Two physics calculation questions and two physics arithmetic problems

Dude, study the V-T image well, and the V-T image can be easily solved for both questions.

In the first question, there is some content missing, either car A suddenly stops at zero speed in front, or car A suddenly slows down in front, but there must be an acceleration (the question does not). The second case requires you to find known conditions, the first one is actually directly drawn to know: B first does a constant velocity (speed in seconds, and then decelerates (acceleration 2 meters per square second) The final velocity is zero, and it does not collide, which is nothing more than the displacement of B is the distance between the two.

The second question is the same image method, but you are missing a sentence (the final speed of deceleration is zero) is nothing more than the displacement of the acceleration stage + displacement of the uniform stage + the displacement of the deceleration stage (in fact, after the image is drawn, it is a trapezoid, the area is 52, the bottom is 16, and the bottom is the time of the constant velocity stage, which needs to be solved) Both methods can solve the first question.

If the elevator arrives in the shortest time, it should be directly accelerated first and then directly decelerated (there is no intermediate uniform process, the image is a triangle with an area of 52).

I didn't give you the correct solution, but I gave you an analysis, which I think is more comfortable than watching the solution process.

First pass: gmm r 2 = mv 2 r

v^2=gm/r=

v=t=2πr/v=2×

The second pass: flat bridge pressure mg=4000

Convex bridge pressure mg-mv 2 r

19200n

It is necessary to make the truck have no pressure on the bridge at the highest point of the convex bridge.

mg=mv^2/rv=

23。(1) Let the gas pressure at the left end be p (expressed by cm mercury column) p+8=p0 (atmospheric pressure) to obtain p=76-8=68cmHg (2) because the temperature is unchanged, so the PV in the gas is constant, and the cross-sectional area in the tube is constant, then PH is constant.

H'=ph P0=68*19 76=17cm24. (1) The ball has gravitational potential energy and kinetic energy at point A, the gravitational potential energy is mgh=20j, and the kinetic energy is 1 2*mv 2=50j

So it has a mechanical energy of 70j

2) From the conservation of energy, we can know that MHL+1 2MVO 2=1 2*MVB 2, we can know that the velocity of the ball at the lowest point b is 2 times the root number 29, and then the ball makes a flat throwing motion, and the speed of the ball moving to the ground in the vertical direction is 2 times the root number 6, and in the synthesis of the velocity, we can know that VC=2 times the root number 35M s

(1) The pressure of the initial gas closure p+8=76 p=66cmHg (2) The length of the closed gas in the left tube of the final state l 19*66=l*76 l=

24.A small ball with a mass of m = 1kg is tied to one end of a light rope with a length of l = and the other end of the rope is tied to the O point at a height of h = 2m from the ground, the ball is pulled to point A so that the rope is straightened horizontally, and the initial velocity of the ball is straight down? g=10m/s^2

1) The mechanical energy of the ball at point A is e0

2) After the ball passes through point B, the rope breaks (without considering the loss of energy), and the velocity vc when the ball falls to the ground

v=30*1000/3600 = 25/3 m/s

Deceleration a = g = =

Safety distance s = s1+s2 = vt' +v 2 (2a) =25 3* +25 3) 2 2* =385 54 meters.

If the braking deceleration is A, the car must be shouted to ensure that it does not collide when it slows down to the speed of pedestrians.

The time required to decelerate to the same speed as the pedestrian t=(v2-v1) Zheng Baohao a = 15-5) a=10 a

Reaction time t'=

S person = v1 (t + t').

Car s = V2T'+(v2^2-v1^2)/2a)

S people +30 s car.

v1(t+t')+30 ≥ v2t'+(v2^2-v1^2)/(2a)

5*(10 a+04) +30 stuffy 15* +15 2-5 2) (2a).

50/a +2 + 30 ≥ 6 + 100/a

50/a ≤ 26

a ≥ 25/13 ≈

v=,s1=,s2=vsquared (2*.)

s=s1+s2=

When the slag brakes are started, the distance from the cluster beam is s=30-(15*Therefore, 15 squares - 5 squares = 2asa=

First of all, the speed is all withered feast converted into meters and seconds.

The reverse acceleration is not as big as a=kinetic friction factor*g=

The distance is based on the formula vt 2-v0 2=2*a*sv0=0m s vt=30

How is there no quality of the car.

