Physics questions in the first year of high school. Thank you for the process

Finding the final velocity Let the initial velocity vo=8m s reach the slope [final velocity] vm s is obtained by v -vo =2ax v=16m s time to=20s This question is the problem of catching up, when the speed of the two is the same 16m s, the debris flow has not caught up, it will never be able to catch up, and there is the shortest distance x

Then when the speed is the same, the time of the vehicle is v=at, t=32s, and the time taken by the debris flow is [32+1]s

Vehicle displacement x=1 2at =256m plus 240m in front of the debris flow x1=256+240=496

Debris flow displacement x2=240+(32+1-20) 16=448mx1 is greater than x2 to escape danger.

The shortest distance is x1-x2=48m

If you don't understand, mention it.,Handwritten.,If it's helpful, please landlord.。

1) Let the rate be v, v 2-v0 2=2as, v0=8m s, a=, s=240m, and v=16m s

2) Debris flow downhill used (v-v0), the car went 1 2at 2 = 1 2 *, when the car and the debris flow speed is equal, the distance between the two is the smallest, you can judge whether the two will meet, v vapor = 16m s, t2 = (, s debris flow = 16 * 13 = 208m, s steam = v steam t + 1 2at2 2 =, because 》208m, so the debris flow will not meet the car, the shortest distance is 256-208 = 48m

1）v^2-v0^2=2ax

v^2=v0^2+2ax=64+2*

v=16m/s

2) The time taken by the debris flow to slide to the low slope is t

T1=(v-v0) a=(16-8) is obtained by v=v0+at, and then there is a uniform motion with a velocity of 16m s

When the speed of the car and the speed of the mudslide are the same, the distance is the shortest. The service time is t2 car v = at t2 = 16

The distance traveled by the car is x1=a2t2 2 2=

The total time taken by the car, plus the driver's reaction time, the total time is t3 = 33s debris flow distance x 2 = vt4 = 16 (33-20) = 208m From the calculation results, it can be seen that the car is out of danger. The closest distance between them is x1-x2 = 256-208 = 48m

Solution: The spacecraft and satellite velocity change to v The time it takes to make this change is t and also know the thrust f spacecraft mass m1

From Newton's second law, f (m1+m2) a

a is the common acceleration of spacecraft and satellites.

Again according to the acceleration formula a v t

It can be obtained from the above two formulas.

m2＝f*△t /△v - m1

That is, the result you want.

For the whole using Newton's second law, there are:

f=(m1+m2)a---

Since δv = a δt ---

Synopsis : m2 = f δt δv-m1

The potential energy of the rammer from the highest point to the stop decreases: mg(h+h) resistance work is equal to the decrease of potential energy:

fh=mg(h+h)

Average resistance f=mg(h+h) h=mg(1+hh).

(1) Let the time of free fall be t, then the height of free fall is gt 2 2 The speed after falling t seconds is gt

So 266=gt 2 2+[(gt) 2-4 2] 2*2 solution t=3s

2)h1=gt^2/2=45m h2=266-45=221m（）（3）t1=3s t2=(3*10-4)/2=13st=t1+t2=16s

The descent of the athlete is divided into two phases, the first stage is a uniform acceleration with an acceleration of g, and the second stage is a uniform deceleration with an acceleration of 2, assuming that the free fall time is t, then the speed when the parachute is opened = gt, and the speed when reaching the ground is 4, then the descent time after the parachute is (gt-4) 2

The free fall distance is: gt square 2, and the descent distance after opening the parachute is: ((gt) square - 4 square meters) (2*2).

The sum of the two distances is 266 meters, take g = 10, and solve t = 3 seconds.

The descending height when the parachute is opened: gt square 2 = 10 * 9 2 = 45 meters, and the ground clearance at this time is 266-45 = 221 meters.

Descent time after parachute opening (gt-4) 2=(30-4) 2=13 seconds, so the total time spent in the air is 16 seconds.

The time of free fall of the athlete is t seconds, at which point the velocity v=gt=10t and the drop height h1= .

The time of uniform deceleration and descent t=(gt--4) 2=5t -2The height of the decline during the uniform deceleration descent is h2=

h1+h2=h 25tt--4+5tt=266 t=3 seconds.

h2=25x3x3-4=221 meters.

Total time = t + t = 3 + 3 x 5 - 2 = 16 seconds.

According to the title, the maximum braking distance is 24m, and the final speed of braking is 0, which can be regarded as a reverse uniform acceleration movement.

According to 2ax=v 2

2*8*24=v^2

v=8√6 m/s

It can be seen from the magnitude of the pressure equal to the support force.

Let the magnitude of the supporting force be f, the magnitude of acceleration be a, and the magnitude of gravity is mg known, a=70m s 2

Solution: f-mg=ma

So f=m(a+g).

Substituting the known into it yields f=80m (g takes 10).

So we know that f mg=8

So the astronaut puts 8 times more pressure on the seat than its own gravity.

