Doubts about physics, questions about physics

Let me tell you if there is a way to determine the relative movement trend.

Hypothetical method. Assuming that the contact surface is smooth, if the object will be in relative motion, there will be a tendency to relative motion, and this relative motion direction is the direction of the relative motion tendency when there is friction. The direction of the static friction force is opposite to this direction.

If smoothness can also remain relatively stationary, there is no tendency to relative motion. There is no static friction.

Let the slope angle be x degrees, and the sliding force (the component of gravity) is sinxg (formula), g is the gravitational force of the object, which is unchanged, and when the surface is a horizontal plane, sinx=0, the sliding force is 0, so there can be no friction. On a sloping tabletop, there must be friction as long as the slope angle is not 0 and the friction coefficient between the table and the object is not 0 (no substances with a friction coefficient of 0 have been found so far).

Friction is the opposite of the applied force, and on a horizontal table, gravity and support counteract it, and it will be stationary even if there is no friction, while an inclined surface is not.

Please read the concept of friction carefully.

Note that friction only exists when the object is in motion or has a tendency to move (and of course, in a rough plane).

The first one, you're talking about being at rest on a horizontal plane, so that the object is only supported by gravity and the horizontal plane, so that the forces are balanced.

This is an interpolation, and the force balance will appear in a state of stationary or uniform linear motion, so many problems need this implicit condition.

The second one is on an inclined plane, okay, now let's analyze the force on this stationary object on an inclined plane.

The first is gravity.

Then there's the support.

But the direction of the support force and the gravity force are not in the same direction, how to balance this?

At this time, I have to think that there should be another force --- friction.

Direction to **? Of course, it is the opposite of the tendency of the object to move.

What about the size? I don't know if I've learned the Cartesian coordinate system method, but according to the Cartesian equation, you can calculate it.

The latter has a tendency to move if it is not a frictional force, the latter is subject to a sliding force, and the former is a balanced force of gravity and support.

When the table tilts, gravity breaks down into a vertical downward force and a forward force, and because it is stationary, the force in front of it is balanced by friction.

cubic decimeters = cubic meters = 10,000 cubic centimeters.

m total = v = cubic meters * kilograms per cubic meter =

m = 25-5 = 20g

m leftover v = 20g 10,000 cubic centimeters = grams per cubic centimeter = 2 kilograms per cubic meter to give points.

CM total = v = 10 cubic decimeters * kg per cubic meter = 25g

m = 20g

m left v=20g 10 cubic decimeters = 2 kg per cubic meter.

This is the displacement time image, and you can see clearly that the slope represents the velocity, and the slope is constant, which is the constant velocity, and the car is moving at a uniform speed in the second.

And v=Forget it, I can't see your diagram which is b???

The diagram that you draw is a displacement time image, and according to what you say, b is stationary, and the drawing is wrong, and it has a big impact.

V-T Diagram. at v=. The displacement of a is less than the displacement of b. (known by calculation).

So t >.

So there is.

The proportional relation is for a uniform linear motion with an initial velocity of 0, and t denotes the equal time interval.

The end of t, the end of 2t, the end of 3t ......The instantaneous velocity ratio is v1:v2:v3......vn=1：

2：3……n derivation: v1=at v2=a 2t .

VN=a NT within t, 2 t, 3 t ......The ratio of displacements.

x1：x2：x3……xn=1²：

2²：3²……n derivation: x1=at 2 x2=a(2t) 2 x3=a(3t) 2....

xn=a(nt)²/2

3. The first t, the second t, and the third t ......The ratio of displacements is x1:x2:x3....xn=l：3：5……(2n-1) derives: x1=at 2

x2=a(2t)²/2-x1=3at²/2x3=a(3t)²/2-x1-x2=5at²/2...

xn=a(nt)²/2-x(n-1)-.x2-x1=(2n-1)at²/2

(1) v=at, the first s is v=a, and so on, v=2a, v=3a, and the 1:2:3:n(2) of these comparisons is given to t(i) as the time taken for the ith identical displacement.

s=at1^2/2

2s=at2^2/2

3s=at3^2/2

t(1)=t1

t(2)=t2-t1=(√2-1)t1

t(3)=t3-t2=(√3-√2)t1

t(1):t(2):t(3):-=1:(√2-1):(3-√2):-

First of all, we must know what kind of movement it is, whether it is a uniform velocity, a uniform acceleration ,...

