Urgent High School Physics Problems, High School Physics Problem Solving Urgent .

The support force is the reaction force of the pressure of the force object to the force object, its work is only related to the displacement of the force object in the direction of the force, the work done by the support force is only the work done to overcome the pressure, and the mechanical energy is the sum of the gravitational potential energy and the kinetic energy, and the two kinds of work are not necessarily related, for example, on the conveyor belt, the support force does not do the work, but the friction force does the work, so that the gravitational potential energy of the object increases, so that the mechanical energy increases (the object is in a stationary state before and after the work, that is, the kinetic energy change is zero), and on the vertical elevator, The work done by the supporting force is equal to the amount of change in the potential energy of gravity, i.e., the amount of change in mechanical energy (the object is also at rest before and after the work is done), therefore, there is no necessary connection between the two.

If the system is only supported by gravity, then the support force is equal to the amount of change in mechanical energy, and if there are other forces, it is not equal.

Depending on the specific force, the change of mechanical energy is equal to the work done by the external force of the whole system + other energy conversion (such as its own friction loss, such as thermal expansion, etc.).

Work done is a physical quantity that measures the transformation of energy, and the amount of change in mechanical energy is equal to the work done by other forces other than gravity.

w=fscosa; The key is to see if there is a component of displacement in the direction of support; a is the angle between f and s. Strictly by definition.

Sometimes you can, sometimes you can't. Can you tell us the specifics?

To be stationary, the amount of frictional force required needs to be equal to the centripetal force mw 2r=4The maximum static friction that can be provided is umg=3.

So it can't be still.

To be stationary, the centripetal force MW2R is 3 max. The solution is w=

If the free fall time is x, the speed is gx when the parachute is opened, and the falling time after the parachute is opened is gx a, which is 4x

Column equation 1 2*g*x 2+(gx*4x-1 2*a*(4x) 2)=420

where 1 2*g*x 2 is the free fall distance, and gx*4x-1 2*a*(4x) 2 is the falling distance after the umbrella is opened.

Just solve the x.

The explanation of item A is upside down, it is precisely because of the deceleration movement that the tension is reduced, I guess you back to this analysis because you ignore the quality of the parachute, the parachute also has mass, it also does a uniform motion at the beginning, and then it also decelerates the movement, you ignore the parachute like this, he is very aggrieved.

Parachute: 5555555, why ignore me!

In fact, it can be understood equivalently that two blocks are connected by a rope, so that there will be no mistake in thinking and then analyzing the parachute, it does a deceleration motion, and the external force is upward, while the other forces remain the same, it can only reduce the pulling force.

Parachute: I can only reduce the pull to slow down, I'm easy!

This is a test of your understanding of tension in parachute lines.

The tension in the rope belongs to the elastic force, within.

Newton's third law allows us to see that the pull of the re-entry capsule on the rope is equal to the pull of the parachute on the rope.

Because the moment the rocket breathes fire breaks this balance, the rocket no longer needs to maintain the original force balance by pulling the rope, that is, at this moment, the rope has begun to relax, and the tension becomes smaller, so the tension provided to the return capsule is reduced.

Generally, it is mixed with the elastic force in the spring, as long as the spring is deformed, it has a powerful effect. As long as the rope is not tight, there is no force to act. Basically, I understand it this way, and I can cope with almost all the exam questions.

1.You say, "At constant velocity, gravity is equal to drag plus pull." The pull is originally upward.

At this time, it was subjected to a reaction force generated by a slow version of the rocket jet, and this reaction force is upward, "all right, the tensile force here is the elastic force generated by the elongated deformation of the rope, and the rope here can be equivalent to a spring that has been elongated, and the return capsule is subject to the tensile force t of the spring and the gravity g of the return capsule itself, and it is equal when it moves at a uniform speed, that is, t = g."

2.In order to decelerate, at this time it is subjected to a reaction force generated by a buffer rocket ejection, and this reaction force is upward, that is, it is equivalent to having a force to support the return capsule upward, and the elongation of the spring is shortened, and the tensile force of the spring becomes smaller, and a part of the reduced tensile force is borne by the reaction force mentioned above.

The rocket begins to jet, and the instantaneous recoil force acts directly on the return capsule.

If you think about it, if a rope is pulling on a heavy object, and then suddenly an upward thrust is applied to the heavy object, the rope pull must decrease.

A option, in fact, the moment of the jet, the deceleration is reduced by the speed of the return capsule, not including the parachute. There is a connection between their plates, and there is a buffer belt. At this moment, the parachute is subjected to three forces, its own gravity, air resistance, and the pull on the return capsule.

This pulling force is equal to the gravity of the re-entry capsule minus its air resistance minus the force generated by the jet, so it is less than the tensile force before the jet (between gravity minus air resistance), because it occurs instantaneously, causing the "gravity" of the re-entry capsule to become smaller, and the original straight rope is instantly deformed.

The buffer rocket is to slow down, and the speed decreases while spraying, so the friction of the parachute is reduced. Don't think about it so complicated, you are considering whether the pull force is the sum of the two pulls, or an increase. As for what you said, if it doesn't change, it doesn't hold, because it has changed.

You kid. Don't go to school anymore, I don't understand such a simple question. I don't look at the title, but I know to choose A

Nangong Moyan is right, 100 points is too much for others!

