High school physics, master enter!! Urgent! High school physics questions, master progress

Kilometers = 108 1000 meters (3600 seconds) = 30 meters and seconds.

After 10 seconds, when accelerating from a standstill to 108 km, a = (30-0) 10 = 3 m per quadratic second.

Resultant external force = ma = 3000n

n = 1kg 1m s 2 = 1000g 100cm s 2 = 10 5g * cm s 2 = 10 5 dynes.

The force that causes an object with a mass of 1 kilogram to accelerate 1 meter without a quadratic second is 1 Newton, which is the definition of Newton.

3. Before the stone below hits the ground, the two stones are in free fall, and the distance between them is meters, with or without a rope, so the rope is of no use here.

The stone below fell from a height of meters, time = root number (2h g) = 1s

The stone above falls from the meter, time =

Phase difference seconds.

1) V=108km H=30Ms, then A=30 10=3Ms 2F=Ma=3000N

2) 1 Newton = 100 * 1 cm * 1000 * 1 gram 1 second 2 = 10 5 dynes (f = ma).

3) The distance between the stones below and the ground h=

When landing, the stone velocity is v

v^2=2gh,v=10m/s

Let the time difference be t

Then, the solution is: t=

v=108km/h=30m/s，m=

The resultant external force f=ma=

1 ox = 1kg * 1m s 2 = 1 * 10 3g * 100cm s = 1 * 10 5g * cm s = 1 * 10 5 dynes.

by h=(1 2)gt2

Get t=(2h g) 1 2

The time difference is obtained.

t=(2*.)

Solution: For this problem, first analyze the force on ball B, which is subjected to gravity (if it is said that the ball is generally considered gravity, and my later analysis can also explain why gravity is considered), the electric field force, and the Coulomb force between the two balls. Because the electric field force is constant, and the Coulomb force is directed at the ball A along the direction of the rope, and the direction is constantly changing when the ball B rotates, so if there is no gravity, then the ball B cannot move in a rapid circular motion, so there is gravity and the gravity and the electric field force are balanced.

In addition, the magnitude of the electric field is not given in the problem, so the electric field force should not be needed when solving the problem, that is, it must have the force to balance the electric field force, which is gravity. For ball b is solved by Newton's second law: k*(q 2 l 2) = m*(v 2 l), and the solution is:

v=q*root number[k(ml)].

The B-ball is negatively charged, so the electric field force generated by the uniform electric field experienced by the B-ball is vertically upward. And the title says that it is an operation of circular motion, so the magnitude of the centripetal force is constant. Therefore, the magnitude of the electric field force generated by the electric field received by ball B is equal to the gravitational force of ball B, so the centripetal force is equal to the electric field force generated by ball A to ball B.

fn=fm*v-squared l=k*q-squared l-squared.

m = root number (k*q square m*l).

1. Judge and build a physical model first

Before pulling, the sheet is subjected to two forces, the pressure of the small object, and the support force of the tabletop, both of which are elastic forces

When the wooden board is pulled, the block has a displacement, which means that there is sliding friction, elasticity, and sliding displacement between the sheet and the table top and the block, which are the two conditions for sliding friction, and then calculate: the friction between the block and the wave plate = the gravity of the block * = mg The friction between the sheet and the table = (Gravity of the block + the gravity of the sheet) * = 2mg The friction force on the sheet = The friction between the block and the wave plate + The friction between the sheet and the table = 3mg If you look at the problem solving process carefully, although it is a simple physics problem, But the problem-solving process I wrote is very useful, and you have to build a physics model before doing any physics problem, which is very helpful for you to learn physics in the future.

If you don't understand anything, you can ask me again.

Sheet positive pressure on desktop 2 (sheet gravity block gravity), friction force 2 is generated between sheet and tabletop; The positive pressure of the block on the sheet 1, and the friction between the plate and the block 1. A total of 3. You must be careful when doing this kind of question, and remember to analyze any force...

