Two high school super basic physics questions, thank you

1.Neither, because the object moves in a uniform circular motion around the ground, as the height increases, the kinetic energy decreases and the potential energy increases, but the sum of the two is getting bigger and bigger. So if the velocity decreases suddenly, the total energy decreases, and at a lower altitude it can still move in a uniform circular motion.

The speed increases resolutely.

2.This is a rule commonly used in physics (as far as theoretically).

Near-Earth – Imagine that the achievement is on the ground, so gravity does not do work.

The same goes for light rods, light ropes - regardless of gravity.

Mass - no volume, hope useful! Give it a good one.

1. Neither of the two, the orbit will change due to the change of kinetic energy, and then run to a new orbit.

Sudden decrease in velocity - decrease in kinetic energy - decrease in total energy - decrease in orbital radius.

Sudden increase in velocity - increase in kinetic energy - increase in total energy - increase in orbital radius.

In fact, satellites are basically elliptical orbits, and their speed and distance from the earth change at any time, rather than falling to the ground or flying away as soon as the speed changes.

2. Satellite launch must overcome gravity to do work, don't consider whether overcoming gravity to do work is ignored for the purpose of calculating certain quantities?

If his velocity suddenly increases, then the radius of his orbit around the Earth will become smaller, that is, he will be close to the Earth, and when he reaches a certain distance, he will keep the radius of his orbit unchanged and move in a circle. If his velocity suddenly decreases, then he keeps drifting in the direction of the outer Earth.

The second sentence is wrong!

1: Yes, as the speed gets smaller, the gravitational force between him and the earth is greater than his centripetal force, and the excess force pulls him towards the earth. As the velocity increases, the gravitational force is less than the required centripetal force, and he runs outward.

2: When the near-earth satellite is launched, it is necessary to overcome gravity to do work, and the kinetic energy also increases, relying on the chemical energy of rocket fuel.

1.When the velocity decreases, it will move towards the Earth, but it will move in a circular motion at a lower altitude. On the contrary, the velocity increases and it moves outside the earth, and continues to move in a circular motion at another higher height.

2.Do you mean a near-Earth satellite launch?

The ball is subjected to two forces when it is in motion, the pulling force of the rope and its own gravity. These two forces provide the centripetal force of the pendulum and the force of acceleration in the tangential direction, and the magnitude and direction of these two forces are constantly changing at different velocities of motion of the pendulum. The "equilibrium position" does not refer to the bottom point, but to the middle direction.

The component of the resultant force in the direction of the rope cannot make the single pendulum return to the middle position, but the component of the vertical direction of the resultant force makes it return to the equilibrium position, so its "recovery force is the component of the resultant force in the tangent direction of the arc of the ball motion", and its recovery force is also constantly changing.

"Quarter cycle" does not refer to "quarter cycle" in the strict sense of the word, but to "quarter of the time". When he moves from the two highest and lowest points, the distance is an amplitude; When the ball moves towards the highest point, its distance is less than one amplitude (the speed at the highest point is the slowest) and when the ball moves from one side to the other side of the equilibrium position, its distance is greater than one amplitude (because the velocity at the lowest point is the largest). You can understand it in conjunction with a sine diagram.

Hope you can figure out these two questions.

1。The two components of the single pendulum force act differently, the force in the tangential direction changes the magnitude of the single pendulum, while the component that is perpendicular to it changes the other direction. The force that is perpendicular is not work, it is always perpendicular to the direction of motion.

The recovery force of a single pendulum is to do work, and it can be distinguished from this direction. The tensile force of the single pendulum rope is changing, and its net force with gravity changes from time to time, and it is not constant to point to the equilibrium position. It can also be proved by physical formulas that the work done by a single pendulum from the highest point to the lowest point is the work done by the recovery force, and the force in the vertical direction does not do the work, that is, the force in the tangent direction does the work.

2.The second problem can be observed from the image. From the equilibrium position to the peak position is an amplitude, from the equilibrium position to both sides of the sequential take 1 8t that it is greater than the amplitude, this section is the section with the largest velocity, so the displacement difference is also the largest.

On the contrary, take 1 8 t from the crest or trough to both sides, that is, the 1 4 t with the smallest velocity, and the dislocation difference is the smallest.

For fragments with non-upward velocity in the vertical direction, the maximum radius is horizontally thrown, l2=1 2*gt 2, r=vt, and r1max=v ((2l2) g).

The vertical velocity is upward and does not reach the ceiling, i.e., v1 2 2g<=l1, v is set to be v1 in the vertical direction, and v2 in the horizontal direction, v1=gt1,2gs=v1 2,s=v1 2 2g, so 1 2g(t2) 2=v1 2 2g+l2,t total=t1+t2= (v1 2 g 2+2l2 g)+v1 g,r=v2*t total, because v1 2+v2 2=v 2, so r=( (v1 2 g 2+2l2 g)+v1 g)* v 2-v1 2), where 0=l1, v is set to be v1 in the vertical direction, v2 in the horizontal direction, and v3 when the vertical velocity is reached to the ceiling, then 2gl1 = v1 2-v3 2, v3 = (v1 2-2gl1), l1 + l2 = v3t1 + 1 2gt1 2, gt2 = v1-v3, t total = t1 + t2, r = v2 * (t1 + t2), because v1 2 + v2 2 = v 2, so r = (t1 + t2) * v 2- v1 2), replace t1 and t2 with v1, and then find the extreme value of r3max

The largest of r1max, r2max, and r3max is the radius.

This question is difficult, and this is a schematic diagram of the meaning of tangential velocity and normal velocity. The elastic collision means that the tangential velocity to the ceiling remains unchanged, and the normal velocity becomes an equal reversal under the elastic force of the ceiling.

According to the conservation of momentum, we can know that the tangential velocity is unchanged and the direction is opposite, because the whole system is not subject to external force in its tangential direction, while the normal direction is subject to gravity, so the momentum is not conserved. Perfect elasticity here refers to the absence of energy loss after a collision.

1 2gt*t=h, find the running time, r=vt, get the radius.

Categorical discussion, do not touch the ceiling and touch the ceiling. Let an angle be arranged separately and find the derivative and the extremum.

d. Stand still.

Using the right-hand spiral, it can be seen that the direction of the magnetic field generated by the current is perpendicular to the outward side of the paper and parallel to the direction of the current in the wire.

Then the wire will not be subjected to force, so it will not move.

I don't want to give more points to these questions, and I'll do it.

There is by s=1 2at 2

s=9 gives t=3

then when rushing up the inclined plate, v=v0-at=4

This step is wrong, s=1 2at 2 is conditional on the initial velocity or end velocity is zero, here obviously has not decelerated to zero, should be calculated with v0 2-v 2=2*a*s1.

Yes, first of all, to be clear, velocity is a vector. And because the y-axis only has two opposite directions, plus and minus, it can only represent the two opposite directions of v. Imagine that the angle of V to 90 degrees, how to reflect it in the direction of the Y axis?

But if it's a rate-time image, it doesn't have to be a straight line, because the rate is a scalar. But there is no such question in high school physics.

No. The amount of velocity change is v. A greater change in velocity means greater v, not a greater change in v. For example, v is constant for the same time in the case of uniform acceleration.