The pulling force you are talking about is actually the air resistance of the umbrella, which is determined by the speed, and the smaller the speed, the smaller the air resistance.

In fact, this is a relative BAI problem. Squirting instantly.

The speed is not reduced, it's just.

DAO gradually decreases, so at that moment we can think, since the speed does not change, then the upward resultant force does not change, and the jet has an upward force, the air resistance must not change, so the pulling force decreases! Attention, it's for that moment!

Note that option A's condition "start jetting" means that the return capsule is in equilibrium at this time. Therefore, when subjected to an upward reaction force, the tension of the paracord decreases. It doesn't seem that you can't do this question, it's just that you're sloppy. Remember it later, good....

At the moment of ignition of the rocket, the acceleration of the parachute is zero, and the return capsule decelerates in motion, and they have a tendency to move in opposite directions.

This means that when the output power of a power supply is p, there are two possibilities for external resistance, R1 and R2.

The relationship between R1 and R2 and the internal resistance R of the power supply is: R= R1R2 Of course, when the output power of this power supply is maximum, the external resistance is equal to the internal resistance R of the power supply, and there is only one value.

For example, when the electromotive force of the power supply is e=6V, the internal resistance is r=1, and the maximum output power of the power supply is the external resistance r=1, p large = i 2*r=(e r+r) 2*r=(6 2) 2*1=9w.

If the external resistance is other resistance, the output power of the power supply is smaller than 9W.

If the output power of the power supply is 8W, what is the external resistance?

p=i^2*r=(e/r+r)^2*r

8=（6/1+r)^2 * r

2r^2 -5r+ 2=0

r1=2 ω r2=1/2 ω

This does not have two values for r, and r = r1r2

The output power p is related to the power supply electromotive force e, the internal resistance r and the external resistance r, p=e *r (r+r).

Then for a given power supply (both e and r are determined), its output power is only related to the external resistance.

For a given power supply, if its output power in a circuit is p, then there are two possibilities for the external resistance of the circuit (there is only one possibility for the external resistance at maximum power, that is, it is equal to the internal resistance), and the external resistance in both cases can each give the output power of a given power supply to p.

For example: the power supply electromotive force is 10V, the internal resistance is 1, and its output power is 9W, then let the external resistance be R, then.

9=10²*r/（1+r）²

9r²+18r+9=100r

9r -82r+9=0 finds r1=9 and r2=1 9

These are the two different external resistors corresponding to p.

The initial formula can be transformed as follows.

p=e²*r/(r+r)²=e²*r/

When r=r, there is a maximum output power p=e *r 4rr, and the external resistance r1=r2=r.

R=R1*R2 under the root number, which means that the power supply internal resistance is equal to the power supply internal resistance under any output power p (except zero output power) P corresponding to the power supply corresponding to the root value of the power supply, for example, in this example, r = 9*1 9=1 under the root number.

As for why, I don't know.

p=i^2*r

u/(r+r)^2 *r

u/(r^2+2rr+r^2) *r

U (r+2r+r 2 r) internal resistance is the maximum value of a certain r+r 2 r: r 2=r 2 that is, r = r (maximum value formula) the denominator is the largest, p is the smallest, so in addition, there must be two resistance values of resistance r1 and r2 (r2=r 2 r1); I don't know if I made it clear!

Select AD analysis "When the magnet is close to the coil, the magnetic field of the induced current generated by the coil should hinder the approach of the magnet, so the magnetic field of the induced current is above the n pole, and the magnet and the coil are repulsed, at this time n mg;

When the magnet is far away from the coil, the magnetic field of the induced current generated by the coil should hinder the distance of the magnet, so the magnetic field of the induced current is above the S pole, and the magnet and the coil attract, at this time n mg, so a is correct.

According to the "come and reject" magnet has been to the left, the coil has been to the right, so the friction force on the coil has been to the left, so d is right.

a, in the process of moving to the center of the coil, the downward magnetic inductance line increases, the induced magnetic field is upward, mutual repulsion, n is greater than mg; In the process of moving away from the center of the coil, the downward magnetic inductance line becomes less, and the induced magnetic field is downward, attracting each other, and n is less than mg.

ad.According to the law of stunned times, when the magnet is close, the coil has a tendency to move away from it, so the magnet will go down to the right so the elastic force is greater than the gravity, when the magnet is far away from the coil, the magnet is subjected to upward and right forces, and the magnet has a tendency to move upward to the right, so the elastic force is less than gravity, and the friction force is to the left.

Find the friction force f, fcos37-f=ma....1) Find the displacement again, s=at 2 2....2)

f does work, wf = fscos37

fDo work wf=-fs

Gravity and support do not do work.

The resultant force w=mas

Do the math yourself. Hope it helps.

Displacement s=1 2at 2= friction f=fcos37-at=48-20=28jTensile work:w=fscos37=60*1*frictional work:wf=28n*cos180*1m=-28jSupport force work:

wn=0Total work: w=w+wf=48j-28j=20j I wish you happy studying

Equations are laid out using Newton's laws or the laws of kinetic energy.

n=(200-60sina)n=164n.The direction of motion and the direction of force are different, so neither the supporting force nor the gravitational force do work.

See the picture below, I'm so tired I'll send you the ones that are off the angle.

I adopted it for the sake of hard work.

The speed is 2UE M under the root number, and since there is no diagram, you have to deviate from the direction by yourself!