A high level of physics. A physics in the first year of high school

Choose A, B, C

Because the stem does not say whether acceleration is the same as velocity or opposite.

If it is the same, there is an acceleration in the same direction and the velocity increases until the acceleration is reduced to 0 and the velocity reaches its maximum. Choose A

Similarly, if it is the other way around, there is a reverse acceleration and the velocity decreases until the acceleration decreases to 0 and the velocity reaches a minimum. Pick B

According to the definition of acceleration: the rate of change of velocity versus time. Since the acceleration decreases, the possibility of choosing C by A must be selected, and D can be ruled out

a, b, and c are in the same direction as the acceleration, then a is true.

If it is reversed, then b is true.

The amount of change in velocity t = rate of change of velocity = acceleration so it becomes smaller.

You can analyze this roughly in the direction of acceleration a, and one is the same as velocity, so it's an acceleration motion with less acceleration.

The other is the reversal of acceleration and velocity, which is the deceleration motion with reduced acceleration.

So I chose ABC, because the acceleration became smaller, and the VT-VO became smaller and smaller, so there was a C option.

The abc positive acceleration gradually decreases, and the velocity is still increasing because of this acceleration. Negative is the opposite. It's just that the rate of increase in speed is getting smaller and smaller.

Select b c d

Because the acceleration becomes smaller. Then its speed must be reduced.

ac velocity decreases to 0, acceleration can't be 0, so velocity is certainly not getting smaller and smaller.

a, b, c because they do not tell the object that the motion is reversed, if it is the same as the acceleration, then a pair, vice versa then b pair, c because the acceleration gradually decreases, so the rate of change of velocity is getting smaller and smaller.

Let the atmospheric pressure be p0, and the air pressure p left at the left end (20+18)s=p0 20s=nrt. Get p left = 20p0 38

According to the principle of the communicator, the pressure of the left end to the bottom drops p0(1-20 38) +18cmHg, then in order to balance, the right side should also drop the same pressure, and the air pressure on the right side remains unchanged, and the pressure is provided by the drop of the liquid level. So the right side descends 76 (1-20 38) + 18 = 54 cmHg

Ideal equations. You're a high school freshman.

The diagram of the force shows that the tension of the rope is along the direction of the rope, and the pressure on the AB rod must be in the direction of BA, otherwise the rod will rotate!

fab=gtan60°=

F rope = 2g = 40n

There should be a lack of conditions, is C directly above A or at the level of the rod If so, the force analysis of B shows that the effect of force BC and force AB is upwards 20N, that is, the resultant force of the two is upwards 20N The triangular principle of distance force can know that the rope tension force is 40N, the rod pressure is 20, and the number is 3N

Let the tension of the BC rope be t, and the thrust of the ab rod is n, orthogonal decomposition.

cos60t=g=20

xin30t=n

So t = 40 Newtons.

n = 20 Newtons.

v1^2-v0^2=2as/n;

v2^2-v1^2=2(a+a/n)*s/n;

v3^2-v2^2=2(a+2a/n)*s/n...

vn 2-v(n-1) 2=2(a+(n-1)a n)*s n left addition = right addition to get.

vn^2-v0^2=2(na+(n-1)a/2)*s/nvn=[v0^2+(3n-1)as/n]^

How to increase the acceleration of uniform linear motion?

I saw that I saw the wrong picture at first, so I corrected it.

The amount of change in the energy energy of the gas is composed of two parts, one is the work done by the outside world to the gas w, and the other is the heat q transmitted to the gas by the outside world, if the algebraic sum of w + q is positive, the internal energy increases, and vice versa decreases.

Since it is an adiabatic system, q = 0. The second factor we do not consider the following workmanship, when the rotation is up, the piston does centrifugal movement, the air volume increases, and the pressure decreases.

Because the pressure of the enclosed gas on the piston is consistent with the direction of movement of the piston, the gas in the piston cavity does positive work on the piston, or conversely, the external gas does negative work on the enclosed gas, W is less than zero, so the energy of the enclosed gas can be reduced. Pick B

There are two possibilities, one: the bevel is smooth, and the other is not smooth.

A false: If it is smooth, the work done by gravity is equal to the work done by overcoming f, indicating that the external force does not do work on the object, and the block has a uniform velocity. If it is not smooth, the friction will slope upwards and the block will slow down and slide.

b pair, by three forces, gravity, support force, f, illustrate smooth, i.e., uniform velocity, i.e., the net force is zero. C false: when the angle between f and the inclined plane is equal to the angle between gravity and the inclined plane, the supporting force is zero, that is, the oblique splitting does not give force, that is, the friction force is zero, so there is no left.

D right: This question has always been done under the premise that there is an initial velocity and the ground is not smooth. Look at a special case of an isosceles right angle, the coefficient of friction is 1, and the friction and pressure are offset horizontally.

c a block may do an accelerated downward slide; It should be decelerated and sliding, and the inclined component of gravity g and the inclined component of f cancel each other out, but the object is also affected by the frictional force f that is upward in one direction, so it decelerates downward. A false.

Block b may be acted upon by three forces, and the net force is zero; It is known by A that its reasonableness is not zero. B false.

c. The oblique split is subjected to the friction of the ground, and the direction must be horizontal to the left.

The diagonal split after d removal of f may not be subject to the friction of the ground. At this time, the inclined component g1 of gravity g is greater than the frictional force f on the inclined surface, and the object moving downwards has an oblique force to the right. The inclined plane component g1 of gravity g is equal to the frictional force f on the inclined face, the object does not move, and the small object has an oblique to the right force on the oblique split.

The work done by gravity is equal to the work done by overcoming f, that is, after f decomposition and gravitational decomposition, the magnitude of the object is equal in the direction of motion (i.e., along the inclined plane), and the block is not subject to friction; As shown in Figure f, the direction deviates from the vertical direction to the right, and the third force is required to balance according to the triangle law, and the supporting force exists;

Analyze 4 options: a The object is balanced by forces along the inclined plane, and there can be no acceleration;

b The object is subjected to 3 forces and has no motion perpendicular to the inclined plane, so it is balanced and the net force is zero.

c The supporting force on the object has a left horizontal component, so the object has a force to the right of the inclined plane, and the inclined plane is balanced by the ground friction to the left.

After d removes f, the inclined plane also gives the object a supporting force, and the inclined plane is also under the pressure of the object, so there must be ground friction.

Answer: BC

The motion is decomposed into two sub-motions along the rod and the vertical rod, and part of the fulcrum motion is VSIN, but the thin rod should be regarded as a circle with C point as the center and C to the tangent point as the radius, the radius of the circle is R Tan, and the angular velocity is VSIN (R Tan).

i.e. vsin tan r

Hope it helps.

The angular velocity is equal to the linear velocity divided by the radius, and the thin rod rotates on the axis at point C, so the radius of the angular velocity is: rcot

The linear velocity in the problem is equal to the velocity you find: v'=vsin

Then the angular velocity is: w=vsin rcot =vsin tan r

This problem is the problem of the synthesis and decomposition of velocity: Analysis: The velocity of grasping the tangent point and moving horizontally to the right can be decomposed into two sub-velocities along the direction of the rod and perpendicular rod.

The vertical partial velocity in the period is VSIN, so the angular velocity is VSIN (r tan )=vsin tan r