High School Physics Compulsory 1 Two Simple Questions

In fact, gravity is a virtual force, which does not actually exist, it is just people's feelings; The gravitational force experienced by the object is equal to the vector difference between the gravitational force experienced by the object and the centripetal force required for the object to rotate with the Earth; The magnitude of gravity is also equal to the magnitude of the ground support force f on which the object is placed on the level ground, and it is considered that the gravity and support force f are equal in magnitude and in opposite directions; Such a treatment is limited to not considering the statics and dynamics of objects on or near the ground at the time of the Earth's rotation.

Gravitation is directed towards the center of the earth, but only at the equator and gravity at the poles points to the center of the earth. The gravitational force of an object at the equator is equal to the difference between the gravitational force and the centripetal force required for the object to rotate with the Earth, and the gravitational force at the poles is equal to the centripetal force.

You thoroughly understand the opposite, the attraction of the earth to the object is gmm r 2, which is decomposed into a vertical downward gravity and a horizontal centripetal force, and the gravitational force completely acts as a gravitational force at the polar regions, so gravity is vertically downward all the time, but the magnitude is not necessarily equal to the gravitational force gravitational force is directed towards the center of the circle, and you will know it in the chapter on gravitational force.

When the earth rotates, the object on the earth rotates with the earth due to inertia, so the gravitational force has nothing to do with the rotation of the earth, so the gravitational force is related to the gravitational force of the earth, and the direction is downward, in the same way, b is correct.

First of all, we must understand the definition of gravity: the force that is applied to an object due to the attraction of the earth is called gravity. The direction of gravity must be straight upwards. These are the original words of the book!

What is called vertical downward, vertical downward is a direction, and the vertical downward direction is the direction of gravity, which is a matter of mutual definition, which is a regulation.

I think so too, what does your teacher say.

The first 5 seconds are the time, and the 3rd second is the moment.

Instantaneous speed is the limit of the average speed, for example, the speed of the car within 1 second, the speed within 2 seconds, these are instantaneous, so it is instantaneous speed.

Average velocity is the ratio of displacement to time over a certain period of time, and average velocity is a vector quantity.

The velocity is the magnitude of the velocity.

The average rate is the magnitude of the average speed, which is equal to the time it takes for a distance to travel longer than that one.

VT charts and XT charts are very simple, as long as you can figure out the Cartesian coordinate system.

Acceleration is the ratio of the amount of change in velocity to the time it takes for this change to occur (acceleration = end velocity - initial velocity time to complete this velocity).

If the final velocity-initial velocity is 0, then the acceleration direction is the same as the positive direction; If the final velocity-initial velocity < 0, then the acceleration direction is opposite to the positive direction.

Just memorize a few formulas for calculations. Average velocity = displacement time, average velocity = distance time, acceleration = (end velocity - initial velocity) time.

These teachers have talked about it in class, and you should listen carefully; It's also in the book, you can check it out

The first thing to know is that the hour is the number of hours on the clock. Then time denotes the interval between two moments. Like you get out of bed.

This is the moment. 40min per class, this is the time. Then the first 5 seconds, obviously, is a time, and the third second is a moment.

Instantaneous velocity is the velocity at which an object moves to a certain moment or position. The magnitude of the instantaneous velocity is called the rate. The average velocity represents the ratio of an object's displacement to the time it takes to complete it.

The average rate is the magnitude of the average velocity. Note that this average speed is not an average speed. ps：

The key is to do the problem, find the correct displacement when doing the problem, and the time of this displacement, the two are OK to divide, regardless of what is halfway and how. Acceleration indicates how fast or slow the velocity of an object changes, note that it is not an increasing velocity. The acceleration in the same direction as the initial velocity is acceleration, and the opposite is deceleration, abbreviated as:

In the same direction, it is added. The direction is reduced. Compute.

Acceleration = (end velocity - initial velocity) divided by the time t to complete this acceleration or deceleration. It's acceleration. It's all in these books.

The total take-off mass is 461t, take-off thrust, which can be used to launch small manned spacecraft. Try to calculate the acceleration of the rocket during takeoff. (take g=10m s2).

