Speed! Solve a high school physics problem

The gravitational force of an object on the ground is 160N, the gravitational constant is 10N kg, so the mass is 16kg, when the pressure is 90N, the acceleration is obtained by dividing 90 by 16, and the satellite is accelerating a=5N kg

As the rocket accelerates vertically, the acceleration due to gravity is a.

Because a = (gm) (r squared) gm = the square of the earth's radius multiplied by the acceleration of the earth's surface, the earth's radius is 6,400,000 meters, the acceleration of the earth's surface is 10n kg, and the substituted r=25,600,000 meters minus the earth's radius.

25600000-6400000=19200000=1920000 km.

The object is subjected to gravity on the ground by 160n, and the mass of the object is 16kg, which can be obtained from Newton's second law'=ma brings in data a=g 2 to get g'=(1 16)g, and the law of gravitation gives gmm r2=g 2=gmm r2, and the result is that the height of the satellite above the earth is 2 with the radius of the earth.

One, the cosine theorem, and two, the graphing method, as you gave the graph, has a scale, but this is not accurate.

Third, extend FF, pass the O point to its extension line to make a perpendicular line, intersect at the K point, then the angle of KF2O is 30 degrees, OK=sin30°of2, F2K=COS30°OF2, in the right triangle OKF, Pythagorean theorem, OK square + FK square = OF square (FK=FF2 + F2K), so that you can find OF.

Derived from the cosine theorem.

f=(90^2+120^2-2*90*120*cos150)^1/2=203n

When the angle is 150 degrees, replace cos150 with cos 30 in the above equation

When the ceiling fan is stationary, the boom tension is equal to the ceiling fan gravity, f=g

When the ceiling fan rotates, because the wind has an axial force f2 on the fan blades upward, f1+f2=g, the boom pull force f1=g-f2 g when the fan rotates

So f1 f=g

Classmates: 1The first case is static equilibrium. In the vertical direction, there is only gravity and the pull of the suspension rod on the fan.

2.The second case: the fan is turned on, and you feel the wind below. If the fan exerts a force that causes the air to move downwards, then the air has a reaction force on the fan (in the direction of the fan).

3.There are 3 forces in the vertical direction:

Downward gravity g

Upward pull f'

Upward air reaction force f''

f'' + f '= g so f' < g

I think the ceiling fan is buoyant upwards when it turns, and the buoyancy cancels out part of the gravity, so the pull of the rope is smaller.

There's something that looks like a bamboo dragonfly, and I forgot, isn't it just a way to fly. I think so, I don't know if it's right.

When at rest, the gravitational force is equal to the tensile force f, because the two forces are balanced, after turning, the wind is blowing downward, that is, the fan exerts a downward force on the air, according to Newton's third law of action, the fan will be subjected to the upward force of the air on the fan, and there is an upward force plus the pulling force is equal to the gravity of the fan (two forces balance), so the gravity is greater than the tensile force at this time.

It's f'>f=g!

When the ceiling fan is stationary, the force balance of the ceiling fan is f=g

When the ceiling fan rotates, the component of the pulling force in the vertical direction is equal to the gravitational force, then the pulling force is of course greater than the gravitational force. f‘>g

The component of the tensile force in the horizontal direction provides the centripetal force of rotation.

So f'>f=g

f=g。。Needless to say, put this? (At rest, there are only two forces in the vertical direction, g and f.) g=f

An electric fan is similar to a helicopter, and the rotation of the fan blades creates an upward lift. Because the fan is blocked by a roof, it can't fly. The sum of the upward lift and the pull of the suspension is equal to the gravitational force (which is the case after the start of the rotation) and balances it. Lift force + f' = g g is greater than f'

f=g, you get the idea, right?

When you stand under the fan, do you feel the wind? So why f'

g = f because the two forces balance when the ceiling fan rotates creates buoyancy and makes the tension smaller.

To give you a detailed answer, see illustration.

Personally, I think that the answer to this question can certainly be done like "Qingqing Pinhua Man", where the work done by thrust is converted into heat energy and gravitational potential energy.

So w=mgh+ mgl+(mgcos)*h sin. But there is a simpler way to express it: the small slider is released from rest at point A, slides down the track, and finally stops at point C, indicating that in this process, the work done by gravitational potential energy is equal to the work done by friction, that is, W resistance = mgh, then in turn the work done by friction from point C to point A should also be equal to mgh, and because the gravitational potential energy from point C to point A increases mgh, therefore, the work done by reasoning w = 2mgh.

If it's a multiple-choice question or don't miss it, I hope it will help you!

You have to notice that it is pushed slowly, so the kinetic energy of the object is negligible.

The work done by thrust is converted into heat energy and gravitational potential energy.

So w=mgh+ mgl+(mgcos)*h sin.

1) The acceleration of the horizontal object at the beginning of the object a1 = 0 * g = 1m s 2a1 * t1 = 4 is solved to obtain t1 = 4s, s1 = , that is, after 4 s, the object is just open and the speed of the horizontal conveyor belt is the same.

On the inclined conveyor belt, there is the following relationship:

a2=g*sin37°- 1*g*cos37°=4m s 2s2=4*t2+ solution t2=2s

Therefore t=t1+t2=4s+2s=6s

2) If 2=, then g*sin37°< 2*g*cos37

Therefore, the object still moves at a speed delay of 4 m s on the inclined plane.

The time of movement on the inclined plane is t3=16 4=4s, so t=t1+t3=4s+4s=8s

Good luck!

1) Acceleration at the level: a = mg m= g = 1 acceleration up to 4m s: t1 = 4s s=at*t 2 = 8m on the inclined plane, the force is divided into clusters and Sun analysis can be obtained

n=mgcos37 a'=(mgsin37-μ1mgcos37)/m=4m/s2

2) 2= then 2mgcos37>mgsin37, so there is no relative slip, t'=16/4=4s

The total time is 8s

1. Select B

160km/h=400/9 m/s 140km/h=350/9 m/s

The two moving bodies are references to each other, and the velocities are added together, so the length of the second train is:

400/9+350/9)× m

2. There is one place in this question that is easy to make mistakes. That is: A is in the first (posterior) half of the [displacement]...

And B is in the first (second) half of [time]...If you mistakenly think that both are displacements, or that both are time, you will be mistaken.

Let the displacement of the whole section be 2s

Then the total time for A to walk the entire section of displacement is expressed as:

S 40+s 60 pass score 5s 120

Then, the average velocity of the entire displacement of segment A is:

v A = 2s (5s 120) = 48 km h, and the time taken for B to travel the whole section of displacement is 2t

then the entire displacement can be expressed as.

t×40+t×60=100t

Then according to the average velocity = total displacement total time gets:

v B = 100t 2t = 50 km h

So V-A