A reward of 10 points, simple high school physics

It's different to do this uniform acceleration linear motion experiment: electromagnetic dot timer (6V). EDM dot timer (220V) Every two dot intervals, yes.

Take a count point for every five points: (s).

Take a count point every five points: (s).

Every five points are used as segments (s).

Hehe, as far as I can see, so.

Take a counting point for every five points, that is, from the point at the beginning to the fifth point as a counting point.

Take a counting point every five points, that is, the point from the beginning and the sixth point are a counting point.

Every five points are used as a segment, and the same as .

The first is the same as the third. , for the counting point) a". "Add 4", "for a paragraph."

The second is equivalent to,. , a paragraph has 6 points.

It's not the same. 1) is one of the 5 points to count.

2) Take a counting point from the first point, and then take a counting point every five points.

3) It is a 5-point segment, and it is not clear which point it is to.

Take one counting point every five points", the time interval between the two counting points.

Take a counting point every five points", the time interval between the two counting points.

Every five points as a segment", the time interval between two counting points.

Note. The selection of counting points is as follows: the dots on the paper tape, the time interval between each adjacent two points is because the frequency is 50Hz, 1 50=, for the convenience of research, every five intervals is because each interval is, for a research unit, as a counting point, the interval between the first counting point and the second counting point is, which is convenient for calculation.

The above statement is inaccurate, "every five points take a counting point", "every five points as a paragraph" both refer to: the interval between five points is only four that is the representative, that is not convenient to calculate; "Take a counting point every five points" in the interval of five points is six intervals, that is, the calculation is not convenient.

Imagine that you are in the current. You come out of the positive pole of the power supply, and it doesn't take long for you to come across a fork in the road, so you split into two, one of which you walk along one road and then S2 in turn, and the other you walk along the other road and L, and converge with the first one. Then you continue to walk forward and go back to the negative side of the power supply through S1.

The circuit diagram is just a way to draw the route.

The right end binding post of the bulb and the right end binding post of the motor are connected together, the left end binding post of the bulb and the right end of the switch S2 are connected together, and the left end of the switch S2 is connected to the left end binding post of the motor, so the motor and the switch S2 are connected in parallel with the bulb after being connected in series, that is, M and L are connected in parallel.

(1) The charge accelerates uniformly in a uniform electric field, and let its initial velocity when entering the second belief be v, which is squared by 2as=v, s=, a=f m=qe m, so v=10000m s

Since the direction of the magnetic field is inward, the velocity is perpendicular to it, so the charge moves in a uniform circular motion, and the radius r=, the centripetal force f=qvb=mv squared r, and b=(2) When the charge comes out of the fourth quadrant and enters the first quadrant, the velocity is upward, and the electric field is left, so it does a similar flat throwing motion, and the acceleration a=qe m

This makes it possible to calculate that it enters the second quadrant with coordinates (0,1) and enters the magnetic field for the second time at 45 degrees.

The speed of the river has no effect on the boat.

Mr. Hoho has talked about this question.

The puppy runs for the same time as A and B walk.

That is, 12 (5+3)=

So the puppy's distance is 6*

The puppy finally stops at the place where A and B meet.

That is, 5* away from the departure place of A.

So the displacement of the puppy is.

Because the time for the three animals must be 12 =

So the puppy also walked the way.

1.Rope tension: f=m(a+g)=5kg x ( 51n People are rubbed by the ground: f=fsin37*=

2.The upward pull of the rope on the person: f. =fcos37*= human gravity: g=mg=70kg x 10m s*=700n

g=f。+n The pressure of a person on the ground n=