A simple physics question, but it seems that I got it wrong in the exam

Solution: 1) From the law of gravitation, f leads = g*m*m (r+h) 2

And the satellite m moves in a uniform circular motion around the earth, and the centripetal force it experiences in the direction of f=m*v 2 (r+h), since f leads = f, so g*m*m (r+h) 2=m*v 2 (r+h), and the solution v=[gm (r+h)] 1 2) is gm (r+h) square. It can be seen that the linear velocity decreases with the increase of the satellite's height above the ground.

2) Since v=2 (r+h) t, [gm (r+h)] 1 2)=2 (r+h) t

Solution: t=2 [gm (r+h) 3] (1 2).

As can be seen from the above equation, the smaller h is, the smaller t is, and when h = 0 (close to the ground), t is the smallest, and this minimum t is small = 2 [GM r 3] (1 2).

We also see that the gravitational force on an object on the ground mg=gmm r 2, so gm r 2=g, so t small = 2 (g r) (1 2).

Substitute g = into the above equation: t small = 2 seconds = 84 minutes and seconds.

Now that you already know, I'm going to give you the simple way to write it.

Get: v=gm (r+h) open root number.

It is obtained: t=2pai*(r+h) 3 gm (the open root number after *) when h=0, the period is the smallest.

The answer is to bring in the numbers.

The centripetal force surrounding the satellite is provided by the gravitational force between the Earth and the satellite. Just grasp that.

1.decreases with increase in H;

2.The t solved by gmm r 2=4m and mg=gmm r 2 is the minimum period

You're looking for the minimum, so t is the smallest when h is zero!!

This kind of problem is just a formula.

It's still very basic to figure this out on your own.

Eat the formula thoroughly

1.Set of formulas: As for you not getting a perfect score, it's your language problem. Just practice more.

and g = gm r 2

That's it!!

The higher above the ground, the lower the linear velocity.

Dude, you know that the answering process is not easy to type, so I'll tell you the idea, okay?

Problem 1: From the beginning to the first time A collides with B, in this process, A travels a distance of L A is subjected to friction, friction does negative work, B is also affected by A's friction against him, but less than B's maximum static friction, so B does not move, and a conservation formula for energy conversion is given to obtain the velocity of A before impact, and the impact between A and B is regarded as the conservation of momentum, and a momentum conservation formula is used to obtain the velocity of B.

Problem 2: Start from the time b stops moving, a continues to move and then stops, and the kinetic energy of a a is converted into frictional heat generation in the process, so um(a)gs =1 2 m(a)v2 where v is the velocity sought.

I'm sorry, let's make do with it, the process understanding is more important than anything else, and I'm done.

The isochronous nature of the free string motion of the b a ball is 2* root number (r g) and the b ball is the root number (2* r g) according to the free fall

cAccording to the periodic formula of a single pendulum, it is 1 4*2*pai*(r g)=pai 2*root number(r g).

Compare the b-ball to the first landing.

I think it's B, which is kind of an energy conversion problem.

Method 1: It can be solved with the help of the kinetic energy theorem and the conversion relationship of gravitational potential energy.

Method 2: Use momentum to solve.

The isochronism of the free string motion of the ball A is 2* root number (r g) and the ball b ball is the root number (2* r g) according to the free fall

cAccording to the periodic formula of a single pendulum, it is 1 4*2*pai*(r g)=pai 2*root number(r g).

Compare the b-ball to the first landing.

Hey, let's discuss it if you have any questions!

There is nothing wrong with this question, the speed of the water refers to the speed of the other side of the water, after the hat falls into the water, because the hat is relatively stationary against the water, so the speed of the drift is also the speed of the water on the other side. On the other hand, the topic clarifies that the speed of the ship relative to the water remains the same, but the speed of the ship to the shore changes when the speed is against the water and the speed of the current, one is the speed of the ship against the water plus the speed of the current, and the speed of the ship against the water is the difference between the speed of the ship against the water and the speed of the current. The boat catches up with the hat downstream at point A, which includes the distance of 30 minutes of drifting with the hat in the current and the distance it takes for the boat to turn around and catch up with the hat, and the distance for the hat to continue drifting.

When all movements are based on the shore as a reference, the column formula is more complicated, so a simple solution is suggested, when the water is stationary, the speed of the boat is the same.

There is something wrong with this topic.

According to this topic, after the straw hat enters the water, it should drift with the current, and the speed should be equal to the speed of the water, that is, it is already known that the straw hat has drifted for kilometers after falling into the water for 30 minutes, and the water velocity is directly calculated.

There's nothing wrong with that

This refers to the fact that the velocity of the boat is constant relative to the still water.

Hello! Which physics book is this? I've learned so far, and I don't understand it, thank you!

2. The reflected ray and the incident ray are separated on both sides of the normal The reflection angle is equal to the incident angle and the incident point In order to obtain the general law of the relationship between the reflection angle and the magnitude of the incident angle 3. The reflected rays are found in the same plane.

4 Definitely. The rest is right.

6. (3) The angle of incidence, the angle of reflection and the normal are in the same plane.

The first void in the seventh question is certain.

I can't help you with anything else, I'm sorry, I suddenly feel like I'm going to go back to make up for physics.