Physics in the first and second years of high school is easy to test, experimental questions and cal

1) Wire frame rotation cycle: t=360t 120=3t rotation angular velocity is: =2 t=2 3t

The speed at which the magnetic inductance lines are cut is:

v=ωlcos60°=πl/3t

Therefore, the instantaneous value of induced electromotive force is: e=blv=2bl l 3t=2 bl 2 vertical number 3t

2) Mean induced electromotive force: d dt=l 2(1-cos60°) t=l 2 2t

The RMS current value is i=b 2 l 2 v Fenghu 2 (r+r), so the silver fiber blocking rate of electric work = i 2 r=b 4 l 6 r 2 18t 2 (r+r) 2

Solution: v1 = 90 km h = 25 m s

t1=2min=120s

s1=v1t1=25×120=3000m

t2=1min=60s

s2=2km=2000m

Total t = T1 + T2 = 120 + 60 = 180s

S total = S1 + S2 = 3000 + 2000 = 5000mV flat = S total T total = 5000 180

A: The average speed is.

I hope it helps you, and if you have any questions, you can ask them

I wish you progress in your studies and go to the next level! (*

The average speed is equal to the total distance traveled divided by the total time. The total distance is equal to 90 times two-tenths plus two kilometers equals 5 kilometers The answer is equal to 5 divided by three-tenths of sixties So it's 100 km h

Solution: The average speed is the total distance traveled divided by the time. s=v1*t1+2=90/30+2=5km v=s/t=5/(3/60)=100km/h

Which brain-dead person made up this question?

The internal resistance of the voltmeter is 750, and even the voltmeter with internal resistance can be made, which is really crazy and incredible......

Well, no way, the question still has to be done:

At the highest point, t+mg+qe=mv1 2 l, and the condition for passing through the highest point is: the pulling force of the rope on the ball t=0.

From the highest point to the lowest point, column kinetic energy theorem, qel+mgl=mv2 2 2-mv1 2 2

Nadir point, against the ball row Ox II, t1-mg-qe = mv2 2 l

Don't mind, although you are giving the test paper, it is not excluded that the test paper may also be misprinted, and it is also possible that the paper is cut incorrectly.

It would be nice if the title was changed like this:

18.When the plane is 224m above the ground, the athlete leaves the plane and does a free fall in the vertical direction. (Horizontal velocity is considered 0).

After a period of exercise, the parachute is immediately opened, and the athlete decelerates and descends with average acceleration after the parachute. For the safety of athletes, athletes are required to land at a maximum speed of no more than 5m s. (g takes 10 m s2).

Seeking: (1) What is the minimum height from the ground when the athlete is in the umbrella? How high does it equate to a freefall when you land?

2) What is the shortest time an athlete can spend in the air?

Specific answer: I don't understand, the answer is correct, I think the landlord should be fine.

There are follow-up questions, there are detailed explanations, and of course, it is better for the landlord to make it himself.