Ask for high school physics problems, ask for advice, and expect an analytical process

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Solution: (1) Let the driving speed of the car be v, then the reaction distance = v * t =

2) Braking distance = (v-0) * v (10 * 2) = (v 2) 20

3) The car from the discovery of the situation to the car stopped running brother mu of the distance collapse dust touch = > = 60m, so x is at least 60m

Analysis: The distance from the patrol car to catch up with the target vehicle l = 2000 + the distance in the form of the target car.

It takes a total of x seconds.

The patrol car l consists of the distance at acceleration + the distance traveled at a constant speed, and the column formula l = 50 2 10 + 50 (x-10).

The distance traveled by the target vehicle is l2=35x

Summary: l=2000+35x = 50 2 10+50 (x-10) to get x=150 seconds.

This is a chase problem, there is a routine.

Let's see if the car has caught up with the target before it reaches its maximum speed. Acceleration a=50 10=5 The displacement that has been driven when the maximum velocity is reached is x=v 2 2a=250m

Less than 2000m, which means that in order to catch up, it is necessary to drive for a while after reaching maximum speed.

So let's take the total time t, it takes 10 seconds to reach the maximum velocity, so after that, we open t-10 seconds at the maximum velocity, and the total displacement is 250+50*(t-10).

The displacement of the target during this time is 35*(t-5) The relationship between the two is: 250+50*(t-10)-2000=35*(t-5) T can be found with a little calculation.

The image is really not clear, especially the data, if the data is wrong, please fix it yourself When u=1000v, the electric field force = gravity, just straight forward. Therefore, f=eq=uq d=mg, get q=mgd u (m can't see clearly, please calculate it yourself, and the following is directly expressed by q. )

In the electric field, the net force of the particle is f=eq+mg, that is, it does a flat-like motion under the action of this force.

Since the initial velocity is known, then the flat throwing time is t=l v, and the vertical height of the flat throwing is d 2, then the acceleration is a, then.

1 2*at 2=d 2,a=d t 2=dv 2 l 2=a can be up or down. Therefore.

f=ma, you can take two values, up and down.

If upward: e1q-mg=ma

If downward: mg-e2q=ma

Solve e2, e1, u=ed. You can get the range of values of u.

Note that since a = 8, which is less than gravity, it is unlikely that e will be in the same direction as g.

v averages the velocity at half the time elapsed in the displacement, so x t2 - x t1 = a*(t1 + t2) 2

Deformed vested.

Question: Analyze the elastic force experienced by object A?

Answer: The elastic force is generated by the mutual extrusion of the first phase of the objects, and the extruded object should restore its deformation to the powerful effect on the object it squeezes, and this force is called elastic force. The direction of the elastic force is that the squeezed object is directed towards the extruded object, i.e. it will be perpendicular to the contact surface, and the rope and spring will follow the direction of the rope and spring.

1) A elastic force on the ground facing it vertically upward. The objects on the right have no elastic effect on A, because there is no squeezing between them.

2) The ground has a vertical upward elastic force on A, and the rope has an elastic force diagonally to the upper left along the rope to A.

3) The left semicircle has an elastic force diagonally to the right and above the right for A, and the rope has an elastic force diagonally to the left for A.

(1) Because f=umg=100n, a1=f m=5 of the slider and a2=f m=1 of the flatbed

2) When the slider is consistent with the speed of the flatbed, it is relatively stationary. Slider v=v0-a1t. Flatbed v=a2*t, when v0-a1*t=a2*t, t=, flatbed s=1 2*a2*t 2=, slider s=v0*t-1 2at 2=, because, greater than 3, it will slide out.

In the shortest time, drive in the direction of the vertical riverbank; The time is 100 4 = 25s and the displacement is 25 * 3 = 75m, and the square (75 2 + 100 2) = 5m

To minimize the time, the direction of the combined velocity should be directly opposite the shore, that is, the bow of the ship should be oblique upwards.

Question 12 can be calculated directly using the deformation formula of Newton's second law of motion.

vt+ brings in the data and you can find v.

Question 13 According to the depth of the well of 125 meters, you can find the time required for a small ball to fall freely, and use h= to find t=5 seconds The question says "When the 11th ball falls", that is, the 11th ball has just fallen, and it has fallen 10 balls, and it takes 5 seconds to throw 10 balls, and the interval between each ball is seconds, and when the 5th ball starts to fall, the third ball has fallen for 1 second, and h= to find that the falling distance is 5m, then the distance between the 35 balls is 5m (according to x= it can be known that when the acceleration is the same, the same time has elapsed The displacement is the same, so the distance between the two balls will not change at 35, and it will remain 5 m all the time).