Solving Physics Problems A master comes and solves a physics problem

It should be the resultant force of the horizontal component of gravity and the pulling force of man acting as a centripetal force.

Later, it may be similar to the principle of *** take-off, the angular velocity of the ball rotation is large enough, and there are some people"fluttering", this force is in the same straight line as the person's gravity, just counteracting the gravity, making the person's pressure on the ground zero.

mg+mg=mω^2l

[(mg+mg)/ml]

where mg is the child's gravity, which is equal to the force of the rope pulling the child. Equal to the force of pulling the ball downward.

About: I would like to ask, what force provides the centripetal force? If the gravity of the ball and the pull of the rope are the force of the ball, then what is the force that makes the pressure on the ground zero?

Answer: It is the gravitational force of the ball and the pull of the rope that provides the centripetal force.

The action of force is reciprocal (Ox three).

People pull the ball, and the ball pulls people. The pull force of the ball on the person is equal to mg. The force of the person is balanced by two forces. So the pressure on the ground is zero.

The ball is subjected to centripetal force f, upward, rope pull force t1, downward, gravity mg, downward;

The person is pulled by the rope t2, upward, the ground elastic force n, upward (here n 0), the gravity mg;

f=mw2l (2 is exponential).

t1+mg-f=0

t2+n-mg=0

n=0t1-t2=0

So w= [(mg+mg) ml].

Is it the gravity of the ball and the pull of the rope that provide the centripetal force, t1=t2 is equal in magnitude and opposite in the direction?

The pulling force of the rope and the gravity of the ball provide the centripetal force, the action of the force is to pull the ball with each other, the ball pulls the rope, and the rope is driven to pull the person, so that the pressure of the person on the ground is 0

According to the formula, elasticity f = stiffness coefficient k * this variable x

Then, let the original length of the spring be l

f1=k*（l-l1）

f2=k*(l2-l)

Add the two formulas. f1+f2=k*(l1+l2)

then, k (f1+f2) (l1+l2).

f1=kl1

f2=kl2

Add the two formulas. So the stiffness coefficient k f1+f2 (l1+l2).

Set the cross-sectional area of the rod s. When the rod enters the water for h length, the buoyancy force is p1gsh.

Gravity of the stick p2gsl. When the rod enters the water, the resultant force of gravity and buoyancy gradually decreases due to the gradual increase of buoyancy, and the rod moves with reduced acceleration. The velocity reaches its maximum until gravity is equal to buoyancy.

p2gsl=p1gsh，h=p2*l/p1

If H is the abscissa and the buoyancy is the ordinate, it is a straight line, and when it enters the water at the height of H, the work done by the buoyancy is the area under the line:

Work done by gravity: mgh=p2gsl*h

According to the kinetic energy theorem:

Find v=( 2*l*g 1) (1 2).

Your question is probably copied incorrectly, it should be 2< 1. The density of the rod must be smaller than that of the liquid, otherwise even if it is completely immersed in the liquid, the net force of the rod will still be downward, and it will always be accelerated, and there will be no maximum speed.

The motion process was analyzed: the rod fell initially as it was accelerated, and the acceleration gradually decreased with the increase of buoyancy. When buoyancy is equal to gravity, the stick gains maximum velocity; After that, the buoyancy force will begin to exceed the gravitational force, and the motion process will become a variable deceleration motion.

When the velocity is maximum, the gravitational force is equal to the buoyant force, and if the rod extends into the liquid surface x, and assuming that the rod has a cross-sectional area s (uniform straight bar), then there is mg = 2slg = 1sxg, so x = (2 1)l. The following is an analysis using conservation of energy: at this time, the kinetic energy of the rod is equal to the work done by the gravitational potential energy minus the work done by buoyancy.

The work done by gravitational potential energy is the work done by gravity acting on the distance x, m*g*x = ( 2 1)mgl.

The work done by buoyancy is variable, but we can notice that the magnitude of buoyancy is proportional to the distance traveled, so it is a uniform acceleration process when the distance of motion is taken as the reference, so the work done by buoyancy is (1 2)*m*g*x = mg( 2 1)l 2, so 2mgl 1 - 2mgl 2 1 = (1 2)mvm 2

The solution yields vm = (2gl 1). Proven.

Idea analysis: before the removal of force f, the system has a common upward acceleration a, after the removal of force f, due to inertia, a and b balls are still moving upward, and the elastic force of the spring does not change abruptly, so the force of b does not change, and its acceleration is still a; However, the force of a changes and the acceleration changes abruptly.

Analysis: At the moment when the force f is removed, the force on b remains unchanged, so the acceleration is still a, the direction is upward, and the analysis of the b ball by Newton's second law obtains fn-2mg 2ma, so the spring elastic force fn 2m (g + a) to a ball is obtained by Newton's second law fn + mg = ma, so the acceleration of a ball is a = 2m (g + a) + mg m = 3g + 2a

Answer: 3g+2a ; a

Excluding air resistance, the two balls will move in a constant upward and uniform straight line, and since the starting velocity is the same, the velocity is s

to: Upstairs.,But the ball shouldn't be without force or balance at that time == Although I'm not very good at it.,But it's better to be blinded.。。

I guess I want to use Newton's second law. According to f=ma,==then what to do...

a=（f-3mg）/3m

t t, I'm sorry, I can't do it.

I guess it's a-g and a-2g... =v="

At this time, there is gravitational acceleration.

(1) km) (hours) = 18km h

2) After calculation, the average speed of bus No. 22 is 18km h, and the speed of bus No. 74 is 36km h, so bus time No. 22 is (, bus No. 74 travel time is (, two meet for (hours, Yue.

The average speed of bus 22 is 50m s

Bus No. 74 takes 450s in 450s than No. 22 in 450s

(1) Find the first derivative, and you can get the function of velocity over time. Finding the derivative one more time gives you a function of acceleration as a function of time.

2) Substitute t=2 into the above formula.

3) Substituting t=1 and t=2 into r, r(2)-r(1) to get the displacement in the second second, and dividing by time (1s) to get the average velocity.

The problem gives the equation of motion of the particle system of the object, the derivative of this equation for t is a function of velocity and time, and the second derivative is a function of acceleration and time, and then substituted for time to solve.

2g*sin53t = 4, i.e., solve the equation.

2. According to the question, there are t+t1=t,v0t1=lcos53,1 2gt1 =lsin53, and the equation is solved.

t=seconds, v0=3m s, process diagram.