Difference between traction power and vehicle power

1. For the process of locomotive starting with constant power: P = FV, since P is constant, the traction force F is inversely proportional to the speed v. therefore, when the speed increases, the traction force will inevitably decrease, resulting in the decrease of acceleration A. that is to say, the locomotive will do the variable acceleration movement with the decrease of acceleration. When the acceleration is equal to 0, When the speed reaches the maximum, the locomotive starts with the acceleration gradually reduced, and then with the uniform speed. 2. When the locomotive starts with the uniform speed, the traction force remains unchanged due to the constant acceleration A. according to P = f * V, with the increase of speed, the power increases. Therefore, when the power increases to the rated power, the uniform speed first makes the uniform speed, After that, the acceleration is gradually reduced and then uniform

What is the significance of the extensive use of electric energy for human energy utilization

The use of electric energy has brought great advantages to the rapid development of human society

One of the most important energy sources is electricity, which is widely used in our daily life

Electric is one of the most important energy resources, which is widely used in our life
This kind of level one always doesn't solve the problem. I won't answer in the future

If x = 1 / 2T, y = 1 / 3T, x-6y + 3 / 4 = 0, then t=

x=1/2t,y=1/3t,x-6y+3/4=0
arcsinx-x
1/2 t - 2 t +3/4=0
The solution is t = 1 / 2

If x = 3-2t and Y-5 = 3T for the system of equations of X and y, then y is represented by an algebraic expression containing x, and the correct result is

t=(3-x)/2
Substituting Y-5 = 3T
Y-5 = 3 / 2 (3-x)
That is, y = 19 / 2-3 / 2x

Finding the first derivative dy / DX and the second derivative of the function y = y (x) determined by x = t + T ^ 2, y = T-T ^ 3

y/x=(t-t^3)/(t+t^2)=1-t
Then t = 1-y / X
By taking x = t + T ^ 2, we can get the following result:
x=1-y/x +1+y^2/x^2-2y/x
That is, x ^ 3 = x ^ 2-xy + x ^ 2 + y ^ 2-2xy
And then it did, right?

It is known that x = 2T + 3, y = 3t-1 (1) denote y with an algebraic expression containing x (2) denote x with an algebraic expression containing y

(1) if x = 2T + 3, then t = (x-3) / 2 is substituted by y = 3t-1, y = 3 (x-3) / 2-1, so y = 3x / 2-11 / 2
(2) if y = 3t-1, then t = (y + 1) / 3 is substituted by x = 2T + 3, then x = 2 (y + 1) / 3 + 3, so x = 2Y / 3 + 11 / 3

Given the equation system 4x + 3Y = 3T + 2x-y = 1-2T, please use the algebraic expression containing x to express y

Multiply the formula on the left by 2, and multiply the formula on the right by 3
Finally, y = (5-14x) / 3

Known: x = (1-T) / (1 + T), y = (3-2t) / (2-3T), please use the algebraic expression of X to express y

x=(1-t)/(1+t)
1-t=x(1+t);
t=(1-x)/(x+1);
Substituting t into y algebra,
After simplification, y = (5x + 1) / (5x-1)

If x = (1-T) / (1 + T) y = (2-3T) / (3-2t), the algebraic expression of X is used to express y

x+xt=1-t
t=(1-x)/x+1)
3y-2yt=2-3t
t=(3y-2)/(2y-3)
(3y-2)/(2y-3)=(1-x)/(x+1)
3xy+3y-2x-2=2y-3-2xy+3x
5xy+y=5x-1
y=(5x-1)/(5x+1)

Difference between traction power and vehicle power