Explain the individual options of several senior one physics questions

Example 1, A, Example 2, and B is actually a flat throw, which is a uniform acceleration motion, but not a linear motion, which is also a trap.

Example. 3. Decompose the velocity v.

The velocity along the rope direction = the velocity of b, then vb = vcos decreases with the right motion and vb increases but cannot be uniform.

Track 4 is the same as Example 3.

The speed of the car can be broken down into two, one is the speed along the rope and the other is the speed at which the rope is going around the pulley.

The velocity along the rope is equal to the speed of the car xcosa, and a is the angle between the rope and the horizontal plane.

The speed of the car is constant, as the car moves, A gradually decreases, so the speed along the rope gradually increases, and this speed is the speed at which the weight rises.

Acceleration can be calculated by derivation, I don't know if the landlord has learned, but it can be understood that the maximum speed of the heavy object A is the speed of the car, so the weight A is not infinitely accelerated, and the acceleration will gradually approach zero, so the acceleration is decreasing.

So choose B

Example 1is a, because the force is concave downwards, so it is a

Example 2is a within 2 seconds, the velocity in the x direction is uniform, and the y direction is 0, so a does not change; And 2 to 4 are x constant y change, a change. If you don't know the acceleration, you choose B if you don't know the concept.

I see you say b you don't know how to explain, but it's really hard to put it into words. I'll write a formula, you see, no, a (v1, v2), t is v, is the root number (vx 2, vy 2), so b is not right.

3 You will know the force when you draw, and you will know it in this way for each of them.

a f m and f is not constant, uniform acceleration motion is wrong.

4 This is the same as 3 If you don't know this type, you need to make up for the analysis. In this drawing is difficult not to say, in fact, it is simple to draw the car to the right and the rope angle changes with time, so you can know, the farther the car is the smaller the angle between the rope and the horizon, and the mathematics tells us that the rope upward speed increases faster and faster is a pair. Math poor you can think of starting the rope speed less than the car speed, infinity, rope up speed equal to the car speed.

a,,a is a circular motion, which is impossible when the force is constant, b is a flat throwing motion, d is an oblique throwing motion, a lot, too lazy to do it.

Do you want to understand the meaning of vectors?

To know the decomposition of vectors?

Knowing these two points, these questions are easy to do!

1.According to the centripetal force formula, the formula is listed: bvq=v 2m r, and v=bqr is obtained because of the relationship between distance, speed and time

t= r v (in radians), we get t= m bq. From this equation, it can be clearly seen that the motion time of particles under the action of Lorentz force in a uniform magnetic field is only related to the angle of motion, and has nothing to do with the radius and velocity. Therefore, the angular range of the particle's flight is determined, and the time range required for its flight is determined.

Now make a straight segment in the direction of perpendicular injection through the point p, and the radius will be perpendicular to the Lorentz force because the Lorentz force provides the centripetal force. i.e., b, v, q, m, and the radius is just different lengths on this line segment. The angle of particle motion will be different if the radius is different, so it needs to be discussed on a case-by-case basis

Case 1: The radius is small, and the particles fly out of the ox axis, at this time =5 3, i.e. t=5 m 3bq;

Case 2: When the radius increases, the particles fly out of the OY axis, and at this time, 2 3< = <=4 3, that is, 2 m 3bq<=<=4 m 3bq.

In both cases, m 6bq is not included in the range, so the d term is wrong.

You start by following my prompt to make a diagram. 1.According to the left-handed rule, the center of the circle in which the particle moves in a uniform circular motion is on the ray PA.

According to the radius of the particle motion, r=mv qb the greater the speed of the particle motion. The larger the radius. Therefore, an infinite number of circles or arcs can be made on the PA.

According to the periodic formula t=2 m qb of charged particle motion, the time of particle motion is independent of velocity. You take another point of b above a and make an arc over pb. This arc is an inferior arc that is smaller than a semicircle

The time of the particle motion t = 360*t = 360*2 m qb = m 180qb ( is the central angle of the inferior arc).

Knowledge by plane geometry: the angle BPC (C at a point in the direction of the arrow) is the chord tangent angle. This angle = half of the angle BPC greater than 60 degrees. Either BP would not be able to have an intersection with the y-axis.

According to this question. t =θπm/180qb= πm/6bq

The solution is =30 degrees and pushes it down like this. The chord chamfer angle BPC is 15 degrees. Less than 60 degrees, this is impossible.

The particles move in a circular motion in a magnetic field, and the figure is roughly drawn out and can be calculated by calculating the period.

Important: Enter at a 30-degree angle and go out at a 30-degree angle. The period can be calculated by the degree of the central angle of the circle, multiplied by 360-2 (180-90-30).

Since the radius is always perpendicular to the direction of incidence, try it yourself.

The time elapsed for the particle to move in a magnetic field when it is ejected from X-ray is 5 m 3bq

When the particle is emitted from y, the geometric relationship knows that the central angle of the circle is always 120°, so the time elapsed by the particle in the magnetic field is always greater than m 6bq

1. First find the time t when the bag falls, use the free fall formula h=(1 2)gt2, substitute h=go in, and find t=. Because the car and the schoolbag were originally moving in a straight line at a uniform speed, the schoolbag maintained inertia forward after the car braked, and the schoolbag went forward in the time from braking to the schoolbag falling, s1=vt=16. After the car brakes, it does a uniform deceleration movement, and it moves forward in time S2=VOT+(1 2)AT2=16*.

