High School Physics Questions Rush Rush Thank you! If the answer is correct, the score will be added

Landlord, a physics problem like this can be regarded as one object when two objects are in the same motion, so it is not easy to make mistakes in thinking. Your question is a bit difficult, the brother upstairs is right, but it may be a bit difficult to understand, unless it is solved like this now. I won the silver medal in the National Physics Olympiad more than ten years ago, but unfortunately the certificate was missing, and the silver medal was left.

Since they are all standard units, and the units are difficult to play, most of the following calculations omit the units, but I believe the landlord knows what it is.

First, sin37°=, and then the elastic force of the spring at rest is calculated as (4+8)*10*N.

Let the acceleration be a, and at the moment of separation, there is because q is no longer subjected to upward elastic force.

f constant -8*10*sin ) 8=a

Then (f constant -48) 8=a

Let the shape variable of the spring at the moment of separation relative to the initial moment be x, where x is equal to the travel of p and q, and since the initial velocity is 0, x=1 2*a*

x = while p is still accelerating a at the moment of separation

At that time, the magnitude of the resultant force of p is the elastic force minus the force parallel to the downward slope of p.

The spring has a force of 72-600x at the moment of separation

The force parallel to the downward slope of p is 4*10*

then there is (72-600x-24) 4=a

Substituting x= to obtain.

72-12a-24) 4=a solution a=3

Before the summary, then f constant.

f constant-48) 8=3 solution f=72 N, as we all know, f after constant force is also the maximum.

The initial f is the minimum value, because the system is in equilibrium at the beginning, and f is the net force of the system at the beginning.

So there is. f (4+8)=3 The solution gives f = 36 N, which is the minimum value of f.

The maximum value of Af is 72 N and the minimum value is 36 N

0 (m1 + m2 equilibrium point) c separation point a (m1 equilibrium point) b (free elongation point).

－x－－－x1－－－x2－－－

At the beginning, p and q move together under the combined action of spring force and f, and have the same acceleration a.

When moving to x, the upward acceleration of m1 under the spring force is equal to the upward acceleration of m2 under the action of f, at which point separation begins, m1 acceleration begins to decrease (close to point a), and m2 acceleration remains unchanged (both acceleration). f Change from variable force to constant force. Thus there is:

Let the start co-acceleration time t, then we have: 1 2at = x

Point A position: (M1+M2)GSIN -M1 GSIN =K X1

M1 acceleration at separation: k(x1-x)=m1 a

The minimum value of f is the maximum spring force at the beginning of the pull, so f can be the smallest: f1=(m1+m2)a

The maximum value of f is the time of separation, when the spring force is minimal, so f reaches the maximum: f2-m2 gsin = m2 a

Solution: a=3m s

x=x1=f1=36n, which is the minimum.

f2=72n, which is the maximum.

The two upstairs were right, I was wrong.

The minimum is 24, the maximum is 72, you first set the distance of the pull to x, the acceleration is a, according to x=1 2at*2 ====》x=, and then analyze the force of p at the time, the acceleration at this time is still a, and then the acceleration decreases, separated from q, so that you can list the formula (m1+m2)gsin -kx-m1g=m1a, combined with the upper sub to find a=3m s*2 f maximum is after f=m2a+m2gsin =72n The minimum is that it starts from rest and accelerates diagonally upward, f=m2a=3x8=24n

In fact, they are all very basic questions......

This question examines how to determine the motion of an object based on a V-T diagram.

a If the velocity of the object is positive, it moves upward, if the velocity of the object is negative, it moves downward, and the velocity shown in the diagram is always positive, indicating that the object is always moving upward.

b According to the area of the figure enclosed by the V-T diagram is the displacement of the object, then the displacement can be obtained by finding the area of the trapezoidal figure shown in the diagram.

c The change trend of velocity is directly judged according to the trend of the V-T diagram, and the velocity first becomes larger, then unchanged, and then decreases.

d Since the velocity varies uniformly, it means that it is a uniform variable speed motion, and according to the motion formula δv=aδt of the uniform variable velocity, the acceleration of the three segments can be obtained.

In summary, BCD is the correct solution. Over

It's a very simple question! The velocity curves are all for you :

Analysis: The movement of the elevator.

1. Accelerating upwards, the speed from 0 to 6 takes 5 seconds.

2. Move at a constant speed of speed 6 for 10 seconds.

3. Do a deceleration movement upward, the speed is reduced from 6 to 0, and it takes 10 seconds.

A: Wrong, keep moving upwards, and the downward velocity is negative.

B: Yes, the displacement is the trapezoidal area.

C: That's right, the graph line is already very obvious.

d: Yes, acceleration is the change in velocity divided by time, which is realized on the graph as the slope of a straight line.

This question should be clear at a glance when you look at the picture.

But one thing that is not rigorous is that the last acceleration should be negative.

The ox axis search coincides with an electric field tent line of q, indicating that q is on the ox axis.

The electric field force of the tentative charge at points a and b is in the same direction as the positive direction of the x-axis. The charge at point A is positively charged and point B is negatively charged. Based on these conditions, it can be judged that:

q is negatively charged, somewhere between a b.

Electricity of the same sex repels each other, and electricity of the opposite sex attracts. Assuming that q is positively charged and is on the left side of a, then the slippage of the force side of a is to the right and the direction of force to the left of b, which contradicts the known. Using a similar reasoning, it is not difficult to conclude that q is between a b and is negatively charged. ）

a b 3 meters apart. Let q be r meters away from a and 3-r meters away from b. Rule.

ea = 40 n/c = kq/r^2

eb = n/c = kq/(3-r)^2

ea/eb = 16 = 3-r)^2/r^2

4 = 3-r)/r

4r = 3-r

r = m.

