In the first year of high school physics problems, you will advance and seek speed

v0 = 72km/h = 20 m/s

The total braking time of the car is t = v0 a = 20 5 s = 4 s

Therefore, the magnitude of the car brake 3s without speed v3 = v0 - at3 = 20 m s - 5 3 m s = 5 m s

Bringing in data from s = v0t - 1 2at yields 30 = 20t - 1 2 5t

Solution : Time taken through 30m t = 2s , t'= 6 s (greater than the total braking time, rounded).

The car has stopped moving at the end of 4s, and the 6th is at rest, and the end velocity is zero, which can be obtained from 0 - v0 = -2as: the total brake displacement s = v0 2a = 20 2 5 m = 40 m

That is, the displacement of the car from the start of braking to 6s is 40 m

72km/h=20m/s

a=-5m s 2, take the negative value.

vt=v0+at0=20-5t

t = 4 sec stop.

v3=v0+at=20-5×3=5m/s

s=v0t+at^2/2

30=20t-5/2t^2

t=2, t=6 (rounded).

3) The displacement at the end of 6s is the displacement of 4s.

s=v0t+at^2/2

40 meters.

Hello, v0=72km h=20m s

a=-5m/s²

1)v1=v0+at1=5m/s

2)2as=vt²-v0²

vt=(400+2(-5)30)^(1/2)=10m/st=(vt-v0)/a=2s

3) It stopped at 4s.

s=v0t+1/2at′²

t′=4ss=40m

Hope it helps!

The difference is that the condition of question 1 is the same time before and after, and the displacement is the same for question 2. 1: Let half the time be t, then, (8t+4t) 2t=6m: let half of the displacement be s, then, 2s (s 8+s 4)=16 3m s

Question 2 Let the displacement be 2x

The 2x (x 8+x 4) average velocity is the displacement ratio time.

2x is the total displacement.

x 8 is some time ago.

x 4 is a later period.

Question 1. The set time is 2t

8t+4t)/2t

8t is the front section displacement.

4t is the posterior displacement.

2t is the total time.

△x=△x=l λ/d

The wavelength of this monochromatic light =d x l=

Refractive index n=c v='

The wavelength of this light in the anti-reflection coating is:'=

The thickness of the anti-reflection coating should be at least taken'/4=

x=l c bring it in and count it yourself.

Wavelength 1 Wavelength 2 = n

Wavelength 2 2 don't understand and ask me again.

Solution: x= dThe wavelength of this monochromatic light =d x l=refractive index n=c v='The wavelength in the anti-reflection coating is:'The thickness of the anti-reflection coating should be at least taken'/4=

When the car slows down to 4m s, it will not hit the dog without hitting it.

Calculate the distance traveled by the car first.

Acceleration a=(0-v0) t'=-20 4=-5m s 2 reduced to 4m s time t2=(v0-v) a=16 5=constant velocity section s1=v0*t1=

Deceleration section S2=V0*T2+

s=s1+s2=

s dog = v * (t1 + t2) = 4 * (

S > s Dog +33

Crashing into a dog with a speed time image is very simple!! If there is an intersection, it is a collision! If it's too complicated to draw, don't draw!

It's hard to type!

This is the driver's reaction time when the car does average speed driving, the 4 seconds of the car to do the average deceleration, v0 is 20m s, v end is 0m s, (0 + 20) x4 2 = 40 this is the distance of the car to do the average deceleration, 10 + 40 = 50m is the total distance of the car, this is the distance the dog walks, 18 = 33 = 51m 51m 50m (the car originally had 33m away from the dog, the car walked a total of 50m spent, the dog is also running.) ）

Let the collision time point be k, and the driver's acceleration is 20 (; The displacement of the driver is n:20* +20*(; The displacement of the puppy is m:4*k so n=m+33; After substituting and calculating, see if there is a real number solution (and it is in between), and if there is, it means that it has been bumped.

The only case when looking at the same speed.

a=20m/s/4s=5m/s^2

When v=4m s.

The displacement time of the car is 16m s 5m s 2=

10m+The displacement of the dog at this time is 4m s*

distance so the car pressed over the dog.

It is easy to establish a coordinate system and draw a V-T image of the car and the dog.

1) As can be seen from the problem, the direction of the Lorentz force is upward, and the positively charged body moves to the right by using the left-hand rule.

2) The object can slide against the lower surface of the horizontal insulating board.

qvb≥mg

v≥mg/qb

The size v should not be less than mg qb

3) By the kinetic energy theorem: wf = 1 2 mvo 2 - 1 2 mv 2

then it can slide along the direction of the horizontal velocity of the insulating surface to the right

The size v should not be less than qvb>=mg

When the car slows down to 4m s, it will not hit the dog without hitting it.

Calculate the distance traveled by the car first.

Acceleration a=(0-v0) t'=-20 4=-5m s 2 reduced to 4m s time t2=(v0-v) a=16 5=constant velocity section s1=v0*t1=

Deceleration section S2=V0*T2+

s=s1+s2=

s dog = v * (t1 + t2) = 4 * (

S dog +33 crashes into dog.

It's very easy to use the speed time image! If there is an intersection, it is a collision! If it's too complicated to draw, don't draw!

It's hard to type!

The beginning of the object is subjected to four forces:

friction force f, along the direction of the inclined plane, up and down unknown;

The inclined plane supports the force n, and the vertical inclined plane is upward;

gravity g, straight downward;

f1=2n, down along the inclined plane;

f2 = 10n, up along the inclined plane.

If the object is at rest, the frictional force experienced by the object is equal to the other combined external force except the frictional force, and the opposite is true.

In the direction along the inclined plane:

The gravitational component GSIN37°=20*, is downward along the inclined plane.

gsin37°+f1=12+2=14n ＞ f2=10n

The sum of the external forces other than the frictional force along the inclined plane is 14 - 10 = 4n, and the downward direction is along the inclined plane.

So the friction force f=4n, along the inclined plane.

Synthesis: f = GSIN37°+F1 - F2 = 12+2-10 =14N].

If f1 is removed, the total external force on the object is 2n and the direction is upward (opposite to the direction of the equal magnitude of f1 removed) [comprehensive formula: f = gsin37° -f2 = 12-10 = 2n].

First of all, the gravitational force is divided into the force 12n downward on the inclined plane and the force 16n downwards on the perpendicular inclined plane, because the object is stationary, the resultant force is zero, you can list such an equation: 2n 12n 10n frictional force, so the friction force is 4n, when it is zero, the same as above, the force analysis result is 12n-10n=2n downward direction, I used my mobile phone to call, it took up my own study time, I sure it is correct, I must do it.

The question only says the magnitude of the velocity, then the velocity at the end of the 1st s and the velocity at the end of the 2s may be in the same direction or in the opposite direction, v1 = 3m s v2 = plus or minus 4m s, so using them and the time between them, you can find a with two values 1 and -7, so the certain answers are wrong, c is right, as for the d answer, no matter which a I use, I did not get 14m, naturally it is not right. The displacement of the object within the 2s may or may not be.