A high school physics problem Tell me why

The answer is B, a uniform acceleration of the decline. It should be the second time that the distance between the two rods is close, you can take the cross-section of the cement bucket for force analysis, and you will find that the positive pressure between the cement bucket and the rod is reduced. So the friction of the second time is less than the first. Of course, the decline should be accelerated.

The inclination angle of the support force has changed, there is no diagram, it is inconvenient to explain, and it is prompted that lz is drawn from the direction of the cross section of the barrel for consideration.

Note that the component of the pressure in the vertical direction must be equal to the gravitational force, NCOS MG. So the smaller the angle, the less pressure. Of course, the friction is also smaller.

The distance between the two rods is reduced, and the direction of the force of the cylinder sliding downward remains unchanged, but the angle between the pressure on the rod and the angle of gravity decreases, so the friction force should be increased, so D should be selected.

This is a fantasy in my mind, and you draw a picture to see if that's the case.

As for learning physics, the most important thing is to fully understand the theorems, know the various elements of the theorem, and simulate the theorems in your mind when you can learn them, so that you can not only exercise your spatial imagination ability, but also better understand the theorems.

Then the learning of any subject should be summarized and reflected, which is also the most important thing in any learning.

B should be selected, because the support force of AB and CD on the "cement cylinder" passes through its center (axis), when the distance is reduced, the angle between the two support forces becomes smaller, so the dynamic friction force also becomes smaller, so it will accelerate and slide down.

I also think D is right.

My method of analysis is this, I cut a cylinder vertically into countless slices, and the flakes are infinitely thin. Then, the flake closer to the center should be heavier, and the heaviest is the one with the diameter.

At the beginning, a thin slice is pressed on each stick, and the mass of the flake is set to m1, then the pressure is m1gcosa and then the stick is moved closer to the center, and the mass of the flake pressed on the stick is m2. From the above analysis, it can be seen that m2 is greater than m1, so the pressure of the second time is also greater than that of the first time.

The pressure is high, and the friction increases, so it should not be moving.

At this time, my personal understanding is not correct, and I hope to verify it.

d. Because the contact angle between the tube and the wooden stick becomes larger.

Therefore, the pressure of the pipe on the stick becomes greater (draw the force analysis diagram), so the friction force increases.

So it is only possible to be still.

Let's learn physics and mechanics.

When you learn about magnetism and light, you will know what life is, and it will be extremely dark.

The most important thing to solve a mechanics problem is to analyze the force analysis.

This question is no exception, but unfortunately I can't illustrate at level 1.

Remember to draw a force analysis for each key moment.

Then the formula has to be very skillful.

Don't be nervous in my first year of high school, I couldn't do it in my first year of high school.

Actually, everyone is the same.

I feel that the gap between high school and junior high school is too big.

It's much better after the first year of high school.

With a little confidence, everything will be fine.

Three years, the deer will die, and it is not certain who will do it.

a.bar. Nothing has changed. The friction has not changed. (The friction area has not changed, the friction coefficient has not changed, and the angle has not changed.) ）

The position of the nth carriage passing by the person shows that the displacement of the train is nl, because the train starts at rest and moves in a straight line with uniform acceleration with acceleration a, then let the elapsed time t, then the displacement of the train is 1 2

at 2, so we get an equation: nl=1 2at 2 unknowns t = 2nl a under the root number

When the cockpit falls to a position of 50m above the ground, because it is still in a weightless state, there is no pressure on the hand, there is no feeling, when the cockpit falls to a position of 15m above the ground, because the brake has begun, so the shot put is in an overweight state, and the acceleration is set to a

When braking is started, the speed is v

Then: v 2 2g = s-h

After starting braking, V 2 2A = H

Substituting the data yields: a=120 7

m s 2 set the hand force to f

then f-g=g g*a

Solution: f

The maximum allowable current of the variable resistor is I0=4A, and the maximum current of the variable resistor shunt circuit and the load circuit is 4A, that is, IMAX=4A

Load resistance r=40

So the maximum output voltage UABMAX=Imax*R=4*40=160V shunt circuit minimum resistance: R1min=UAB IMAX=200 4=50 Maximum value of variable resistance in series with the load circuit: R2max=100 Minimum output voltage:

uabmin=uab*r/(r+r2max)=200*40/(40+100)=8000/140=400/7≈

The maximum permissible current is I0 = 4A, and the minimum value of variable resistance shunt: RMIN = UAB I0 = 200 4 = 50

The maximum value of variable resistance shunt: Rmax=140, because the maximum current is 4 A, the minimum value of the resistance of the series part with the load is 10, so the resistance change range of the parallel part with the load is 50 140, and the resistance change range of the series part of the load is 10 100, so the voltage shared by the load is UAB=IR. The minimum value is 200 140 * 40 = 400 7V

The maximum value is 200 50 * 40 = 160 V.

The output voltage varies from V to 160 V.

The maximum voltage value is that when A slides to the top, the load voltage is 200V, considering that the maximum current allowed by the variable resistance is 4A, so the minimum value of the variable resistance below the contact A is; r=u/i=50ω

Then the remaining part of the variable resistor is connected to the power supply in series with the load, so the load sharing voltage is u=UAB*40 (150-50+40)=400 7

Therefore, the possible range of UAB variation: 400 7 UAB 200

Force analysis, for A: Because A does not move, and A is only subjected to friction and tension of the rope horizontally, so fm= mg

To B: mg (i.e., the pulling force of the rope) and the frictional force f2 of B act as the centripetal force of B. At this time, B moves in a uniform circular motion and does not slide.

If you don't understand, you can contact me again.

The centripetal force of B's rotation is provided by the tension of the rope, even if the centripetal force exceeds mg but does not exceed mg, A is still stationary, and B is still subject to the resultant force of tensile force and sliding friction force to do circular motion, and B's static friction and sliding friction have nothing to do with it, they will still be relatively stationary.

Can you read it? I think it's easy to understand something ......It's so difficult to explain......

I'm also a freshman in high school......

B move, do a circular motion.

The pulling force of the rope against B together with the static friction force experienced by B provides the centripetal force.

The maximum pulling force of the rope against B is mg

When on the ground, the gravitational force is approximately equal to the gravitational force:

g0=mg=gmm r2 (2 is squared).

When accelerating at a certain height h, the equation for the resultant external force is:

n—mg1=ma

In this case, the simultaneous equation of g1 = gm (r+h)2 (2 is squared) gives h = 2g0r2 (2n-g0) under the root number

To find the acceleration of the satellite's present position, let the acceleration of the satellite's current position be g1, the mass of the object is g0 g, the support force of the satellite on the object is n, and the resultant external force of the object is mg 2, then there is mg 2=n-mg1. Find g1=(n-g0 2)g g0Then according to the ratio of the gravitational acceleration on the earth's surface to the acceleration obtained at the high position h, g g1=(r+h)*2 r*2 can be found h.

Let the height be h, and there is n-gmm (r+h)2=mg2

Just solve the h.

Let the total displacement be 2x

The time through the first half of the displacement t1 = x 15

Let the time of passing through the last half of the displacement be t2

x=12(t2)/2+6(t2)/2=9t2t2=x/9

The average speed of the whole journey is:

v=2x/(t1+t2)=

The entire displacement is s

The time of half displacement is t s 2 15 s 30 and the remaining half displacement is s 2 12 t 2 6 t 2 9t , so there is s 18t and t is the remaining time.

The average velocity of the whole process v s (t + t ) = s (s 30 + s 18) v can be obtained by subtracting s