Why is it easy to ride a variable speed bicycle? According to the principle of energy conservation, the total work done from one place to another is the same, so why does riding a variable-speed bicycle feel labor-saving and less tired than riding an ordinary bicycle? What good post?

Why is it easy to ride a variable speed bicycle? According to the principle of energy conservation, the total work done from one place to another is the same, so why does riding a variable-speed bicycle feel labor-saving and less tired than riding an ordinary bicycle? What good post?

Variable speed bicycles achieve the effect of variable speed by adjusting the ratio of the front and rear gears. The larger the front gear is adjusted, the less effort it takes to ride, but the speed slows down. The smaller the rear gear is adjusted, the more effort it takes, but the speed is fast! The front gear is adjusted to the minimum, and the rear gear is also adjusted to the minimum. The riding speed is the fastest, but it is also the most laborious! Variable speed bicycles are like cars

The larger the rear gear of a variable-speed bicycle, the more labor-saving. What is the principle?

Power arm and resistance arm

The point (2, root 2) is on the image of power function f (x), and the point (- 2,1 / 4) asks what value x is on the image of power function g (x) ①f(x)>g(x) ②f(x)=g(x) ③f(x)<g(x)

A function in the form of y = x ^ a (a is a constant), that is, a function with the base as the independent variable, the power as the dependent variable and the exponent as the constant is called a power function
From the point (2, root sign 2) on the image of power function f (x), f (x) = x ^ (1 / 2) = √ X
From the point (- 2,1 / 4) on the image of the power function g (x), G (x) = x ^ (- 2) = 1 / X ²
① When x ∈ (0,1), f (x) > G (x)
② When x = 1, f (x) = g (x)
③ When x ∈ (1, + ∞), f (x) < g (x)

Image crossing point (1) of known power function y = f (x) 2, two 2) , then log2f (2) = _

Let the power function y = f (x) = X α,
∵ its image passes the point (1)
2,
two
2),
∴f(1
2)=(1
2) α=
two
2,
∴ α= one
2.
∴f(2)=21
2=
2,
∴log2f(2)=log221
2=1
2,
So the answer is: 1
2.

If the image of the power function f (x) = XA passes through point (2, two 2) , then f (4) = ___

F (2) = 2A =
two
2=2−1
2, so a = - 1
2, so f (x) = x − 1
2, so f (4) = 4 − 1
2=1
two
So the answer is: 1
two

If the function f (x) = (M2 − m − 1) XM2 − 2m − 3 is a power function and is a subtractive function on X ∈ (0, + ∞), then the real number M = () A. 2 or - 1 B. -1 C. 3 D. 2

∵ f (x) = (M2 − m − 1) XM2 − 2m − 3 is a power function,
M2-m-1 = 1, M = 2 or M = - 1,
If M = 2, f (x) = x-3 = 1
X3, is a subtractive function on (0, + ∞);
If M = - 1 can be obtained, f (x) = x0 = 1, it does not satisfy that it is a subtractive function on (0, + ∞);
To sum up, M = 2,
Therefore, D is selected;