My experience: gravity produces pressure, and pressure produces friction, so the gravity is analyzed first, then the pressure, and finally the friction is analyzed. 〗

That's right, that's right, it's 3 mg, 1 above, 2 below

Solution: Let the distance passed by the wooden block be s, the mass of the wooden block is m, the dynamic friction factor between the wooden block and the ground is u, and the acceleration of the motion of the wooden block is a

According to the inscription, the sliding friction between the wooden block and the horizontal ground is 2n, so umg=2n....1)

a*2s=6m/s-2m/s=4m/s

a=2m/s^2

It is derived from Newton's law of motion 4n-2n=ma.

m=1kg, take g=10m, s2, u=

s=2s*(6m/s+2m/s)/2=8m

a=(6-2) 2=2 s=2 times 2 squared = 4 meters from Newton's second theorem: m=4 2=2kg friction factor = 2 2=1

When using the horizontal tension of 2N, the friction force is 2 N when the horizontal tension force is used to do a uniform linear remote motion on the horizontal ground, the combined external force is 2 N and the acceleration is 2, the mass is 1kg, the dynamic friction factor is 9, and the distance is 9

What is called starting from a standstill, the velocity increases from 2m s to 6m s

Previous question: The velocity of the car is orthogonally decomposed in the direction of the rope and the direction of the perpendicular rope, and the velocity component in the direction of the rope is equal to the velocity of the weight. i.e. v*sin v

Latter question: Since there is no picture, it is speculated that the two objects are connected by a rope and a fixed pulley. Since they are both moving at a uniform velocity, the net force is 0.

For the above object a, there is f t fab mo, where t is the rope pull force on the lower object b, there is t fab mo fb ground mo.

i.e., f t * mg and t * mg *mg mg) to obtain f 4 * * mg

Let the acceleration of the object be a, the velocity to point A is v0, and the time taken to pass through the AB and BC segments is t, then we have:

l1=v0t+at2/2 ……l1+l2=2v0t+2at2………Together:

l2－l1=at2………3l1－l2=2v0t………Let the distance between o and a be l, then there is:

l=v02/2a ……Together:

l=(3l1－l2)2/8(l2－l1)

From the problem, the velocity of b is (l1+l2) 2t, the acceleration is (l2-l1) t 2, the velocity at the intermediate moment between a and b is l1 t, which is obtained by va 2=2ax, xoa = (3l1-l2) 2 8(l2-l1).

(1) The analysis of m1 and spring 1 allows the length of spring 1 to be x1 and the length of spring 2 to be x2, m1+m2)*g=k1*(x1-l1);

m2*g=k2*(x2-l2);

x1+x2 is the total length of the two springs.

2) Spring 1 returns to its original length, there is no deformation, so spring 1 has no force effect on M1, and the number of words is limited.

First of all, it is important to know that the light rope in high school physics is weightless, inelastic, and stresses along the direction of the rope.

The most easily understood balance of forces is this.

The transverse components of the tension of OA and OB cancel each other out.

The longitudinal component of OA and OB tension is added and gravity balanced.

If you understand this, you will understand that as point A gradually moves upwards, the longitudinal component provided by OA will become larger and larger.

Since the longitudinal component provided by OA becomes larger, the longitudinal separation provided by OB will be smaller, while the angle of OB will remain unchanged, so the transverse component of OB will become smaller. As a result, the transverse component of OA, which cancels out the OB transverse component, becomes smaller.

As a result, the longitudinal component of OA is getting larger and larger, and the transverse component is getting smaller and smaller. It will inevitably go through a process of first increasing and then becoming smaller, or first getting smaller and then increasing.

When the OA is parallel to the ground, the gravity of the lamp can only be balanced by the upward pull of the rope because the OA cannot provide an upward pull force. So the tension on OA is mg*tan

From experience, it is clear that it first becomes smaller and then larger.

If you have time, try to make a system of equations along the lines of force equilibrium.

Suppose the angle between OA and the ground is ob and the angle between the ground is (where is a constant).

Solve the relationship between the magnitude of the total force on OA and the function of , and the answer will be clear at a glance.

Solution: (1) a=4 2r t2=

2) Solution (1) When the object is at the equator, the gravitational force decomposes the centripetal force of the object with the rotation of the earth and the gravitational force of the object, and when the gravitational force is equal to the gravitational force of the object, so if someone weighs 700n at the north pole, then his effective weight at the equator g1=g-ma=700-(700

3) If the two stages are slightly flattened, then r is larger, and the believed acceleration a at the equator is larger (2) the answer g1 is smaller.

1 Calculate how many meters the equator has turned in the last hour, and divide the square of this value by the radius of the earth.

(1) The centripetal acceleration of an object at the equator a=r =r(2 t) = (2) He is at the equator.

700=g+ma

Effective weight g =

3) In fact, the earth is slightly flattened by the poles and the two stages are slightly flattened, then r is larger and the believed acceleration a at the equator is larger (2) The answer g1 is smaller.