Xiao Ming and Xiao Li move from village a to village B at the same time. Xiao Li's speed is 4 km / h, Xiao Ming's speed is 5 km / h. Xiao Li arrives 15 minutes later than Xiao Ming To find the distance between villages a and B, it is best to use the equation,

Xiao Ming and Xiao Li move from village a to village B at the same time. Xiao Li's speed is 4 km / h, Xiao Ming's speed is 5 km / h. Xiao Li arrives 15 minutes later than Xiao Ming To find the distance between villages a and B, it is best to use the equation,

Let the distance be x kilometers
Then Xiao Ming is x / 5 hours
Xiao Li is x / 4 hours
15 minutes = 1 / 4 hour
So x / 4-x / 5 = 1 / 4
Take 20 on both sides
5x-4x=5
x=5
A: the distance is 5km

Xiao Ming and Xiao Li travel from village a to village B at the same time. Xiao Li's speed is 4km / h. Xiao Ming's speed is 5km / h. Xiao Li arrives 15min later than Xiao Ming to find the distance between the two villages

The distance between villages a and B is x km
x/4=x/5+15/60
x/4-x/5=1/4
x/20=1/4
x=5
The distance between village a and village B is 5 km
This question our teacher said! Absolutely correct! I am also a junior one!

Xiao Ming and Xiao Li walk from village a to village B at the same time. Xiao Li's speed is 4km / h, Xiao Ming's speed is 5km / h, and Xiao Li arrives 15min later than Xiao Ming
How to find the distance between village a and village B? Set Xiaoming x hours to village B. why do I pick up 1 every time

4*15/60=1km
1/(5-4)=1h
5*1=5km

Experimental problems of Newton's second law
In the experiment of verifying Newton's second law, draw the obtained data into A-F diagram, and the result is that the straight line is not the origin. Why are the diagrams made by two students with the same set of data different?

Newton's second law is verified by using the inclined board car. Because there is no way to remove the error caused by friction, the car's running board should have an inclination angle. In order to balance the friction with the car's own gravity, friction = car weight m * Sina, a is the inclination angle, so there are two cases of the origin of A-F diagram

Given that the area of a circle is 2 π cm ^ 2, find the radius of the circle. Given that the area of a square is 11cm square, find its side length

Radius = 2 π / π = 2 (R & # 178;)
Radius = √ 2
Side length = √ 11cm
I hope I can help you. I wish you progress in your study. If you don't understand, please ask me. If you understand, please adopt it in time! (*^__ ^*)

A light spring with stiffness coefficient k = 100N per meter, the original length is 10cm, one end is tied with a small ball with a mass of 0.6kg, and the other end of the spring is the center of the circle
Let the ball do uniform circular motion on a smooth plane, and its angular velocity is 10 rad per second?

Assuming that the spring extends x, the radius of circular motion is r = 0.1 + X (10cm equals 0.1M)
Therefore, the spring force provides centripetal force, that is, KX = m ω ^ 2 * r = m ω ^ 2 * (0.1 + x)
Take in the number, solve x, and calculate KX

Four mixed operation questions, like 50 × 2 △ 2 + 2-2, 200 answers. Thank you
Urgent need

50×2÷2+2-2=50 10×2÷5+3-6=1 36×10÷40+50-9=50
12×2÷3+16-3=21 17×20÷10-10+10=170

Two pieces of wood a and B with mass m ` m are stacked on a smooth horizontal table. The dynamic friction coefficient between a and B is μ. Now a horizontal tension f is applied to B to make a and B move to the right relatively still. The displacement is s. in this process, what is the work done by force F to object B? What is the work done by friction force to object a?
Suppose a is above:
Because f acts directly on B, the work done by F on B is w = FS
The friction between a and B is f = UMG
Then the effect of friction on a is W1 = FS = umgs
But aren't a and B relatively static? Aren't they static friction? So isn't friction f?

A and B are relatively static, and the static friction between them is neither UMG nor F, which should be solved by Newton's second law
Suppose a is above, AB has the same acceleration: a = f / (M + m)
The resultant force of a = friction: that is, f = ma = MF / (M + m)
The work of friction on a is: w = FS = MFS / (M + m)

In the known arithmetic sequence {an}, Sn is the sum of its first n terms, let A6 = 2, S10 = 10. (1) find the general term formula of the sequence {an}; (2) take the second term, the fourth term, the eighth term from the sequence {an} , 2n In order to form a new sequence {BN}, try to find the first n terms and TN of the sequence {BN}

(1) Let the first term of the sequence {an} be A1 and D respectively, then a1 + 5D = 2 (1) 10A1 + 10 × 92d = 10 (2) simultaneous solution (1) and (2) obtain A1 = - 8 and d = 2, so an = 2n-10 (n ∈ n *) (2) BN = A2N = 2 · 2n-10 = 2n + 1-10 (n ∈ n *), so TN = B1 + B2 + +bn=4(1−2n)1−2-10n=2n+2...

The car moves at a constant speed of 20 m / s, and the acceleration after braking is 5 meters per second. Then what is the ratio of the car displacement within 2 s and 6 s after braking?

To solve this problem, we only need to pay attention to the trap of whether the car is stationary or not. From v = at, we can get that the longest deceleration time is 4 seconds, so the object is still moving at the end of 2 seconds, s = vt-1 / 2at ^ 2, which is 30 meters. At the end of 6 seconds, the object has already stopped, s' = V ^ 2 / 2a, which is 40 meters, so the ratio of two displacements is 3:4