye!

That is, the distance from the eye to the actual eye chart is meters!

Answer: First of all, 5m is the distance from the eye to the mirror plus the distance from the mirror to the eye chart. In this way, it can be concluded that the distance from the mirror to the eye chart is 7m, and the person is in the middle of the mirror and the eye chart, so the answer can be concluded by subtracting it.

Haha, m, 1. The distance from Xiao Wang's eyes to the eyes in the mirror is.

Rice. 2。The distance from the phantom image of the eye chart in the mirror to the flat mirror is.

Rice. 3。According to the symmetry, the distance from the realistic eye chart to the realistic Xiao Wang's eye is. Rice.

The eye chart is on the same side as the person in the mirror, and the distance between the mirror and the eye chart is, the distance between the person and the mirror, so the distance between the person and the eye chart.

The distance from the plane mirror to the virtual image is meters.

The distance from the plane mirror to the actual eye chart is.

So the distance from his eyes to the actual eye chart is 2,

Why didn't I learn this in junior high school???

Problem solving process: 1) After pouring the liquid, the mouth of the barrel is just level with the horizontal plane, and the pressure of water on the bottom of the metal cylinder is the pressure of the water at that depth, that is.

p= water * gh = pa (h is the depth of the bottom of the metal cylinder from the water surface at this time) 2) before pouring the liquid, two-thirds of the volume is immersed in water, at this time buoyancy f1 = 2n;

After pouring the liquid, the mouth of the barrel is exactly level with the water level, that is, the entire volume is immersed in water, and the buoyancy can be known according to the Archimedes' principle.

f2=3n.

Therefore, it can be seen that the weight of the liquid in the metal cylinder is 1n, and the mass of the liquid is m=, which is known by =m v.

m/v=

Two-thirds of the volume of the metal cylinder is submerged in the water alone, and the mouth of the barrel is level with the water surface after the liquid is added. So the mass of the liquid is one-half that of a metal cylinder, i.e.

2.The height of the liquid in the barrel h=125 10=

p = density * gh density = p gh = 3000 (10 * grams per cubic centimeter.)

The buoyancy f1 = g (gravity) = 2n, and the buoyancy force when the liquid is added: f2 = 3 2 * f1 = 3n

The pressure at the bottom of the cylinder: P=F2 S bottom, i.e.: P=3 10=At the same time, F2=2N+G liquid=2N+V liquid* liquid=3N, then:

V liquid * liquid = 1n liquid = 1 10 * 125 (I forgot a little bit about the unit conversion, check it yourself, sorry).

When two-thirds of the volume enters the water, the buoyancy is equal to the gravity is 2n, and the volume is multiple of the original water volume when all the water enters the water (1=2 3*, so the buoyancy is 2n*, that is, the pressure on the bottom surface of the water drum is 3n, and the pressure p=f s=3n; The total weight of the liquid and the barrel is 3n, the liquid weighs 3n-2n = 1n, the mass is 100g, the density = 100g 125 cubic centimeters =.

Solution: 1, when lifting 3 first stove envy slate g, slate = 300n; G round = 50n; g pallet = 200n

Machine-debating efficiency = useful work total work =

2. Slate velocity v1=5m s; The speed of the rope beater v2 = 10m s The tensile force of the free end of the rope f = 1 2 (G slate + G wheel + G supporting plate) = 625N; Power p=f*v=6250w

If the human weight is 700N, the maximum tension force at the free end of the rope can only be 700N, so the maximum gravity at the pulley end is 1400N

n Number of slabs = pcs.

So the number of slabs is 11.

Maximum mechanical efficiency =

=w has a file line w total = (100 * 3 * s) (50 + 200 + 100 * 3) s)) =

p=w t=((200+100*10+50)2) line: 4s=620w

The most important lead is large = w has the maximum w total maximum = (100 * 11) (200 + 100 * 11 + 50) =