The distance that the schoolbag travels more than the car is the distance between the schoolbag and the back of the car (shelf): s=s1-s2=

2. Because A and B both maintain a uniform linear state, the resultant external force of A and B is 0. Therefore, f1 = 0, because there is no other force in the horizontal direction above A, if f1 is not 0, it is impossible for A to have a uniform velocity. Similarly, f2 is equal to the tensile force f, and the two forces are just offset by one before and one behind, so they can move forward at a uniform speed.

Select d, the two waves in the two waveforms of A and B represent the transmitted and received signals respectively, and it can be analyzed that the time t is 4 1 10 to the negative fourth power, and the original distance between the aircraft and the radar x is half of the time multiplied by the speed of light, so the time when the aircraft flies from the original position to the radar is 2 1 10 to the negative fourth power, and then use the Pythagorean theorem to find the distance of flight in the air, and then use the displacement time and velocity relationship to find 300, which is an old problem, the key is to draw a physical model.

To analyze the stones and the car separately.

1.In order to make the minimum force of the object move under the condition of fmin=umg=5n>2n1), the wooden block is not moved, and the trolley and the wooden block are regarded as a whole and subjected to a force of 2n. Then the friction between the trolley and the wooden block is equal to 0, and the acceleration of the trolley and the wooden block is the same as a=f (2+2)=

2) When f=20n, you can judge the friction force f on the wooden block when it slides'=5n, the net force of the block is 20-5=15n

a=15/2=

The trolley receives the resultant force.

For 5n, the acceleration is 5 2=

2. F=UMG=5N, the direction of the closing force 20-5=15N of the stone is to the right, and the car is also subjected to the friction of 5N, to the right (the same direction as the force of the stone).

The acceleration of the stone block is a1=15 2=

The acceleration of the car is a2=5 20=

When the stone reaches the front edge of the car, the whole stone is one longer than the distance of the car.

The answer hall can be found by listing the clear state hidden (1 2)a1*t 2-(1 2)a2*t 2=1, and t in this solution g=10).

In the first question, you can draw a force analysis diagram.

The second question is the pressure of B on C, and the other is the support of the ground for C.

1. Because the triangle makes the teeth of the zipper separate to both sides, the teeth that were originally occluded are easily separated by the triangle, and the two rows of teeth bite very tightly without being broken.

2 A is pushed, and its horizontal support has only the frictional force that B gives it, so the force is transmitted to B, and B is transferred to C, so C is subjected to the horizontal force, and there is also the pressure of B on it, and the answer should be C

A and B are under pressure on him, you might say, isn't there still pressure from C and F? Because in order to generate elastic force, it must be touched and squeezed. C did not make contact with him, so there was no force on A.

f does not act on a, so there is no force on a. Only B is in contact with A and squeezes, so there is only one force.

The acceleration is -3m s

Indicates that the magnitude of acceleration is 3m s, unchanged.

A negative sign indicates that the direction of acceleration is opposite to the set positive direction.

When the V direction is positive, A and V are reversed, and the object decelerates.

When the V direction is negative, A and V are in the same direction, and the object accelerates.

A Not necessarily, the magnitude of the velocity is the velocity, and if the acceleration is in the opposite direction of the reference, the magnitude of the velocity becomes larger and larger

b In the same way, if moving in the reference direction, negative acceleration is equivalent to deceleration.

c acceleration is clearly unchanged.

That's it.

Obviously, the phenomenon of A, it is possible to decrease first and then increase, when the velocity decreases to zero, it is a reverse acceleration, and B is the same, both are wrong! Option c, as can be seen from the question, if the acceleration is unchanged, that is, d is correct.

1.The lowest point velocity is 0 and the acceleration is upward, so f>g

2.The kinetic energy is all converted into the work done by friction, which is naturally 10*5=50.

A flat throw is not a straight line. b Diagonal throwing is not a straight line. c is the definition of uniform acceleration.

1.The velocity at the lowest point is the maximum, and the acceleration is upward, so f>g

2.Conservation of energy = >> The change in kinetic energy is equal to the work done by friction: 10*5=50

A flat toss is a curvilinear motion. b Diagonal throwing is also not a straight line. C is correct.

Uniform circular motion always has a centripetal force. So fn=n-mg, so n mg

w=fs=10n*5m=50j, how much work is done, how much energy is converted

The flat and oblique movements are curved movements, so they are c.

1:d, at the lowest point, that is, the time when the speed of the aircraft changes from downward to upward. So the acceleration is upward. So the force of an acceleration upwards, plus the reaction force of gravity, then f is greater than the gravitational force.

2: w is 50j3:c obtained by w=fs

1. Select D, at the bottom point f=g+f.

2 choose C, EK2-EK1=W=FL

3. Select C, AB is only a linear motion with uniform acceleration in the vertical direction.