So f pushes -mg=ma a=

Force analysis. The spacecraft is subject to gravity and thrust. Because the direction of inference is vertically upward, and the direction of gravity is vertically downward, we can get: f pushes -g=ma and then brings in the data to be solved. Pay attention to the units.

1. Set the maximum speed to V

then the braking time is v 5

Then there is 1 2*v*v 5=

Calculate v=15m s

2. Let the acceleration be a

The deceleration time is 10 a

Then there is 1 2*10*10 a=

Calculate a=125 3m s

3. (1) Set the acceleration time to t

Then there is 1 2*5*t*t=100

Calculate t=2 10s

Then the actual speed when moving to 100m is 5t=10 10s10 10<50

So no. 2) Set the initial velocity to V

then the acceleration time is (50-v) 5

Then there is (v+50)(50-v) 5 2=100 to calculate v=10 15m s

It's all relatively simple, if you need a detailed explanation, please hi me.

The scale bar can be rolled out according to the length of the body, ie.

Scale bar k = the original car length is 12mm longer than the car in the picture above, isn't it just equal to the constant of 50 when the unit is converted?

As for the formula x=50 l, I don't need to say it, it's the scale of reality = scale multiplied by the scale on the graph.

x represents the change of the actual distance, 50 is the scale, k l is the change of the distance on the figure, the original length of the body is known in this question, and the length of the body in the figure can be seen on the figure, isn't the scale out.

yes, that's a matter of proportions, isn't the title for you, the length of the car is and the picture is a 50:1 ratio...

At least that's true when you work backwards from the answer, but it's not very good.

1）x1=2m x3=6m t=1s

Delta x = (3-1)at 2

a=2m/s^2

x1=v0t+1/2at^2

v0=1m/s

x6=v0t6+1/2at6^2=42m

v5=v0+at5=11m/s

v6=v0+at6=13m/s

v pull = (v5 + v6) 2 = 12m s

2) t=10s x1=8*8=64m x2=6*8=48mdelta x=at 2

a=x1=v0t+1/2at^2

v0=3)a=v5/t5=

v4=at4=

xd5=v4t+1/2at^2=

v3=at3=

vpluckd4=(v3+v4) 2=

4)v0=-at4

x1=v0t1+1/2at1^2

The solution yields a=-4m s 2

x last = 1 2at1'^2=2m

x=1/2at4^2=32m

5）x1=1/2gt^2

t=tTotal=xTotal=1 2gtTotal 2=

x2=1/2gt2^2=

xd3 = x total - x2 = 6m

Start from A to B if you have been accelerating uniformly, s=(1 2)at 210=(1 2)a6 2

Then a = 5 9m s 2

And v end = at=5 9*6>2m s

It is not compliant, and the launch should be accelerated first and then at a constant speed (under the action of static friction, the speed is uniform when the speed reaches 2m s).

Let the uniform acceleration time be t, then.

2=ats=(1/2)at^2＋(6-t)2

The solution yields t=2s, a=1ms2

If you want the shortest time, you need to ensure that it has been accelerating uniformly (the displacement is certain, only the speed is increasing, in order to reduce the movement time), to ensure that it has been accelerating evenly, that is, when the object moves to B, it just begins to become a state of uniform motion, and the speed is as large as the belt at this time, which is the critical point, that is, the minimum speed of the belt, and a = 1m s 2 and a certain rule.

s=v 2 2a (the initial velocity is 0, the end velocity is v, that is, the belt velocity) is solved to obtain v=2 times the root number 5m s

If it has been accelerating, the acceleration is 5 9, and the velocity is greater than 2 after 6s, which is not compliant.

If the acceleration is first followed by a constant velocity, the acceleration is negative and does not meet. So the title is wrong.

It should be accelerated first and then at a constant speed, at least two five meters per second.

You go and see, it must help.

Physical wolf pack.

