A few physics questions, I hope that well wishers will help and ask for the process.

1. The range of water overflowing from the cup is 0--10 grams;

2. Greater than or equal to 10 grams;

3. 0-10 grams of overflowed water;

Alcohol = Alcohol Alcohol Calculate the volume of alcohol spilling out, in fact, calculate the volume of the metal block v.

According to mwater= water v, the value range of the overflowing v water is greater than or equal to 0 less than or equal to v alcohol.

In this way, the range of values for v water can be calculated.

1.(1) When the amount of water in the cup filled with water is small, it can be achieved without overflow at all, so at this time M overflow = 0

2) Considering that the density of alcohol is less than that of water, then the maximum mass of the spilled water can only be achieved if the metal sinks to the bottom in both alcohol and water, i.e., both are submerged, and the volume of the two spills is equal (at this time the cup also needs to be filled with water).

v overflow = v substance = v overflow alcohol = m alcohol alcohol = 8g m overflow = water v overflow = 1g cm3 10cm3 = 10g The combination of the two can be obtained, and the mass range of overflow water is 0g m overflow 10g2Analysis: The minimum floor for overflow water is:

The glass is "filled" with alcohol, and the metal block floats in both alcohol and water (when the spill is equal in volume, the mass of the water is minimal, thinking that the density of water is greater than that of alcohol).

At this time, because the object is always floating m overflow = m overflow alcohol = 8g If the container is large enough and the mass and volume of the metal block are large enough, the mass of the overflow water can reach infinity.

That is: M overflow 8g

3.The amount of water is small enough to overflow the water level to 0

If the container is large enough and the mass and volume of the metal block are large enough, the mass of the spilled water can reach infinity.

i.e.: m overflow 0

The wind was blowing from west to east, and the speed of the wind was the same as the speed at which he was running.

It seems to be based on junior high school knowledge, and there is such a name.

Relative motion. of, frame of reference.

When using him as a reference, he should be regarded as stationary, and relative to the second object, such as the ground, at this time the ground is backwards, and this is the reason for the choice of reference, understand!

The wind speed is 18km h!The direction is from west to east! Because it is in a vertical falling state. So it's relatively static! So it's the same speed as Li Ming!

6.(1) CDAB (4) Small, I am strong in physics in the third year of junior high school. There is no picture in the small question! You give me bounty points first, I'm helping you! Thank you.

cab you don't have a picture! The density is on the high side.

Looking down, the reading is greater than it actually is.

Hello, according to the conservation of energy, the starting mechanical energy is mgh+mv2 2 (we choose the ground as the reference potential energy point).

Then the mechanical energy of landing is kinetic energy, and the gravitational potential energy is 0, so kinetic energy = mgh + mv 2 2 choose d

Option D, according to the law of conservation of mechanical energy, the kinetic energy in the horizontal direction does not change 1 2mv0 2, the gravitational potential energy in the vertical direction is converted into kinetic energy, and the kinetic energy is mgh, so the kinetic energy when landing is 1 2mv0 2+mgh

d According to the conservation of energy, the ground is selected as the zero potential point, and the initial mechanical energy is 1 2mv0 2+mgh

The last mechanical energy is the kinetic energy (the gravitational potential energy is zero).

I can't see the diagram.

But the big difficulty of this question should be D.

When the stone rises and falls to the height of h above the ground after being thrown, the horizontal velocity does not change, the vertical velocity does not change, and the direction is opposite, [this is the key] ......

Regardless of the resistance, then use the conservation of energy to solve the problem, and the kinetic energy thrown out is converted into gravitational potential energy, and the original gravitational potential energy is added.

Choose d, from the perspective of energy conversion, when throwing, by the kinetic energy decreases, the gravitational potential energy increases, when the highest point is reached, the kinetic energy is zero, at this time from the highest point begins to decline, the potential energy decreases and the kinetic energy increases, when returning to the height of the throwing point, the obvious gravitational potential energy is converted into kinetic energy (1 2mv0 2), at this time the gravitational potential energy continues to convert kinetic energy, the increased kinetic energy is obviously the reduced gravitational potential energy, how much gravitational potential energy is reduced? Obviously, it should be mgh, so the kinetic energy is also increased mgh, so the kinetic energy when it reaches the ground should be: 1 2mv0 2+mgh

d first analyzes according to the theorem of conservation of energy:

The energy before throwing is gravitational potential energy: mgh and stone kinetic energy 1 2mv0 2

Excluding the air workhorse. That is, there is no loss. That's mgh+1 2mv0 2

Pick D. According to the law of conservation of energy, the kinetic energy at the beginning is ek=1 2mv0 2, plus the gravitational potential energy ep=mgh, which is 1 2mv0 2+mgh

d Zero potential energy in the horizontal plane. According to the law of conservation of energy, the initial kinetic energy is one-half mv squared, and the potential energy potential: ep=mgh so d is chosen

Let the speed of the vehicle be V, the distance between the vehicles when car A honks is x, V sound = 340m s, after t = 4s, the sound emitted by A meets B, vt + V sound t = x, pass t'=20s, A meets B, vt'+vt'=x，x=1511m=

Car A honked the horn and the distance between the vehicles was l kilometers.

4 (v + v) = l v is the speed of sound 4v = 20 (v + v) = 40v

v=v/10

l=4(v+v)=44v/10=44*34 mthankyou

s=(v sound + v car)*4

S = 2 * V car * (20 + 4) = 48V car

That is: (V sound + V car) * 4 = 48V car.

4V sound = 44V car.

V-car = V-sound 11

s=48*v-sound 11

Let the speed of the vehicle be x, the distance between the two cars when car A whistles is s, V sound = 340m s, after t = 4s, the sound emitted by A meets B, then s=340*4+4*x sound propagation process, car B is also moving, and the sound emitting point remains unchanged.

After t'=20s, A and B meet, s=2x*(20+4) can be understood as: the clock starts when the horn is sounded, at this time both cars are at the starting point, and a total of 24s have experienced to meet;

x=340/11；s=