I m in my third year of junior high school, and I hope the big brothers and sisters will help me ans

Hello, it has something to do with the lever, because the center of the whole arm is the midpoint of the balance of the two forces, and it is also the reason for the common point force;

Problem one, because the green dot and the blue dot are diagonal, according to the reason of gravity, they will be clockwise, and the green dot happens to be the upper part of the force arm, and the force cannot be balanced, and your green dot and the blue dot happen to be opposite, so they will fall down with the blue dot;

Problem 2, under the premise of the same force, the position of the red dot is more advantageous than that of the green dot, and there is a certain advantage in the balance of force, so it will not fall relatively speaking, but if you increase the force violently, it will fall, and the blue dot is the center, but if your force is slowly increased when it is moving, then the bottle will not fall, because the friction consumes part of the force during the movement, which plays a role in trying to balance. If you push directly, it is because your thrust is greater than the force balanced by the red dot, so it will tipple;

The third problem is that due to the friction force consumed in the process of movement, it has an obstructive effect, so it will become slower and slower; Because the bottle is always subjected to a gravitational force that is downward, and the force you give is not downward, it can be solved according to the principle of force synthesis, and you can first analyze the force on the bottle and then draw the conclusion of your problem.

Hope it helps, thank you.

The following points should be taken into account when the bottle moves under force: the position and size of the force point, the friction area of the friction point, the size of the air and friction area resistance, the inertia force, and the center of gravity of the bottle;

When the red dot is stressed, the magnitude of the force is just enough to make the air resistance of the bottle and the resistance of the friction point the same and equal, the center of gravity of the bottle remains unchanged, and the bottle translates under inertia without falling.

When the bottle is stressed by a large force, the air resistance of the bottle and the friction resistance of the friction point are unequal, the bottle will tilt, the center of gravity of the bottle will be shifted, and the bottle cannot be translated and falls down under inertia.

Yes, it has to do with the principle of leverage.

Push the green dot, the fulcrum is the blue dot, and the moment (force multiplied force arm) is larger. Because there is friction, the blue dot will not slip, and if the gravity of the bottle is relatively small, the blue dot may slip and it is not a fulcrum.

The bottle is only subjected to the friction of the table to the left during the automatic deceleration and gliding to the right, and if the bottle is high enough, it may turn to the right, because the center of gravity is high and the stability is weak (the bottle is not at the same speed up and down and causes it to turn over).

Haha, I won't reply to the question,,, it's not very clear.

I like your kind of thoughtfulness and diligence very much! There is always a reason to find it! Physics is in life! Stand up!

Take a look at the center of gravity of the bottle.

O great god! How so many.

I lazily wrote it with Zen and gave me the general idea.

1) Prove EHO and DGC congruence.

eh=dg eo=dc heo= gdc (the last two are equal and come out with a parallel line).

Then you can get oh=cg

OHE= CGD, subtract the spring by 180° to get OHG= CGH, and get OH parallel CG

So OH and CG are parallel and equal, so OHCG is a parallelogram.

This kind of proof topic is annoying to write proof that trembling dust, but in fact, the idea is quite simple.

2) Yes, the dg remains the same.

Because it is a perpendicular line that C leads to both sides, the OECD is a rectangle that is constantly established.

oc is the radius of this thing, which is also the diagonal of the rectangle.

oc=de=3 is certain, so dg=1 3de=1 should be, that's right, welcome to correct.

Series circuit: u=u1+u2; i=i1=i21) r1=u1/i=2/

2) p2=u2*i=(6-2)*

The ammeter has the maximum current.

u=i(r1+r2)

6=R2=0 (When R2=0, the circuit current will not exceed the maximum value of the voltmeter by 3V

then i=i1=3 r1=; That is to say, to ensure that the electro-dust imitation silver pressure gauge does not exceed the range, the maximum circuit current is.

6=r2=10ω

This party banquet means that R2 cannot be less than 10, and the voltmeter value will not be greater than 3V. r2≥10ω

When r2=10: u2=ur2 (r1+r2)=3v;

p2=u2^2/r2=9/10=

When r2=50: u2'=ur2/(r1+r2)=6*50/60=5v；

p2'=u2'^2/r2=25/50=

Let the vertex formula y=a(x+1) 2+4, bring c(0,3) into a(0+1) 2+4=3, and the solution will be a=-1, so the analytical formula is y=-(x+1) 2+4=-x 2-2x+3, so that y=0, and the solution will be x=-3 or x=1, which is discussed in two cases.

y=ax^2 bx=c？？You're not mistaken? It's c.