Very urgent physics! Three charged balls with mass m are placed on an insulating and smooth horizontal surface, and the distance between them is l, which is far greater than the radius of the ball. The charged quantity of ball a is 2q, and that of ball B is - Q. now a right horizontal constant force F acts on ball C to keep the distance L between the three balls and move in a straight line, then the electric charge of ball C is____ The constant force F is____

Very urgent physics! Three charged balls with mass m are placed on an insulating and smooth horizontal surface, and the distance between them is l, which is far greater than the radius of the ball. The charged quantity of ball a is 2q, and that of ball B is - Q. now a right horizontal constant force F acts on ball C to keep the distance L between the three balls and move in a straight line, then the electric charge of ball C is____ The constant force F is____

Horizontal constant force to the right f?

2. The Shenzhou-7 manned spacecraft was successfully launched at 21:10 on September 25, 2008, carrying three astronauts to realize the spacewalk. The small satellite carried by Shenzhou-7 was successfully released at 19:24 Beijing time on 27th. This is the first time that China has carried out the accompanying flight test of microsatellite on the spacecraft. After the release of the small satellite, it gradually left the spacecraft at a slow speed, In order to catch up with the orbital module, the Beijing flight control center will control the following small satellites to gradually approach the orbital module and finally realize the flight around the orbital module A. It should be accelerated from the lower track B. from the higher track C. It should be accelerated in the same orbit from the orbital module D. no matter what orbit, just accelerate

The spaceship should accelerate from the low compasses, do centrifugal motion, and dock with the spacecraft from the elliptical orbit to the higher circular orbit

How to do problem 3.4 on page 56 of mathematics book Volume 2, grade 5, people's Education Press

3.2740mm=2.74m 1525mm=1.525m 25mm=0.025m
2.74×1.525+(0.025×1.525+2.74×0.025)×2=4.39175
4.4.25×2×8=72

People's Education Press: math exercise 22.3 No.1,

Let every branch grow x small branches. From the meaning of the question, 1 + X + x 2 = 91, X / + x-90 = 0, X ₁ = - 10, X Ψ = 9 ∵

People's education press, the seventh problem of Exercise 1.1 on the fifth page of mathematics in junior high school

At that time (17:00), the temperature was 7-4 = 3 ℃. After 7 hours, it dropped 4 ℃. At that time, the temperature (0:00) was 3-4 = - 1 ℃

1、 A freight car departs from the station to make a uniform acceleration linear motion history. The attendant stands next to the front section of the first carriage. The first carriage passes by him for 4 seconds, and the whole train passes by him for 20 seconds. The length of each compartment is equal to the length of the car joint How many cars are there in this train 2. How long does it take for the last car to pass by him? 2、 When the car was driven on the Yanping straight road at 5 meters per second, the acceleration obtained after braking was 0.4 meters per square meter 1. How long does the car stop after braking? 2. How long does it take? 3. How long does it take for the brake to slide 30 m? 4. What is the sliding distance in 2.5 seconds before stopping? The first question and the second question should be "the last nine carriages". I just got the wrong number.

1. Suppose that the freight car has n cars, because the freight car has a station, so the initial speed is 0, so that the first car passing by the attendant time T 1 = 4 S, the whole train passing by the attendant T 2 = 20 s, s = at 1 ^ 2 / 2 (1) ns = at 2 / 2 (2) from (1) / (2), n = 25, the freight car has 25 cars

An athlete touches the height in the training place. The weight of the athlete is 75kg and the height is 1.80cm. In 0.2S, he pedals vertically with a force of 1500N on the ground. What is the height that the athlete raises? (g takes 10m / S2)

a=F/m=1500/75=20m/s^2
v1=at=20*0.2=4m/s
∵v2^2-v1^2=2gh
∴h=(0-4*4)/(2*-10)=0.8m
The height is 0.8 + 1.8 = 2.6m

The basic process and method of solving the problem of solving high one

Senior high school physics and calculation process is the application process of physical laws. When solving problems, we should first carefully examine the problem, find out the relationship between known conditions and unknown quantities, judge the physical concepts and laws that can be used, and then correctly use the physical laws to list relational expressions, also known as doing equations. The remaining thing is the mathematical problem, and solve the unknown number in the equation

In acrobatic top pole performance, a person stands on the ground with a 20 kg bamboo pole on his shoulder, and there is a 40 kg person on the bamboo pole. How much pressure does the bamboo pole exert on the performer when the person who blocks the trunk slides down at the speed of a = 4m / S ^ 2

20X10+40X(10-4)=440N

(1) a sliding block starts from its rest, and accelerates to slide down from the top of the slope, and the speed at the end of the 5th second is 6m / s. find: (1) the speed at the end of the 4th second; (2) the displacement in the first 7 seconds; (3) the displacement in the third second 2. When a construction team is carrying out blasting task, it is known that the burning speed of the fuse along the fuse is 0.8cm/s. In order to make the igniter run to a safe place 120m away from the ignition point before the fuse flame burns to the explosive, how many M is needed for the fuse. (assuming the speed of human running is 4m / s) 3. The car starts to move in a straight line with uniform acceleration from standstill. On the way, it successively passes through two points a and B 125 m apart. In 10 seconds, the speed at point B is known to be 15m / s (1) displacement of vehicle from starting point to point a (2) time from departure to point a Suffocating ~ National Day holiday for 7 days, I forgot all the knowledge that the teacher taught before

When the velocity is 6m / s in 5S, then a = 1.2m/s ^ 2 (1) the velocity at the end of 4S is v = at = 4.8m/s (2) the displacement in the first 7 seconds s = 1 / 2at ^ 2 = 29.4m is the same as a. (3) the displacement in the third second is s = 1 / 2at ^ 2-1 / 2A (t-1) ^ 2 = 3M, and the direction is the same as a