Senior one physics questions, please talk about the solution method, thank you

Answer: ACA .......That's right.

Synchronous satellites, i.e. with the same period as the Earth's surface, so t1 = t3. From gmm r = m(2 t) r, t r = constant value. Near-Earth satellites have a smaller radius r than geostationary satellites, so the period t is also smaller.

So t1=t3>t2

b……Mistake. The mass of the satellite and the object is not told, so the centripetal force cannot be judged.

c……That's right. Only the gravitational force of the satellite provides the centripetal force, so a=gm r, and the geostationary satellite r is large, so a is small, so a1a3

I hope it helps you, if you have any questions, please ask o( o haha

1.Compare cycle t first

The equation is established by the periodic formula of the centripetal force and the formula of the gravitational force, there is.

gmm/r^2=m4π^2r/t^2

m is the mass of the celestial body and m is the mass of the planet.

Simplified: t=(4( 2)r 3 gm) (1 2)From the above equation, we know that the size of t is only related to r, and the larger r it is, the larger t is.

Therefore, t1>t2 (the above equation only applies to objects moving around ** celestial bodies, not objects on the ground).

There is t1 which is a geosynchronous satellite, so there is: t1=t3

A right. 2.Compare centripetal force.

For satellites: f=gmm r 2

So f1r3

Hence f3f1>f3

B false. 3.Compare accelerations.

For satellites: a=f m=gm r 2

The larger the r, the smaller the a.

There are A1C pairs. The centripetal acceleration of the ground object is a3=4 2r t 2, where r is the radius of the earth and t is the period of rotation of the earth.

A3 is known by comparison< a2d wrong.

Solution: The angular velocities of the two objects at the equator and Beijing moving in a uniform circular motion with the rotation of the earth are equal, both =2 t=2* 24*3600 rad s= rad s

The linear velocity of an object located at the equator in a uniform circular motion with the rotation of the earth v1= r=

The linear velocity of an object located in Beijing moving in a uniform circular motion with the rotation of the earth v2= rcos40°=

Children's Shoes: Hello!

This is a typical problem for us to learn about the kinetic energy theorem and the conservation of mechanical energy. Normally, students will find it difficult to see similar questions, but in fact, after you understand the relevant knowledge of the kinetic energy theorem and the law of conservation of mechanical energy, and master the method of solving the problem, then it is no longer considered a difficult problem.

Moreover, from the perspective of the college entrance examination, the kinetic energy theorem and the law of conservation of mechanical energy are very important and must be examined, and the score is very large; From the perspective of learning physics, energy knowledge is the most essential part of physics, and many phenomena in nature can be understood and explained by using energy-related theories, and will be helpful for you to learn chemistry and biology. Therefore, no matter what aspect it is, students should work this content. Now I'm going to give you a solution to this question:

First, let's analyze the physical process of this problem:

The ball enters the tube at a horizontal velocity v0, and now moves horizontally, then moves in a semicircular pipe in the vertical direction, and finally rushes out of the tube horizontally at a speed; This is the whole physical process of the ball in the east;

So is the nature of its movement in a horizontal tube the same as that in a vertical semicircle? This requires us to analyze the force and the motion state of the ball in the horizontal and vertical directions, and finally carry out the work analysis.

Because when the ball moves in the horizontal direction, the tube is smooth, so the horizontal direction is not forced, and the vertical direction supports the force and the gravity of the ball itself is balanced, so when the ball moves in the horizontal tube, the resultant force is zero, so it does uniform linear motion;

When the ball enters the semicircular tube, because the trajectory of the ball is a curve, it does a curvilinear motion, so the net force is not zero, so the ball moves in a curve in the semicircular pipe, and finally rushes out of the tube at a certain speed. So.

The process of solving the problem is as follows:

Performing the force analysis at c gives n+g=mv 2 r

by the law of conservation of kinetic energy.

So finally v0 = under the root number is the meaning of the number of times.

Since the pressure on the pipe wall at c shows that the centripetal force is provided by it, then, by using the kinetic energy theorem, the velocity at b can be obtained.

It's been three years since I graduated from university, and I really can't remember the physics formula.

This question is actually quite simple, I'm afraid your physics is really bad

1.3v 2as=v*v v=2as v1=2a9s =3v under root.

2.The first process s=1 2at*t is brought in, and the velocity after 5s is v=10m s and the distance traveled is s=25m

The second process s=10*120=1200m s=120 v=10m s, the third process 2as=v*v brings in data to get a=1 s=50 t=10s, then v=25+1200+50 (5+120+10)= I don't know if it's right, I hope it can help you.

The relative speed of the boat and the water remains the same, the boat leaves the water hyacinth for half an hour, and it also takes half an hour to chase back the hoist, for a total of 1 hour for the hoist to move in the water. (For example, if you leave your seat at a certain speed and return to your seat at the same speed in a train, it will take the same amount of time.) ）

Whereas, after the gourd falls in the water, it is relatively stationary with the water, that is, it moves downstream at the speed of water.

So the water velocity is 5400 3600=

Taking car B as the frame of reference, then the problem becomes that B does not move, and the initial velocity of A is 72km h - 10m s = 10m s

The time it takes for car A to stop (relative to B) is t: V - at = 0, t = V a = 10 A

A should stop at 100 meters (relative to b): vt - 1 2 * a * t 2 = 100

10 * 10 A - 1 2 * A * 10 A) 2 = 100, A = M S2 or km H2

Ideally, two cars should go at the same speed.

Answer: (v -v2) t 2 = 100

t=20sa=

1.The force decomposition of F, assuming that the force on Ab is 0 Newton, then the three forces of gravity, AC and AF of A in the figure are balanced, and you can calculate what AF is, if AF is larger than the value obtained, then the force in the vertical direction is bound to be greater than the weight of the object, then the object moves upward, and AB cannot be straightened, so the AF calculated this time is the upper limit of the force.

2.Assuming that the force on AC is 0 Newton, then, the gravity of AB, AF and A, the three forces are balanced, and the magnitude of F is calculated according to the angle relationship, if AF is smaller, then the component AC of AB in the horizontal direction must be greater than the component of AF in the horizontal direction, then AC cannot be straightened, so the value is the lower limit.

According to these two equations, the equation is fsin +t1sin mg=0 .1 fcosθ－t2－t1cosθ=0 ..2 from Eq. 2 gives f = (t2+t1cos ) cos =t2 cos +t1 .

3 From Eq. 1 we get t1 = (mg-fsin ) sin =mg sin -f .4. Bringing Eq. 4 into Eq. 3 gives f = t2 cos +mg sin -f, that is, 2f = t2 cos +mg sin f = t2 2cos +mg 2sin, and merging to obtain f = (t2sin +mgcos ) 2sin cos = t2 2cos +mg 2sin

1.Erasers are fast because the paper is subject to high air resistance during the fall. If pinched, the ball will fall faster because there is less air resistance.

2. h = 1/2 * g * t^2

In free fall, the displacement is equal to g(t 2) divided by 2, so 2 = if there is air resistance, the falling speed is not that fast, so the actual height is less than the calculated value.

Answer: 1. Release a piece of paper and a piece of stationery eraser at the same time. Stationery erasers will fall quickly; Then knead the paper into a very small and tight paper ball and release the eraser at the same time, and the paper will fall faster than the last time.

Because the object will be affected by air resistance when falling (moving) in the atmosphere, the lighter the object, the larger the area of the moving surface, the greater the impact, so the first paper piece falls slowly, when the paper is pinched into a very small and tight paper ball, it is affected by air resistance and becomes smaller, so it is faster than the first time.