Humbly ask for a high school 3 physics summer homework

Solution: This topic is more exciting, so I need to analyze it for you in detail, and you need to use a lot of methods to solve the problem, and what I think is the limit.

According to what you said in the title, the ball is lowered from a height of h meters, and then the velocity becomes 3 4, then according to the law of conservation of energy, the next time the object is thrown up, the height becomes h1

h1=9/16h.(How to calculate it, you think you will too) and so on. h2=9/16×9/16h

h3=(9/16)^3h

And what about time? The time taken for a fall at 5 meters to reach a velocity of 10 m is t0=1s

Then the second time it bounced up and fell, it took 10 3 4 10t1 2t1 = 3 2s

t2=3/2*3/4

t3=3/2*(3/4)^2

And so on. Constitute a proportional sequence with the first term 3 2 and the scale 3 4.

The sum of its n terms and lim(n>infinity) s=a1 (1-q)=6s plus the previous 1s, so it is 7s

I used to do this kind of problem when I was in high school, and by the way, I would like to remind you that there are many things that use mathematics in physics, and there are also a lot of geometry, so you must be good at mathematics. This question is also often asked how far the ball has traveled. Wait a minute.

It's been 2 years since I graduated from college, and 6 years since I graduated from high school.

The ground clearance rate when jumping up is 3 4"...Will it be still? Hard calculations aren't slow, right?

Let the length of the rope l the lower end of the rope reach a when it takes t1, then:

Solution l = only focus on the time it takes for the lower and upper ends of the chain to reach point a).

2。It is necessary to use the calculation of the proportional series, but only the idea is introduced: first, calculate the velocity v0, t0 when the first fall touches the ground.

After that, the speed is reduced to 7 9 from the previous time. Therefore v1=7 9v0,。。 vn=7 to the nth power of 9.

Exercise time t1=(2*(7 9) 10)v0; tn==(2*（7/9）/10)v(n-1).Sum and find the limit so that n reaches infinity.

3。Make a schematic diagram of the force, the corresponding edge is proportional to the force, that is: f:6m= solution: f=4000n

Analyze the force and reaction force on the rope, the tension of the rope by the car and the tension of the rope by the car are equal to the magnitude of the action force and the reaction force.

4.First, find the velocity v1 when the catenary and the velocity v2 when leaving the net, and obtain the velocity v1=-8m s, v2=10m s. by the formula h =, v=gt(Set vertical up as the positive direction).

a=(v2-v1) t to obtain the acceleration of 18 and then calculate the force from f=ma as: 900nDirection: Vertical upwards.

5.First, calculate the fall time of the wall, which has h=, and the fall time is 10s. The person runs for a time of 05=

The speed at which a person runs is at least 100m.

To collide, object A must collide before object B hits the ground!

then v>=(hg 2) (1 2).

For object B to collide with object A during the fall, v<=(hg) (1 2) and v>=(hg 2) (1 2).

At this time, there is no horizontal force, and it is not subject to friction).

Constant velocity, the resultant external force is 0).

I should be good in physics, if you are satisfied, please give me one, thank you.

The support force of the escalator is 420j for the passenger, the gravity force is 420j for the passenger, the friction force is 0j for the passenger, and the external force is 0j for the passenger

Support force for passengers 70 10 6 = 4200

Gravity on passengers -6 10 70=-4200

The friction force is 0 and the external force is 0

4200j、-4200j

w branch = 700 * 6 = 4200j, gravity is the opposite of it is -4200j, no force in the horizontal direction, the net force is 0.

Support force Q1 = 70kg*

Gravity q2=-q1=-4116j

Friction Q3=0

The resultant external force q4=0

Solution: The current of the series circuit is equal, P lamp = i * U lamp, i = 2 amps, the internal voltage drop of the power supply u = 1 * 2 = 2 volts, u slip = 1 * 2 = 2 volts.

1) The voltage at both ends of the motor U2 = e-u inner - u lamp - u slip = 14-2-2-2 = 8 volts;

2) Motor output power p output = 8 * 2 - 2 * 2 * watts, (motor power consumption minus heating power).

3) The total power consumed by the circuit p total = i * e = 2 * 14 = 28 watts.

The lamp works normally, so the current is 4W 2V=2A, the output electromotive force of the battery is 14-1*2=12V, and the voltage at both ends of the motor is 12-2-1*2=8V. The power of the motor is 11*2=22W. The total power consumed by the circuit is 14*2=28W.

If you don't understand, you can ask me.

The cosine of the angle between the inclined plane and the ground is 4 5, and the sine is 3 5, so the component force of gravity g along the inclined plane is 3 5 * g, and the component force perpendicular to the inclined plane is 4 5 * g, in the ascending stage, (1) the initial velocity is 0 uniform acceleration motion, the net force f1 = 100-(3 5 * g + 4 5 * g *, according to the law of conservation of energy, when rising 2 meters, the displacement along the inclined plane is 10 3 meters, at this time, the work done by "net force f1" is converted into kinetic energy, and the velocity can be obtained accordingly.

2) When the final velocity is 0, the net force f2 = 3 5*g + 4 5*g*, when the velocity is 0, all kinetic energy is converted into gravitational potential energy and internal energy, and the rising height and displacement along the inclined plane can be obtained accordingly.

Descent stage: (3) the uniform acceleration motion with an initial velocity of 0, the resultant force f3=3 5*g - 4 5*g*, (1) and (2) the gravitational potential energy corresponding to the sum of the heights raised by the two processes is converted into kinetic energy and internal energy.

In fact, it is not the result, but the force analysis in the process to understand.

MV 2 r=, this can't be broken, A and B are not 40cm apart.

Nails A and B, 400 meters apart 1 meter long string?

Here's a similar question I found online:

Two nails A and B are nailed on a smooth horizontal plane, 20 cm apart, with a thin rope 1 meter long, one end is tied to a small ball with a mass of kilograms, and the other end is fixed on A, at the beginning the ball and the nail A, B are on a straight line, and then the ball begins to do a uniform circular motion in the horizontal plane at a rate of 2 meters per second. If the maximum tension that the rope can withstand is 4N, then what is the time elapsed from the beginning to the break?

The process of this solution is as follows: Solution: Every time the ball turns half a circle, the rope hits another nail that is not the center of the circle, and then uses this nail as the center of the circle to do a uniform circular motion, and the radius of the motion is reduced, but the velocity is the same, (because the pull of the rope on the ball only changes the direction of the ball's velocity).

According to f mv2 r, every time the rope touches the nail, the tensile force (centripetal force) of the rope will increase, when the tensile force of the rope increases to fmax 4n, the radius of the ball to do uniform circular motion is rmin, then there is: fmax mv2 rmin

rmin＝mv2／fmax＝（

The radius is reduced after the rope hits the nail for the second time, and the radius is reduced after the third time it touches the nail. So the rope is pulled off after hitting the nail for the third time, and before that the time of the ball motion is:

t＝t1＋t2＋t3／v＋π（

Conservation of mechanical energy.

The 8s of the upper throw is divided into two sections, the first 7 seconds go up and then return to the original place, and the last second is the same as the downward throwing process (conservation).

Get up movement seconds.

So the initial velocity is 35m s, and h=40m is obtained

The landing time is the same as the free fall time from a height of 40m.

t = 2s under the root number a = root number 8

s=vt=30*2√2=70√2

Let the initial velocity of the ball be v, and the downward direction is the positive direction of the velocity, then the first ball h=v0t+1 2at 2=-8v+320, and the second ball h=v0t+1 2at 2=v+5, v=35m s, h=40m

The third ball h=1 2at2,t=2 2s, distance s=vt=70 2m.