Very urgent A high school physics problem, detailed to clean up by

a b a if the three balls meet. Pair A: The point where it meets C is directly below C The point where it meets B is directly in front of B, so the point where the three balls meet is at P.

b to A: When A meets C, the horizontal displacement and B's horizontal displacement are equal so B is at p.

c In the vertical direction, the vertical displacement of A and C is equal, so when A and B meet, A falls to the ground, so C also lands, so C is wrong.

d This is not necessarily, the point where A and B meet can be before and after p, A and B are sure to meet, but when they meet not and C, they need to consider v, so D is wrong.

a b Consider in terms of relative motion. The horizontal velocity of the flat throwing motion remains unchanged, and the movement time of A and C is the same.

Whether the three balls can meet depends on the size of the velocity and the height of A, the correct answer should be A and B, A and C meet then B must meet at point P, A and B meet at point P then C must also meet at point P, you can calculate the relationship between height and initial velocity according to the meeting of A and B or A and C at point P.

A B A and B have equal horizontal velocities, then they may meet at point P, to the left of point P, or to the right of point P (depending on the height of A), and B is directly below A when A does not land.

A and C are in free fall, and they may meet at point p, or they may meet above point p (depending on the horizontal velocity of A), and A and C fall at the same height, so when A and B meet, C falls to the ground.

If B and C meet the two balls only at point P (depending on the horizontal velocity of B and the height of C), if the three balls meet only at point P (depending on the horizontal velocity of A and B and the height of A and C).

Option D This question tests the balance of forces, when the horizontal speed is constant in A, it is only supported by gravity and the conveyor belt, while in B, because the conveyor belt is tilted, the object is subjected to gravity, support and friction. In this way, the balance of forces can be achieved. In this way, if the object is at rest or moving in a uniform linear line, it is a state of equilibrium, and once the force is analyzed, you will find out what is wrong.

The object in a is not subject to friction, and you can prove that it does not exist by counter-proof.

The object in the da diagram is selected to move in a uniform linear velocity, so the resultant external force is 0, so it is only subjected to gravity and elastic force.

Figure B The object moves in a straight line at a uniform velocity, so the resultant external force is 0, so it is subject to an additional frictional force along the inclined surface.

If an object is at a constant velocity, then its net force is 0, and the two forces are in opposite directions (gravity and support), and if there is a third force, it must not be balanced. Answer Just choose in AB, the component of gravity of object B is in the direction of the inclined plane, and its other component is perpendicular to the inclined plane, this force perpendicular to the inclined plane can only be offset by the supporting force, and the remaining force along the inclined plane can only be canceled by frictional force, so it is subject to gravity, friction, and elastic force. The answer is D

First, perform a force analysis:

The ball is subjected to a supporting force, gravity and an external force f.

fx=fcosθ=3mg/2

gx=mgsinθ=mg/2

then fx(combined)=fx-gx=mg

Then it is obtained from Newton's second law.

fx(合)=ma

a=g is known from the geometric relation.

The displacement from point C to the midpoint of BC is h sin 1 2=h and since its initial velocity is 0, v(midpoint) = 2ax is obtained

v (midpoint) = root number (2ax) = root number (2gh) while f (ab rod hinge at b) = fx (combined) = fx-gx = 3 mg2 -mg2 = mg

At the beginning, the pressure of A = atmospheric pressure P0, Pb = P0 + H2 After pouring a small amount of mercury column with a height of ΔH, the pressure of B increases ΔH, so B is compressed, C plane drops, and B plane also drops. B is compressed, so the C side drops more, and A is correct.

After the B surface falls, the A side rises, and the gas A is also compressed, so the pressure increases, and the difference between the pressure of A + AB = the difference of Cd, so the height difference H2 between C and D is greater, and D is correct.

Select: AD.

In this electric field, the net force of a is 0, which can be seen from the uniform motion of a, the gravitational and supporting forces in the vertical direction cancel out, and the horizontal force cannot be acted upon by the force (because if the force from b is received, there is no other force to counteract, so it can only be that the two objects are smooth between them, and their initial velocity is the same, and a is subject to friction and electric field forces, both of which are balanced). Therefore, after B suddenly disappears and A carries the charge of positive Q, since the two objects are smooth and there is no force between them, A is affected by the electric field force, and B is only affected by friction, so the answer is C.

First, the moment is balanced with o as the point of rotation.

Let OA=xm and the angle of the inclined plane is

mg*oa*cos = mg(1-oa)*cos calculates oa = 4 5m

Then take A as the rotation point.

n*oa*cosθ=mg*ab*cosθ

The solution is n=25n

It's a matter of moment balance. O point is the fulcrum, according to the moment balance condition, that is, the power multiplied by the power arm and other resistance by the resistance arm, the length of OA can be calculated by listing the equation, because only the O point has a force on the light wooden strip at this time, so the magnitude of the force on the wooden strip at the O point is the sum of the gravitational force of the two objects.

Opposite. The P2 field strength is e2-e1

Assuming that AB is positively charged, it is assumed that AB is divided into left and right halves, with the midpoint being C, and the field strength of the AP segment at P1 is small and equal to the field strength of the PC segment at P1, and the direction of the field strength of the CB segment at P1 is to the left. The field strength of AB at P2 is to the right, so E1 is opposite to E2, and E2 is greater than E1.

If the right side 2 of the insulating rod is cut off and removed, it is equivalent to the field strength of the CB segment at P2 disappears, and it can be seen that the field strength of the CB segment at P2 is equal to E1, so the field strength at P2 is E2-E1.

Instead, e2-e1

The pole is uniformly electrified, take the AB midpoint O

The electric field at ap1=p1o point p1 is generated only by the right part of the pole in the direction to the left and the magnitude is e1, while the electric field of p2 is the direction generated by the whole pole in the direction to the right and the magnitude is e2. After the right side is truncated, it is e2-e1.

a Sanya.

Analysis: Compared with the launch of other three places, the main advantages of launching geosynchronous satellites in Sanya are low latitude, close to the equator, large linear velocity with the rotation of the earth, and low energy consumption.

a.Because it is possible to make use of the Earth Autobiography.

Sanya near the equator has the maximum linear velocity, and if it is at high latitudes, additional fuel is needed to bring the rocket to the linear velocity at the equator.

1) The landlord is wrong in the right tube, the piston is in the right tube before leaving the bayonet, and the gas pressure in the right tube is strong.

The state of the gas in the right tube changes to an isochoric process, ie.

Therefore, taking the gas in the left tube as the research object, let its initial state be , ls and t0, and the unstate is , v2, t2, then v2=(l+ )s is composed of gas.

Analysis: Taking the piston as the object of study, t0 is the pressure of the gas at the right end of the gravitational atmospheric pressure and the support force of the bayonet. In which gravity and bayonet support forces are balanced, and atmospheric pressure and gas pressure in the tube are balanced.

So if you want the piston to leave the bayonet, the critical point is that the internal gas pressure is equal to the atmospheric pressure + the piston gravity.

So there is, p1 t1=p2 t2 p0 t0=(p0+mg s) t2 t2=p0(p0+mg s) t0

2) Analysis: When the liquid level on the left side drops h 2, the pressure at the left end of the gas column is atmospheric pressure + piston gravity + h high liquid pressure according to the principle of closed gas and liquid transfer pressure. The corresponding pressure can be found and then solved according to the gas law.

Yes, p1v1 t1=p2v2 t2 p0*ls t0=(p0+mg s+qgh s) t3 t3 can be found t3.