Substitution 40 n c = 9 10 9 n m 2 c 2 ) q ( m) 2

q = 10 (-9) coulombs.

Considering the electronegativity.

q = 10 (-9) coulombs.

Ans: q is negatively charged q = Coulomb. Between a b, a meter away. i.e. the coordinate position is meters.

Won't ask me again. ok?

1 All (1) middle ball force:

The support force of the left ball n, to the upper right;

The support force of the right ball n, to the upper left;

Gravity mg, downward.

The three forces are balanced in the vertical direction: mg=2ncos ..1) (2) The force on both sides of the ball (symmetrical left and right, taking the right ball as an example):

Gravity mg, downward.

The middle ball pressure n, to the lower right;

Rope tension f, diagonally to the upper left.

Three forces are balanced horizontally:

nsinβ=fsinα..2）

The three forces are balanced vertically:

ncosβ+mg=fcosα..3) obtained from (1): n=mg (2cos), substituted by (2), (3) obtained:

mgtanβ/2=fsinα..4）mg/2+mg=fcosα..5) (4) (5) De:

tanβ/3=tanα

So: tan tan = 3

The two sides are symmetrical, and the triangles are similar.

The smooth ball of the two is mg sin(b-a)=n sinb, and the smooth ball of the middle mg sin2b=n sinb

If you can get me to draw a diagram and then you can do a force analysis to determine the angle relationship.

Well, your idea is right, just ask for the horizontal displacement of the raindrops to know the radius r of the ground circle, because r 2 = x 2 + r 2.

The horizontal velocity (i.e., initial velocity) of the raindrop as a parabola is the linear velocity of the raindrop on the edge of the umbrella, linear velocity = = angular velocity * umbrella radius r

There is no balancing friction;

When the mass of the object is constant, the acceleration is proportional to the resultant external force.

The image is not the origin, indicating that there is no equilibrium friction.

Reason for analysis: When a=0, f ≠0, it means that there needs to be a force to balance with f, that is, there is no equilibrium friction.

Conclusion, when the mass of the object is constant, the acceleration of the object is proportional to the combined external force it is subjected to Reason: The image is a straight line.

1: There is no balance friction. 2: When the mass is the same, the acceleration is proportional to the magnitude of the external force.

Hello, first question: The gravitational force of the keg (g=mg) is converted into the pulling force of the trolley, even if the acceleration is zero at the beginning, but f(g) is already there. Second question:

As can be seen from the image, the larger f is, the greater a, and the mass of the object does not change, so we can get f=ma (Newton's second law). This is a small opinion.

This is because the object moves to the maximum static friction.

The bullet energy change is only kinetic energy, and the total energy reduced is: 1 2mv * v = 450j

All these kinetic energy are converted into internal energy, that is, thermal energy, because the target does not increase kinetic energy, potential energy, and the bullet kinetic energy is 0, and the potential energy does not increase. So the total heat generated is: 450J

Half is absorbed by the bullet, which is 225J

The kinetic energy of the bullet is converted into internal energy w=1 2*mv 2=450J, and the heat absorbed is 225J

1. While the spacecraft is moving, the earth is still rotating, and the direction of rotation is from west to east, so the projection point is west.

2. According to the figure, it can be seen that the spacecraft has to pass through the equator twice around the earth, just like the periodic motion of the direct point of the sun The longitude difference between the curves is the period of the spacecraft, (

3. The geosynchronous satellite is located above the equator and is relatively unchanged from the Earth's position, so it must pass the initial position again after a period below the subsatellite,24

I don't know if when the spacecraft reaches the other side of the planet is under the satellite, if it is 8 hours, you think about it yourself)

4. According to Kepler's law, r 3 (to the third power of r) is proportional to t 2 (the square of t), and the ratio of the radius is calculated to be 256

This question is difficult, I think it should be calculated in this way, but it is not necessarily right, when the small block moves down the inclined plane, the large block will be translated to the left, and the force analysis of the small block is carried out, and the inclination angle of the large block is q due to the action of gravity and support force

Vertical NSINQ=MA levels for B mg-NCOSQ=MA.

For the analysis of the force of A, the pressure of gravity, ground support and b nsinq=ma'

And it can be known that B never left A

It can be obtained from the above three equations.

aVertical = g-ahorizontal cotq

B never leaves A, then there is A vertically (A-level + A')=TanQ to get A-level = mg [MCOTQ+(M+M)Tanq] and then bring A level into the second equation to get n =(mmgcos) (m+msin )

The force exerted by object A on B is dealt with in two cases.

1. When B is subjected to the friction force of A on it is greater than or equal to the component of its gravity parallel to the inclined plane: A and B are stationary relative to the ground, and B is in equilibrium at this time, and is rubbed obliquely upward parallel to the inclined plane, and has a supporting force perpendicular to the inclined plane. The magnitude of the resultant force of the two is mg and the direction is vertically upward.

2. When B is subjected to the friction force of A on it is less than the component of its gravity parallel to the inclined plane: the momentum of the two objects A and B is conserved in the horizontal direction, and there is relative motion between A and B, A moves to the left relative to the ground, and B moves to the right, and B has acceleration, which should be treated with Newton's second law and conservation of momentum.

Do you have the right answer? Can I help you analyze what happened?

Well, maybe I want to be simple, since there is no friction, then at the beginning, when B is still on A, the force should be mg (cos@) squared (@是那个a斜面与水平面的夹角), and then B slides off A, and then A is 0 to B

No, the main thing is that wealth is given less.

The force exerted by object A on B: the frictional force diagonally upward parallel to the inclined plane, and the supporting force diagonally upward perpendicular to the inclined plane. The magnitude of the resultant force of the two is mg and the direction is vertically upward.