High school physics questions, I hope to explain clearly

Analysis A is zero when the external force is at rest on the inclined plane, A is acted on by five forces on the inclined plane, namely gravity, support force, spring elastic force, frictional force, tensile force f, when the direction of friction force is along the inclined plane upward, F mgsin37° ffm f spring, f 22 n, when the friction force is down the inclined plane, the minimum value of f is zero Option C is correct

Answer C

If the spring changes from 10cm to 14cm, the spring tension is 32N.

Because the spring length does not change after the force of f is added, it means that the object must still be at rest.

Then f+mgsin37f+25<20+32n

then f<27n

So choose c, if this question is single-choice, you can choose the largest one directly without arithmetic.

14-10) spring elasticity).

50*gravitational component).

So f should be between 2---22n.

cd error.

That's how it should be.

The force force up the vertical inclined plane has zero net force (otherwise a would fly up) and there are two forces that are fixed This brings convenience to the solution: the component of gravity down the inclined plane gsin37 = 30n The elastic force f along the inclined plane upward

kx=(these two forces add up = 2n (different directions).

Therefore, f minimum is 2n and maximum is 22n

As shown in the figure, according to the figure, it can be seen that the total distance of the first honking sound propagation = 340 8 = 2720m, the second time is 340 6 = 2040m, and then calculate ab = (2720 + 8v0) 2 = 1360 + 4v0, cd = (2040 + 6v0) 2 = 1020 + 3v0

The column equation is obtained: ab=cd+8v0+27v0

Substituting data: 1360 + 4 v0 = 1020 + 3 v0 + 8 v0 + 27 v0 solution v0 = 10m s

So the distance from the cliff for the first honk is ab=1400m, and the speed of the car is 10ms

I didn't find a pen and wrote it on a computer, I hope it's right.

Let the speed of the car v1=x when the car honks for the first time and the cliff is l8x+340*8=2l, that is, l=4x+136027s, and the distance traveled by the car is 27x

The second time the car honked its horn, it was s from the cliff

Same as above: 6x+340*6=2s, that is, s=3x+1020, so l-27x-s=8x, 4x+1360-27x-3x-1020=8x-26x+340=8x, 34x=340, x=10m s

The generation launched l=1400m So the car speed is 10m s The car honks for the first time and is 1400m away from the cliff

Solution: Let the cycling speed be V and the sound speed V', the distance is x when the horn is blown for the first time.

Then there is: v*t1+v'*t1=2x

v*t3+v'*t3) 2+v*t2=x-vt1 is combined, and the solution is v=10m s

x=1400m

High school freshmen, don't do this kind of question......

Let the speed of the car be v and the distance of the first whistle is s1, then there should be 2s1=8v+ 340*8, and the distance of the second whistle is s2, then there should be 2s2= 6v+340*6 s1=s2+(27+8)*v 3 equations and 3 unknowns can be solved. This topic should be from junior high school, right? This topic also does not require a drawing.

I don't have paper or pen, so I'll figure it out myself.

There are no pictures...

It is a matter of moment balancing (lever balancing).

The study plate, with a as the axis, must have t*l=n*l2 + (mg 2)*l 2 when it is about to rotate (where n is the pressure of the person on the midpoint of the plate; l is the length of the board, (useless, will be eliminated); t is the tension force on the rope, which is the force of the athlete acting on the rope, the force of the rope acting on the athlete, and the force of the rope lifting up the B end).

For people: n+t=mg (this n is the support force of the board to people, and the n above is the interaction force).

Two formulas of joint force: t*l=(mg-t)*l2 + (mg2)*l2 to eliminate l, and you can solve it.

t=mg/2）

1. For the analysis of the force of the CD rod, the gravity, ampere force, and support force are statically balanced, and the gravity is decomposed into the direction along the inclined plane and the vertical inclined plane, and the component force along the inclined plane direction is equal to the ampere force, and the component force in the vertical inclined direction is equal to the support force.

2. The constant velocity of the AB rod is under the action of the external force F, and the electromotive force is induced during the movement, and the current is in the loop, so that the CD rod can remain static under the action of ampere force.

3. The second question of the third question has been calculated, according to the force balance of the AB rod, the gravity is decomposed, and it is calculated according to the equilibrium conditions.

Still have any questions? **I don't understand it, you ask me again.

(1) Cd is at rest, balanced by force, it is subjected to gravity and the ampere force of AB to it (pointing to AB along the inclined plane). The gravitational force is decomposed along the inclined plane and the perpendicular inclined plane, and the force is balanced, so the component force of gravity is equal to the ampere force.

2) I don't understand what you're asking about The equilibrium state is a state of equilibrium by force, and both the state of rest and the constant velocity are equilibrium states, which can be seen as essentially the same. I think the difference is that one moves, one doesn't move. The solution is the same when analyzing the problem.

3) f is the f of the second question

The work done by the AB rod is converted into the ampere force work and the gravitational potential energy of AB, where the tensile force is equal to twice the MGSIN30 ampere force, there are two, one acts in the AB direction along the inclined plane, and the other acts on the CD direction, along the inclined plane, and the two ampere forces are the same.

It is precisely because the cd is at rest that the ampere force is equal to the component of gravity downward along the inclined plane

The second problem can be the conservation of momentum, or the equation can be equations, because the electric current is the same, the field strength is the same, the length is the same, the ampere force is the same, and when the momentum is conserved, it can be regarded as an internal force, and the momentum is zero, and the velocity does not change. You can also list the equation first f=mgsin30+bil, and the second bil=mgsin30That's the specific question!