Questions about Senior 1 Physics Compulsory 1. Physics savvy people help helpers.

1. If you take an airplane to the west, as long as the speed of the airplane (or other flying objects, of course, other means of transportation on land can also be used) is greater than the linear velocity at this altitude caused by the rotation of the earth, you can see the sun rising in the west.

2. The direction is set by people, and if the direction is changed, the sun will rise in the west.

You can see the sun rising in the west, and that's when you unconsciously use some object in the west as a reference.

Normal: The sun rises in the east.

To see the sun rising in the west, you have to move westward in excess of the Earth's linear velocity, which is the first cosmic velocity, and the Sun should still be rising from the east relative to the ground, but if you exceed the Earth's linear velocity at your point, you are flying towards the place where the sun is already shining, and the sun is behind you (i.e., west), isn't that okay?

Run in the same direction as the sun and you can rise faster than the sun.

There's also the possibility that you'll see it on a plane.

The question ...

Running east faster than the sun.

Frame of reference Earth is fine.

I think it's the horizon as a frame of reference, and remember the answer is this, I did.

It's mainly a matter of frame of reference. You can refer to an object on the ground. If it's a multiple-choice question, you can use elimination.

The lz solution can start with the formula of physics itself, or you can draw a diagram, or even you can list quadratic functions.

To chase and meet is to use a moving object as a reference, for example, a bicycle as a reference, and the frame of reference is similar.

It's like a car chasing a stationary bicycle at a speed of 12m s 120m ahead, so it needs t t 120 12 = 10s

The equation for the second problem is solved, and the first encounter is followed by t, and the second encounter is v1t+1 2at 2=v2t

The algebra value is obtained, t=12s, because t is greater than 9s (the car is stationary when 9s), so this is equivalent to the situation that the car continues to accelerate backwards after stationary, so the real time is t=x, v=81m (6ms)=, if the time is calculated from the beginning, the previous 10s must be added

It seems so.

1.The displacement of the car is equal to the displacement of the bicycle plus 120m

2.Taking the time of encounter as the 0 moment, first discuss whether the displacement of the bicycle is greater than that of the car when the speed is reduced to 0, if it is greater than that, then the time is obtained by the displacement formula of uniform acceleration motion according to the displacement of the two equal, if it is less than, the process is divided into two stages: the car decelerates to the speed of 0 and the car is stationary to obtain the time.

1.In this way, using the bicycle as a frame of reference, the speed of the car is (18-6) meters per second. So the time it takes to travel 120 meters at such a speed is what we require. i.e. 120|（18-6）=10

2.Think about it, if two cars meet again, then the car needs to slow down to a speed smaller than the bicycle, assuming it takes t seconds, then there is 6t = 18t + 1|2 (-2) (t squared). Note that the acceleration is negative because it is a deceleration.

Note that t does not take a negative value, if there is one, it should be rounded.

The last meter per second of seconds.

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 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.

l1=1/2at*t+v0t （a）

l2=1/2at*t+v1t（b）

v1=v0+at （c）

s=（v0*v0）/2a （d）

According to these four expressions, you can find the expressions of s with l1 and l2, see how you solve it!

I offer the middle method, but it is certainly not the easiest solution.

l2-l1=(v1-v0)t=at*t

a=(l2-l1)/t*t

c）->b) l2=1/2a(2t*2t)+v02t (e)a)+(e)->v0=(l1+l2-at*t)*(l1+l2-at*t)/4t*t

With v0,a, you can find s.

The exam is also connected to the Internet through the mobile phone.,To be expelled.,Don't be like this.,Why bother cheating for so many points.,Have the ability to cheat in the college entrance examination.。。。

So you set the first, second, and third acceleration a1a2a3, and the time t of the second and third segments, so that the equation is listed according to the formula!

Since the object falls at a uniform speed, then the force equilibrium in the vertical direction f=mg=kv

So v (mg k).

It should be: let the resistance be f, there is f=m*g and k=f v 2 then v=m*g k under the root number

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.

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.

Atoms are the basic particles of chemical reactions.

The relative atomic mass is based on the ratio of the true mass of any atom to the mass of a carbon-12 atom 1 12, which is called the relative atomic mass of the atom Fe CO2 44 H2SO4 98

The quantity of a substance is the ratio of the number of particles (n) (e.g., molecules, atoms, etc.) contained in a substance to Avogadro's constant (na), i.e., n=n na. It is a physical quantity that connects microscopic particles with macroscopic weighable matter.

It indicates the number of particles contained in a substance.

Unit mol

Avogadro's constant.

The number of particles of any particle in 1mol is Avogadro's constant.

The basic physical quantity of material grain is the amount of matter expressed by the addition of the amount of matter, and the amount of matter is measured as the object.

Strongly ask for extra points, tired.

Wrong, the smallest particle is a quark, and in a chemical change, the smallest particle is an atom. (Because atoms do not change in chemical changes) other problems can be read by yourself^

Alas, the bounty is high, it's good, so the RPG question gives some motivation

Chemical reaction 1/12 of the mass of C12 56 44 98 2:1:2 2:1:2

The question is too long! The bounty is too low!

Hehe, think about it yourself.

Excuse me, are you from Jining?