When just throwing out the object of 200N, the force of the balloon changes, the direction of buoyancy and air resistance is the same, greater than gravity, so the balloon will change from uniform speed to deceleration, until it is 0, at this time, the resistance of air no longer exists, at this time, because the buoyancy is still greater than gravity, so the balloon accelerates upward, in the process of movement, the air resistance appears again, is 100N, then the buoyancy is equal to the sum of the gravity of the balloon and the air resistance, the force received by the balloon is equal, and the direction is opposite, The balloon will then move at a constant speed. What should be considered here is the change in air resistance, when the balloon speed is 0, of course, there is no air resistance.

In this way, when the balloon descends at a uniform speed, it is subjected to three forces: downward gravity, upward buoyancy, and upward resistance.

According to what is known, it can be calculated that buoyancy = gravity - drag force = 1800-100 = 1700n When a part of the object of 200n is thrown, the total gravity of the object g = 1800n - 200n = 1600n buoyancy 1700n

The balloon must float upward, and at this point, the air resistance becomes downward again, because the net force of resistance is always opposite to the direction of motion = gravity - buoyancy + drag = 1600 - 1700 + 100 = 0

So, according to Newton's laws, the balloon rises at a constant speed.

This is a force analysis of two processes, the air resistance is in different directions, and the air resistance is always in the opposite direction of the motion, and both times it is in equilibrium, so on the vertical plane, the forces in the same direction are put together and then equal.

1. The maximum static friction is generally greater than the sliding friction.

2. In the object on the earth, the source of gravity is the gravitational force given to it by the planet, and the direction of gravitational force is directed to the center. In addition, if the objects on the planet move with the rotation of the planet, then they will also be affected by a centripetal force, the centripetal force is small relative to the gravitational force, and the direction refers to sliding towards the earth's axis (only at the equator the direction of the centripetal force points to the center of the earth), and the vector sum of the centripetal force and gravity is the gravitational force. Therefore, the gravity at the equatorial front is directed towards the center of the earth, and the gravity at the non-equatorial is not directed towards the center of the earth, but the deviation is so small that it is almost negligible.

1.Not equal.

2.No. Always pointing straight down is for objects near the ground. After studying the gravitational force, it was found that gravity is a component ...... of the gravitational force

For parallel excitation socks. k=(k1x1+k2x2+k3x3+……knxn)/n(x1+x2+x3+……xn) is in series.

4.Let the mass of the weight in the solid lead rock test be m, and the mass of the trolley is m, in fact, the gravity of the weight is used to replace the fixed force f in the actual movement

But weights have mass, so mg=f=(m+m)aOnly when m is much smaller than m, can it be considered that m+m approximate =m, so that the result will be consistent with f=ma

3. Series: (K1*K2) (K1+K2) Parallel: K1+K2 The relationship between K1+K2 and resistance is a bit like a return to knowledge, Qingxiao but just reversed.

4. The quality of the car is m, and the weight quality is m.

Pair car: f=ma

Pair weights: mg-f=ma

Eliminate f:a=g (1+(m m)).

So f=mg (1+(m m)).

In the experiment, it is believed that f = mg, so f is approximate mg only if m is much smaller than m

1.The former makes the town slightly larger than the latter.

2.No. Because the earth is an ellipsoid.

3.Contrary to the conclusion of the electric year cherry resistance.

The smaller it is, the more accurate the result!

I can't see your diagram, but if it's a multiple-choice question, the CD is definitely wrong, choose ab

The force on the rope is upward...

Choose b rope subject to gravity and buoyancy.