That is to say, to physically skidding all the way to point b, first analyze the first process, he goes through two processes of acceleration and constant velocity, let the friction coefficient be a, the acceleration time t, and the uniform time is t1, and there is a*10*t=2

10=2t1+1/2a*10t^2

t1+t=6

The solution is t=2, t1=4, a=

The coefficient of friction is 1

To get up all the way to speed up.

Meet 1 2a*10t2 2=10

t2=sqrt(20), sqrt means square root in C language and MATLAB (sorry, it's inconvenient to play the root number).

a*10*t^2=sqrt(20)

The conveyor belt speed is at least sqrt(20)m s

v = 2 times the root number 5m s and the first friend's solution process is the same.

Have you finished your condition? Did the speed reach 2m s by the end? If that's the original question, I wouldn't.

1. There is a contradiction in the data of the question, and the mass of the object cannot be obtained. Take a look at the analysis below.

Let the mass of the object be m, the acceleration of the fall is a, the drag force is f, t is 2 seconds, h is 16 meters, and the muzzle velocity is 0

Obtained by h a*t 2 2.

16＝a*2^2 / 2

a＝8 m/s^2

And because the object is subject to gravity and resistance when falling, it is obtained by Niu Er.

mg－f ＝ma

f ＝mg / 10

The acceleration is a 9 g 10 9*10 10 9 m s 2

Obviously, the acceleration values obtained from kinematics are different from those obtained from the forces and the bulls.

Note: If the value of the drag force is known (give the specific value, not a multiple of gravity), there is no contradiction and the mass of the object can be obtained.

2. m=2*10 3 kg, resistance f, muzzle velocity is 0, t1 20 seconds, v1 10m s

In the first 20 seconds, it is obtained by v1 a1*t1.

The acceleration in this phase is a1 v1 t1 10 20 m s 2

Obtained by Niu Er f lead f m*a1

The traction force sought is f lead f m*a1 oxen.

When the traction force is removed, the acceleration of the car is a2

f ＝m*a2

a2＝f / m＝ m/s^2

Let the car pass through time t2 and stop moving, and the distance that can move forward again is set to s2

0 v1 a2* t2.

The time sought is t2 v1 a2 10 1 10 seconds.

It is obtained by s2 a2* t2 2 (with the reverse process).

The distance sought is s2 1*10 2 2 50 meters.

3. When the object slides upwards along the inclined surface, it is subjected to gravity mg, support force n (vertical inclined upward), and sliding friction f (downward along the inclined surface).

The magnitude of the resultant force is f, mg*sin37 degrees f, and the direction is downward along the inclined plane.

F n * mg * cos 37 degrees.

The magnitude of the acceleration is a f and m g*(sin37 degrees *cos37 degrees) 10*( m s 2

The direction of acceleration is downward along the inclined plane.

Let the maximum distance an object can glide upwards is s

0 v0 2 2*a*s.

0＝10^2－2*10*s

The maximum distance sought for an upward glide is s 5 meters.

The first problem is that since the drag force is always 1 10 of the gravitational force, the acceleration of the object is always the same regardless of its mass, so the mass cannot be found according to this condition, because the mass can be eliminated at the end. Therefore, the conditions for this question are incorrect.

The second question: according to the momentum theorem, ft-ft=momentum is the variable, so f*20-2000*20=2000*10 solves f=3000n, which is the traction force (of course, you can also use the uniform acceleration motion to calculate the acceleration and then use f=ma to solve it); After the time t, the car is stationary, and it is easy to find s = 50 meters according to the coefficient of the momentum theorem ft = momentum, and it is easy to know that t = 10 seconds and the distance is s, and it is easy to find s = 50 meters according to v 2-v0 2 = 2as (there are many methods, depending on your preference).

The third question: not specifically written, the idea is as follows: first carry out force analysis, according to the force balance perpendicular to the inclined plane direction to find the support force, this finds the frictional force, and then decomposes the gravity along the inclined plane direction, find the resultant force along the inclined plane direction (the gravity component and friction force are added), and then according to f = ma, find a, as for the maximum distance, directly according to the above v0 2 = 2as, where v0 = 10m s, a is the previous a, directly bring it in you can find out.