There is a physics problem, a high school one, a high school physics problem

Sine theorem: a sina = b sinb=c sinc--> a:b:c=sina:sinb:sinc=2:3:4, let a = 2k, b = 3k, c = 4kRule.

cosc=(a2+b 2-c 2) (2ab)(2 2+3 2-4 2)k 2 (2*2k*3k)c is an obtuse angle. Obtuse triangles.

Obtuse triangles.

Sine theorem. a sina=b sinb=c sinc=2r, so the original formula is a 2r:b 2r:c 2r=2:3:4a:b:c=2:3:4

a +b so for. Obtuse triangles.

a/sina=b/sinb=c/sinc=a/b/c=2/3/4

Therefore cosa = (3 2 + 4 2-2 2) 2*3*4

I calculated b c according to this method

The speed of the opposite train is U, and the relative speed of the two trains is V

Then v=u+10

This 10 is 10m s = 36 h

180/12=v

So v=15

So u=5m s

18km/h

v=s/t=180/12=15m/s=54km/hv-v=54-36=18km/h

So it's 18 kilometers per hour.

I don't know if it's right or not.

The combined speed is 180 12 = 15 meters per second, and the speed of the train is 36 km h=10 m s, so the speed of that train = 15-10 = 5 m s

As the title suggests, 36 h = 10 m s

Let the speed of that train be x, then.

10 + x) *12 = 180

The solution yields x = 5 m s = 18 h

(1) People can run downward to give the upward friction force of the plank, and the magnitude of this friction force is equal to the component force of the gravity of the plank along the inclined plane.

i.e.: f=mgsin. According to Newton's third law, the plank has a frictional force of equal size but downward towards a person, so the net force of the external force on the person is:

mgsinθ+mgsinθ.Hence the magnitude of human acceleration is: (mgsin + mgsin) m

2) To make the person immobile relative to the ground, the plank should have an upward friction force for the person, and the magnitude is equal to mgsinAccording to Newton's third law, man has a downward friction force against the plank, so he runs upward.

The study of the plank, the resultant force of the external force is: mgsin + mgsinSo the magnitude of acceleration is: (mgsin + mgsin) m

Divide by the mass of that object to see which object is the object of study.

The answers to the two questions are reversed and should be reversed. Because in the first question, the plank is not moving, and the person is accelerating downward, the equation is listed for the analysis of the person, and it should be divided by the mass m of the person; In the second question, although the person runs up relative to the plank, but the person does not move relative to the inclined plane, the plank is accelerating downward, so the equation is obtained from the analysis of the plank, which should be divided by the mass m of the plank.

(1) The weight and the spring dynamometer are regarded as a whole, and the system is acted on by two forces, the external force f and the gravitational force (m0+m)g

According to Newton's second law, f and ma.

f－(m0+m)g=(m0+m)a

The acceleration of the system is a=f (m0+m) -g(2) for a heavy object, acted by two forces, the vertical upward elastic force kx and the vertical downward gravitational force mg

According to Newton's second law, f and ma.

kx－mg=ma

We can get k=m [(m0+m)k]f, answer c

Think of the block spring as a whole.

The acceleration is: a=f (m0+m).

Reanalyze the block.

F spring - mg = ma

So F spring = mg + mf (m0 + m).

So spring elongation = f spring k

So the answer is.

mg+mf/(m0+m)】/k.

Select F-(M+M0)G=(M+M0)A of Newton's second law to solve a=f m+m0-g, and then analyze the force on the heavy object, by gravity and spring tension, using Newton's second law, there is t-mg=ma

The solution is t=m(g+a)=m(g+f m+m0-g)=mf m+m0

Consider the block spring as a whole, the acceleration is: a={f-(m0+m)g} (m0+m), and then analyze the spring, m0a=k l, the solution seems to be unanswered.

Hey, why did I calculate that fm [(m0+m)k] for identification.

The tensile force exerted on the weight f=ma+mg a=[f (m0+m)g] divided by (m0+m).

b I remember that the indication of the spring dynamometer is only related to the tensile force on the hook, and the hook is subjected to the upward tension f, so the spring elongation is f k

The maximum friction force above b is greater than the maximum friction force below, so ab must be relatively stationary when moving at a uniform speed.

Taking ab as a whole, the force of the left dynamic friction force of 3 N above C can be obtained, and the force on the pull line of ab is 3 N, and the force on the same rope is equal, so C is pulled by the rope to the left by 3 N.

The pressure below the investigation c is 60 New, so the friction is 6 New, and the direction of movement is opposite, left.

Add it up to 3+3+6=12

At the same time, A and B move. The friction between A and B is 5N, and the friction between A, B and C is 3N, and 3N is less than 5N, so it may be true to keep A and B relatively stationary. And because A, B, and C move in opposite directions, the force can be regarded as a pair of interaction forces along the direction of the rope.

The magnitude of the interaction force is the same and the direction is opposite.

On the same rope, the required tensile force is 6n+3n+3n=12n, so d is chosen.

The friction of the selected D c object by the ground is 6N, and if C is pulled out at a constant speed, it must be subjected to the friction force of 3N given by A and B.

The next thing is the key.

At this time, B must be subjected to the 3N friction force of C to the left, and B must be subjected to the 3N friction of A to the right if it is stationary relative to A.

b gives a pull to the left, and the rope gives a pull force to the right 3n then.

If you want to pull C, add the pull force of the rope 3n A total of 6 + 3 + 3 n = 12n I finished answering the question, I don't know if I am satisfied, my home is in Anshan, Liaoning Province is also a high school year, It just so happens that I am also studying in a middle school (Anshan's) Let's make friends, my common screen name is Sky Pavilion.

When the spring is stable, that is, the combined force of the thrust and spring force provides acceleration for block A, and the spring force provides acceleration for block B.

Because the spring is stable, it means that the speed of the two wooden blocks is the same at all times, so the acceleration of the two wooden blocks is the same.

So the acceleration of a: f m1-t m1 = t m2, here we can express t (spring elastic force) to:

t=（fm2）/（m1+m2）

Therefore, the moment the external force is removed, the spring has time to change in the future, and the acceleration of a is t m1b, and the acceleration of t m2 is still t m2

Just substitute the t that comes out again.

First, find the elastic force f of the spring in the original case, and analyze b to obtain f bomb m2*a

Later, when the external force f is just removed, the spring elastic force remains unchanged, and the acceleration of a is a1 f bomb m1 m2*a m1

b At this moment, the force is unchanged, and the acceleration a2 a

B is only affected by the spring force, so starting with the state of motion B, the force on the spring is A*M2.

Then the external force is gone, and the two objects are only subjected to the spring force, so a1=(a*m2) m1 a2=a(that is, (a*m2) m2).

Concise ha.

The acceleration is the slope of the V-t image, i.e., tan60 and tan30, respectively; The ratio of sizes is 3

For V-t images, acceleration is the slope of a straight line.

The acceleration ratio is tan60°:tan30°=3:1

Acceleration is the slope of a straight line v-t.

tan60:tan30=3:1

Time takes 1 velocity 1 = root number 3

Velocity 2 = 3 of the root number of 3 points