1. Solution:

Analyzing the problem, this situation can be summarized as when the car happens to collide with the bicycle, the motion time of the car and the bicycle is the same when the speed of the two is equal, and the same is set to t, and the formula x1=v1t+1 2at

v2t=x2

x1-x2=x

v2=v1+at

Substituting the numerical values gives two formulas.

10t-1/2×6t²-4t=x

4=10-6t

The solution is t=1

x=32, the falling process can be divided into four sections, which are accelerated falling, falling at a uniform speed, falling by deceleration, falling at a uniform speed, and accelerating the fall.

The acceleration is the acceleration due to gravity g=10m s

Acceleration to a speed of up to 50 ms

According to v0+at1=v

v0=0a=g.

t1=5sx1=1/2gt1²

x1=125m

x2 = x total -x1 - x

x2=2000-125-200=1675mt2=x2÷v2

t2=1675÷50=

t3=x3=（1/2）×（v2+v3）×t3

x3=④x4=x-x3

x4=t4=x4÷v4

t4 = in summary.

t=t1+t2+t3+t4

1.Because the car just can't touch the bicycle, so in the process of braking the car to 0m s, their time is equal, so according to the formula s car = vt + 1 2at 2 gets: t = 2.

And according to the time-dissipation formula v0 2-vt = 2as: s car = 25 3, during this time s bicycle displacement = vt = 8m, so x = 1 3.

When the object just starts moving, the pulling force is equal to the maximum static friction force.

That is, 200*When pulling the object forward at a constant speed, the pulling force is equal to the kinetic friction force.

That is, 200* (k is the dynamic friction factor).

i.e. k 200*

As a physics teacher, I'm really sad for you, not just sad!!

12. The velocity angle is 90 degrees, and the velocity is a vector, then it means that the vector product of the two velocities is equal to 0, and the velocity of the two small balls is 90 degrees, respectively, v11 and v12, v01 is the negative direction, and the tung is disadvantaged

v11 (-v01, gt), v12 (v02, local gt).

v11*v12=-v01*v02+(gt) 2=0, so t=root(v01*v02) g

There is no air resistance, then the horizontal speed of the two balls in the falling process is unchanged, set A to the left is v01, has been to the right is v02, the longitudinal swimming speed is gt, their speed is the horizontal direction of the history of the world and the vertical direction of the velocity, and the net force is the same way of drawing, so as long as the v01 gt=gt v02 is on the line, t=

Root number (V01V02) Limb nucleus Limb G

You draw a picture. 13 too lazy to do it.

Not a lot, the braking acceleration of both cars is the same.

The meaning of the question is that car B is behind and cannot collide with car A, that is, assuming an initial distance s, the final distance is 0, simply put, the braking distance of car A is equal to the distance of car B running for seconds and braking.

The second question assumes a time of constant motion, and the starting acceleration is twice the braking acceleration, that is, the braking time is twice the starting time (can you understand?). If you don't understand this, you can ask again) three times are set, and tell you that the building is high, find the time, you will know the maximum speed (that is, the speed at a constant speed), the shortest is simple, from 0 speed to 0 speed, there is no uniform speed, and the braking time is still twice the starting time.

If you don't know which formula to use, don't practice doing the questions, you can't do this foundation, do the exercises after the book.

If you know which formula you didn't get right, go to a math teacher.

Amitabha! This kind of topic is really a scourge for children!! It's a question from a person with a pot of mushy brains... Amitabha!

Ignore him ...

(11) First analyze the force of the balloon, gravity and buoyancy and force f1 up 4 Newtons, wind force f2 horizontal to the right, rope tension f 360 degrees downward, four forces work together to make the balloon balance f3 * 60 degrees positive Xuan = f1 f2 = f3 * 1 2

The object in front of the "1s" in the title falls ts

Instantaneous velocity in the middle = average velocity.

t1 = t + is the intermediate moment of "a certain 1s", and the instantaneous velocity of t1 v1 = the average velocity of this 1s = 20m 1s = 20m s

Because v1=gt1=g(t+

Therefore t=(v1) g

Total time of fall t=t+1+

The total height of the falling object h=(1 2)gt 2=

If the average velocity within 20m is 20m, then the landing speed is 40ms, and the height can be found to be 320m.

It may be wrong, so be careful!

Use 20m and 1s to calculate the intermediate moment velocity of 20m s, that is, the average velocity. From v=gt, t=2s, that is, the object has fallen for 2s in the middle of this 1s, then the object has fallen for a total of 4s. Based on H=1 2 GT2, H is 80m.

Can be drawn to understand.