Something simple about physics

Here comes the standard answer.

1) There are two ways of signal propagation here: one is transmitted to the debugging room by the studio through the cable, the other is transmitted to the communication satellite by the studio, and then transmitted to the studio by the communication satellite, obviously the distance experienced by the latter transmission path is much longer than the distance experienced by the previous transmission path, and the transmission speed of the electrical signal is equal in these two transmission paths, so the time taken by the latter route is longer than the time taken by the previous way to propagate, so the picture transmitted by the latter channel is always "backward" In the image transmitted by the previous channel, the change in the person's movements on the screen transmitted by the satellite always occurs later than the corresponding change in the picture transmitted by the cable

2) Since the speed of radio signal propagation is equal to the speed of light, i.e., v=3 10 8m s, the time of propagation by the cable is extremely short, negligible, and the time required for the signal to travel back and forth through the satellite is.

Since the propagation time through the cable is negligible, the above is the time of the picture delay

This is the received electromagnetic wave.

t=s/c=2*36000/3*10^5=

1) It is not necessarily impossible to see the image at night when there is thunder and lightning, but the invisibility of the image is due to the electromagnetic interference is too serious, and the sound of the speaker is also interference, and it is the same reason that cutting a file with a wire in front of the radio will cause noise. (2) Distance Speed of light, s v=36000000*2 300000000=

1. When the 10th ball has just finished the unit time, the first one has walked 125 meters, and the unit time is t, then 1 2at 2 to 1 2a (10t) 2 is 1 to 100, so the 10th ball has gone with an interval of .

2. The movement of the third ball relative to the fifth ball is a uniform linear motion with an initial velocity of 25m s, the distance is, that is the initial distance, the upstairs is reversed, the third ball is not the penultimate ball, in fact, there is no jump, you see, after the third ball falls, the fifth just begins to fall, at this time, the acceleration of No. 3 is 10, and the acceleration of No. 5 is also 10, so there is no acceleration of No. 3 relative to No. 5, and at this time, the speed of No. 3 is 25m s, and No. 5 is zero, The velocity difference is a constant value, that is, 25m per second, so it can be seen that the No. 3 ball is a uniform linear motion with a velocity of 25m per second relative to the No. 5 ball, because there is a displacement difference at the beginning, that is, the displacement of the No. 3 ball in the first two time periods, so it should be added when calculating.

It takes t time for the ball to fall to the bottom of the well.

h=1/2gt^2

125=1/2*10*t^2

t=5, that is, it takes 5s for the ball to fall to the bottom of the well.

11 balls, then each interval is.

5s/10=

That is, the time interval between two adjacent balls to start falling is.

4s have passed since the third ball was thrown, and 3sh3=1 2*10*4 2=80 has passed since the fifth ball was thrown

h5=1/2*10*3^2=45

The difference in height between the two balls is h3-h5=80-45=35m

1.From the very beginning of the ball to the last ball just shot, that is, when the first ball just landed, the elapsed time t, h=1 2gt 2=125, the solution is t=5, because the interval between the shots of a ball is equal, then, the time interval between the two adjacent balls is t'，t'=t/10=

2.The third ball dropped 8 t'time, the same goes for the fifth ball to drop 6 t'The height at which they fell at this time was, h3, h5

h3-h5=1/2gt3^2-1/2gt5^2=35m

Let the inclination angle of the inclined plane be , then the length of the inclined plane is l=d cos, and the height is the velocity v 2=2gh of the h dtan free fall

Acceleration down the inclined plane a=gsin - gcos v 2-v0 2=2al

Solution: =v0 2 2gd