Solve the physics problem in the first year of high school it is better to talk about the main proc

Solution: Let the maximum velocity at the shortest time be v, then there is:

v²/2a1+ v²/2a2 =s t min=2s/v

Then there is: t min=2s v=2s 2a1a2s (a1+a2) under the root number = 2 (a1+a2) s a1+a2 under the root number

The velocity at point A is 0.

The final velocity of the acceleration phase is set to v, which is also the muzzle velocity of the deceleration phase.

The average speed of the acceleration phase is (0 v) 2 v 2 and the average speed of the deceleration phase is (v 0) 2 v 2

So the average speed of the whole journey is v 2 , and v 2 s t is obtained

From v 2 2*a1*s1 we get the acceleration phase displacement s1 v 2 (2*a1) 2*s 2 (a1*t 2).

From v 2 2*a2*s2 to get the deceleration phase displacement s2 v 2 (2*a2) 2*s 2 (a2*t 2).

And s s1 s2 , so s [2*s 2 (a1*t 2)] 2*s 2 (a2*t 2)].

The time obtained is the t-root number [2*s*(a1+a2) (a1*a2) ]

With the V-T image method, the horizontal axis is t, the vertical axis is v, and the area is s. Let the maximum velocity be v1, and the time to reach the maximum velocity is t1, a1 v1 t1 a2 v1 (t t1) s v1 t 2 to solve the system of equations, t 2s(a1 a2) a1a2

The 10-storey building is 28m high

This question can be considered to arrive and catch just when it lands.

So the time to rush to the falling position is equal to the time of falling.

t is obtained from h=1 2gt 2

The tenth floor is to say that there are nine flights of stairs, ie.

By free fall knowledge, h=(1 2)*gt2

It can be obtained: t is approximately equal to seconds.

(1) Let the dynamic friction factor be u, then u=kl (m*g)=25n m*

2)f=(m1+m2)*g*

Untie. (1) F-bomb=

ab·ma·g=f shells.

ab=2)f=f bomb + f(friction between b and plate)=

Solution: Let the direction of the electric field be upward along the inclined plane. According to the title, Ball A is below Ball B. Let the inclination angle of the inclined plane be , the electric field strength is qe, and the tensile force o of the string is f

For AB overall, 2QE-3mgSIN = 0

E 3mgsin (2q) is obtained

For ball A, F+QE = 2mgsin, obtained.

f=2mgsinθ-qe＝2mgsinθ-3mgsinθ/2=mgsinθ/2

After the string is disconnected, the net force on the system is zero, and the momentum is conserved, so that when the velocity of ball A is zero, the velocity of ball B is v, then.

3MVO=MV, get V=3VO

For b-ball, the kinetic energy theorem is obtained (qe-mgsin)s=(1 2)m(3vo) -1 2)m(vo).

The distance from the moment the string is broken to the point when the velocity of ball A is zero, the distance that ball B rises is s=

Answer: (1) E 3mgsin (2q).

2）f=mgsinθ/2

3）s=

Isn't there an angle of inclination for an inclined plane?

Solution: Let the maximum speed of the shortest only skin time or the maximum speed of the mountain be v, then there is:

v²/2a1+

v²/2a2st

min=2s v then there is: t

min=2s/v=2s/

2a1a2s (a1+a2) = 2 (a1+a2)s a1+a2 under the root number

(1) Set the maximum speed to V

v = rated power divided by resistance = rated power divided by (2) v = p f f f-f = ma i.e. brought in to get v = 4m s

3 Because f=p v

So there is a=1m s 2

When you look at it, you should note that I did not distinguish between the physical quantities in the three questions, and I should mark the corner mark for not doing it. You can distinguish it yourself. I must have done the right thing, because I am a high school physics teacher, so I quickly added the points. Thank you.

(1) When the maximum speed of the car, it is moving at a uniform speed, and the resultant force is 0.