Overweight status.

The idea is to first determine the magnitude and direction of the astronaut's resultant force according to the acceleration, and from the acceleration of a=70m s 2, we can know that the resultant force of the astronaut f=ma is 70m) upward, and it can be known that it is overweight.

Let the vertical upward direction be positive and the pressure be f pressure.

According to the force analysis, the astronaut is subjected to two forces, one is the pressure of the seat and the other is gravity, and the resultant force is 70m, and the direction is upward.

70m=-mg+f pressure (the direction of gravity is downward, so -mg is used) f=80m=8mg=8g

So the pressure is 8 times the dead weight.

At this time, the astronaut must be overweight.

Because astronauts have upward acceleration.

The amount of pressure on the seat of the astronaut is equal to the support force of the seat on the astronaut (Ox3), and for the astronaut, there are 2 forces, one is the upward support force, and the other is the self-gravity force known by the Ox2 f-mg=ma

Bringing in can be known to be 8 times.

Do an acceleration movement upwards f upwards "g", so it is overweight. a=f m=(f--g) g g=(f-pressure--g) g g=70

The solution yields f pressure = 8g

It is easy to tell that when the rope burns out, the elastic force in the rope is 0, and the spring elastic force is instantaneously unchanged, so the acceleration of a is 0

The total external force of the whole treatment is 3mg

Since BC is relatively stationary, the acceleration of BC is the same: 3mg=m multiplied by 0+mA+ma

Explanation Resultant external force = mass of a multiplied by the acceleration of a + mass of b multiplied by the acceleration of b + mass of c multiplied by the acceleration of c.

It can be solved that a= b c is the same.

At the moment when OC is burned out, the force on each object is as follows.

A turned out to be in equilibrium by gravity and the tension of the spring, and at the moment of burning, the length of the spring did not change, so the tension did not change, so the force of A did not change, there was no acceleration, and the pulling force of B and C was pulled down, and the pulling force was mg

Because B and C are tied together with a rope, and because they are pulled downward, the rope will directly transmit the force at this time, so BC can be seen as a whole.

B and C are originally subject to their own gravity 2mg, the pulling force of the lower spring mg and the upward pulling force of the upper rope OC 3mg, the moment the rope burns out, the upward pulling force of the upper rope OC 3mg disappears, and the other forces remain unchanged, at this time the mass of the two is 2m, the force is gravity 2mg plus spring tension mg, a total of 3mg, at this time the acceleration of b and c is a=f m=3mg 2m=

Pay attention to the difference between the force in the spring and the rope, the spring is the process force, the process is needed, the moment is unchanged, the rope and the rod change instantaneously, so for the instant, the A object is still balanced, the acceleration is 0, when the B object is balanced, the rope tension 2mg is the rope at C, but the tension in B is the rope and the solid conduction, and it disappears instantly, but the downward spring elastic force still exists, so the acceleration of B is 2G, the C object is subjected to an upward 3mg pull, and the B rope is 2mg pull, which are all ropes, so the moment of shearing, disappears, then the C object only has its own gravity, so the acceleration is g

The first question is set to time t=1s, then the air quality passed during this time is m= vs

The kinetic energy it originally carried is (1 2) mv 2, all of which is converted into electrical energy w

Then the electrical power p=w t=(1 2)mv 2=(1 2) vsv 2=(1 2) sv 3, choose b

The second question discusses the elastic potential energy of the spring, and the simplest analysis starts from the work done by the spring elastic force w=k x* x, choose b

Question 3 Analysis 1: The gravitational potential energy of an object decreases or increases, and it is only necessary to pay attention to how much work the object's gravity does on the object.

Analysis 2: If you want to analyze the conservation of energy in this question, then the gravitational potential energy is 100J = overcoming the resistance 30J + the kinetic energy of the object is 70J

The fourth question is to do work on the rope, the kinetic energy of the rope remains unchanged, and the gravitational potential energy will inevitably increase, so the center of gravity of the rope will move upward.

1.Let's take this wind as a cylinder, and at the time t we convert the power of w, and the height is l=vt, so its v=vts knows the density, so m=pvts, so the energy of this wind is v 2, and pvts 2 are all converted. So you divide it by a p=w t, so p=how much I don't count.

Pick b The other questions downstairs have told you.

Analyzing it, a total of 100 J of work was done, and some of them did 30 J of work to overcome resistance. So the rest is converted into kinetic energy 70J So who gave this energy? is the gravitational force of the object, so the gravitational force of the object loses the gravitational potential energy of 100j.

For 4, you need to tell whether the ends are at the same point or different points, and whether they are fixed or slided.

Question 1: 100 percent?? Or am I 2 anymore?

Question 2: CDB comes in order.

Question 3: The amount of reduction in gravitational potential energy = the work done by gravity.

Question 4: The center of gravity is upward, because f does work on the rope, the kinetic energy of the rope does not change, and the gravitational potential energy increases.