Buffer physics questions in the sophomore year, physics questions in the sophomore year

When the seat belt is tightened, the speed of the person v=(2gl) 1 2=10m s From the topic, the action time of safety and people should not be tightened to the lowest point, but should be pulled to the position of the original length of the seat belt, because when the person is pulled up from the lowest point, the seat belt still has force on the person.

In other words, the time taken from the time the seat belt is tightened to the time when the person reaches the lowest point is according to the momentum theorem, if the orientation is in the positive direction, then 0-mv=(mg-f)t Note that gravity mg cannot be missing in the above equation, because f in the momentum theorem is the resultant external force.

Substituting the data yields f=1600n

s=1 2 gt; 2=> t=(2s g) 1 2v=gt=g*(2s g) 1 2=(2gs) 1 2ft=changemv= f=[m*(2gs) 1 2] t=500(n)g, take 10n kg

That is to say, in this second, a 500n resultant force in the opposite direction of gravity is needed to act to achieve the effect of momentum changing to 0.

500+mg=600n

This is the average punch of the seat belt.

In the time of the second of action, the gravitational force does not change the velocity of the object because the height does not change.

ft=mv v is 10m s (no doubt about this) m=60kg t=bring it in.

From the inscription, it is known that the velocity in the horizontal direction is v1 = root number (2eu m) and the time when the electron passes through the plate t = l v1

The acceleration in the vertical direction is a=ee m

then the vertical velocity v2 = at

According to the Pythagorean theorem, the magnitude of the velocity of the electron when it has just left the deflected electric field, v = root number (square of v1 + square of v2).

Direction tan@=v2 v1

Dude, this question is wrong, you correct it and I'll answer it for you. The width and incidence position were not said.

Hehe, if you give me a pen and paper, you should be able to write it, but at the moment it seems that some formulas have been forgotten, just go to the reference book to find similar examples, I can't meet that's it, it works.

This question is in the high school physics book!! Read more books!

This question is very ordinary, comrades work hard!

In 4s, the change in velocity is -2m s-10m s=-12m s, and this quantity is the so-called v, that is, the change in velocity, which is a vector quantity. So the acceleration a = v t = -3m s 2 (note that the minus sign just indicates the direction).

Therefore, the force f=ma=-30n, and the magnitude is 30n.

This is done by the law of conservation of momentum: the impulse of force f in time t = the amount of change in the momentum of the object during that time.

ft = mv2 - mv1 = kg·m s, so the impulse experienced in 35s is kg·m s

i.e.: f = momentum t = 35 = 10 6 n

1.Solve ft=mv2-mv1 with the momentum theorem t=4s v1=-2 v2=10 solution f is 30

2.Solving it with the impulse theorem is the same as the previous problem. The impulse is ft=mv2-mv1 to the seventh power, and the average resultant value is 10 to the sixth power.

1 Ask the teacher, 2 ask the classmates, 3 look at the answers, 4 read them a few times, but which formulas, those theories...

I don't have the elite school alliance with me, so I'd better pass it on.

1 asks the teacher, 2 asks the classmates, and 3 looks at the answers.

This question is relatively simple, you can change the circuit diagram. When disconnected, it is equivalent to R3 and R4 in series, R1 and R2 in series, and then in parallel with R5.

When closed, R1 and R3 are connected in parallel, R2R4 is connected in parallel, and then strung together, and R5 is connected in parallel.

If you don't give the question, how can I answer you?

When using ohmics, we try to keep the lower hand in the middle of the dial, which is more accurate, if the gear is large, the hand will be on the right side of the dial, which is very different from the scale in the middle, affecting accuracy.

Because you can calculate r=ohm directly with p=u 2 r, you can use ohm 1 stop.

When the conductor rod is at rest, the ampere force is equal to the gravitational force.

e=δbs, δt, i=e, r, mg= bil, and the sum of the three formulas can bring in the data.

The area you gave may be wrong, is it possible to have a coil area of 10,000 square meters? So I didn't count it yourself.

b/δt= mgr/bls