The maximum velocity obtained by p f resistance * vm u*mg*vm is vm p f resistance 60*1000 ( m s(2) when the acceleration a1 2m s 2, the velocity is set to v1, and the traction force is f 1 then p f1*v1

f1 f resistance m*a1

v1 p (m*a1 u*mg) 60*1000 (5*1000*2 m s) is obtained

3) When the velocity is v2 6m s, the acceleration is a2, and the traction force is f2 then p f2*v2

f2 f resistance m*a2

A2 [ p v2) u*mg ] m [ 60*1000 6) 5000 1 m s 2

1.When the maximum speed is reached, the net force of the car is zero, and the traction force is equal to the frictional force, so at this time p=f·vm= mg·vm

vm = 12m s

2.According to Newton's second law, the traction force f at this time then f- mg=ma, a=2m s

Then f=15000n, the power is constant, so p=fv finds v=p f=4m s3Again, the traction force at this time f=p v=10000n according to Niu Er, ma = f- mg

a=1m/s²

Hope it helps.

Your main problem is that you don't understand (l-lcosx), but this is the vertical height of the ball's descent. Note that gravitational potential energy must refer to the distance perpendicular. You make a perpendicular line to the middle line of the pendulum at the highest point of the ball, and use the triangular relationship to find that the vertical distance of the ball from the hanging point is LCOSX, so subtracting this vertical distance by l is the distance of the ball from the lowest point of the pendulum, that is, the height of the ball's fall.

To put it simply, when mechanical energy is conserved, it is the gravitational potential energy of the ball that is converted into kinetic energy.

The gravitational potential energy consumes mg(l-lcosx).

The kinetic energy is increased by mg(l-lcosx).

And the original kinetic energy is 0

So at the end of the moment kinetic energy = one-half mv 2 = mg (l-lcosx).

You're just a freshman in high school, and you haven't learned the kinetic energy theorem and conservation of energy, which is a very simple problem if you do.

The ball has gravitational potential energy at the highest point, in the process of the ball falling, the gravitational potential energy of the ball is converted into kinetic energy, the formula for calculating the gravitational potential energy is e=mgh, the lowest point of the ball is 0 potential energy point, h=l-lcosx, due to the conservation of mechanical energy in the whole process, when the ball drops to the lowest point, the gravitational potential energy is all converted into kinetic energy, the formula of the kinetic energy theorem is e=one-half mv 2, this is a formula, just remember it, the teacher will talk about it later, there is mg(l-lcosx)= from the conservation of energyA half MV 2 will do.

This is a very simple physics problem, but your knowledge reserve is not enough, the teacher will talk about it later, and you will understand it by then.

Also, physics is a very interesting subject, I hope you can find fun in it, don't treat it as a subject, and connect it with the reality of life.

Conservation of mechanical energy Ek2+EP2=Ek1+Ep1Mechanical energy is composed of gravitational potential energy and kinetic energy MV2, Mg(L-LCOSX)

Because the gravitational potential energy is related to the height, it can be considered to be 0 at the lowest point, and the gravitational potential energy is 0mv 2+0=0+mg(l-lcosx)Just solve it.

The first step is to select the zero potential energy surface, the mechanical energy in the initial state (kinetic energy, aggravating force potential energy).

Step 4 Do the math yourself.

By the way, your foundation should be hard to make up, find a tutor. It's hard not to go to school.

1.Taking the position of the ball when it swings to the lowest point as the reference plane, then the ball must have a gravitational potential energy at the highest point, the gravitational potential energy is mgh, h should be the distance between the highest point and the lowest point, and according to the geometric relationship h should be l(1-cosx), so the gravitational potential energy of the highest point is ep1=mg(l-lcosx). Because the ball has no velocity at its highest point, the kinetic energy is 0, i.e., ek1=0.

4.It is mainly calculated according to the conservation of energy, that is, kinetic energy at the highest point + gravitational potential energy = kinetic energy at the lowest point + gravitational potential energy.

The velocity v can be obtained by bringing in the potential energy (0, because the change point is used as the reference plane) and kinetic energy (1 2mv2) at the lowest point listed in step 2 and the potential energy (ep1=mg(l-lcosx)) and kinetic energy (0, because there is no velocity) at the highest point listed in step 1