When an object moves in a curve, the physical quantity of a certain change is () A. Gravitational potential energy B. velocity C. acceleration D. combined external force

When an object moves in a curve, the physical quantity of a certain change is () A. Gravitational potential energy B. velocity C. acceleration D. combined external force

A. The condition of curvilinear motion is that the external force and velocity are not in the same straight line, which has nothing to do with the gravity of the object, so it has nothing to do with the gravitational potential energy

1. Which physical quantities are vectors and which are scalars in the first chapter of physics compulsory 1 of senior one? 2. The relationship between acceleration and velocity variation?

Vector: displacement, velocity, variation of velocity, acceleration, force (including gravity, elastic force and friction force)
Scalar: time, distance, quality
2. Acceleration = change of velocity / time
Acceleration is the speed of speed change, and the change of speed is the amount of speed change

Curvilinear motion unit test 1. For an object moving in a curve, the physical quantity of a certain change in the process of motion is a. velocity B. combined external force
2. The trajectory of a particle moving from point O to a under the action of constant force F is shown in Figure 1. At point a, the direction of velocity is parallel to the x-axis, then the direction of constant force F may be along ()
3. About the horizontal throwing movement, the correct one in the following statement is ()
A. Horizontal throwing motion is not uniform variable speed motion
B. The horizontal displacement is only related to the horizontal velocity
C. The flight time of horizontal projectile motion only depends on the altitude of initial position
D. The direction of velocity and acceleration of horizontal throwing motion changes continuously
Where are the answers to this paper?
Where is the answer to this paper? It's not the answer to these questions
Curve motion unit test

The direction of 1B speed will change
Figure 2?
3A horizontal distance depends on height and horizontal velocity, acceleration direction is vertical downward

Cosx SiNx / 1 + sin ^ 2x indefinite integral

Indefinite integral of (1 + x ^ 2) * sin (2x)

∫ (1+x²)sin2x dx=∫ sin2x dx + ∫ x²sin2x dx=-(1/2)cos2x - (1/2)∫ x² d(cos2x)=-(1/2)cos2x - (1/2)x²cos2x + ∫ xcos2x dx=-(1/2)cos2x - (1/2)x²cos2x + (1/2)∫ x d(sin2x)=-(1/2...

Indefinite integral of X / sin ^ 2x

∫xsin^2xdx
=∫xcsx^2xdx
=-∫xd(cotx)
=-xcotx-∫cotxdx
=-xcotx-∫cosxdx/sinx
=-xcotx-∫d(sinx)/sinx
=-xcotx-lnsinx+c.

X * (1 + sin ^ 2 x) / sin ^ 2x indefinite integral

Original formula = ∫ x * (CSC ^ 2x + 1)
=∫ x * CSC ^ 2x + X (separate integral)
Front = - x * Cotx + ∫ Cotx = - x * Cotx + ln | SiNx|
After = 1 / 2 x ^ 2
Remember to add C

What is the definite integral of sin (x) between 0 and 0.5 π
Is it 1?

It's 1
The definite integral is - cos (0.5 π) - [- cos0] = cos0 = 1

When the absolute value of X is very small, SiN x is about equal to X

Expand by series
sin x =x-x^3/3!+x^5/5!-x^7/7!+…
When x tends to 0, the following terms are infinitesimal of X, so SiNx is approximately equal to X

A proof of calculus The upper limit of integral is π / 2, the lower limit is 0, the function is f (x) absolute value sin NX n tends to be positive infinity, and the upper limit and the lower limit are the same as above. The product function is f (x). It is proved that the last definite integral is 2 / π times of the next one
The right side is also multiplied by 2 / π

(0,π/2)∫f(x)|sinnx|dx
Substitution NX = t
=1/n*(0,nπ/2)∫f(t/n)|sint|dt
=1/n*[(0,π)∫f(x/n)sinxdx-(π,2π)∫f(x/n)sinxdx+...+(-1)^k(kπ,(k+1)π))∫f(x/n)sinxdx+...]
For the negative term, the substitution is x-k π = t, thus - ((k-1) π, K π)) ∫ f (x / N) sinxdx = (0, π)) ∫ f [(T + k) / N] sintdt
So the original form
=1/n*[(0,π)∫f(x/n)sinxdx+(0,π)∫f[(x+π)/n]sinxdx+...+(0,π))∫f[(x+kπ)/n)sinxdx+...]
The relationship between K and n
When n π / 2 is even, 2K = n
When n π / 2 is odd, 2K + 1 = n
When n →∞, all finite terms can be omitted. For convenience, 2K = n
simple form
=1/(2k)*{(0,π)∫f[x/(2k)]sinxdx+(0,π)∫f[(x+π)/(2k)]sinxdx+...+(0,π))∫f[(x+kπ)/(2k)sinxdx+...}
=1/(2k)*(0,π)∫sinx{f[x/(2k)]+∫f[(x+π)/(2k)]+...+f[(x+kπ)/(2k)]+...}dx
According to the definition of definite integral K →∞
The ending term (upper limit) LIM (x + K π) / (2k) = t, the beginning term (lower limit) = Lim X / 2K = 0
Integral interval length = LIM (K π + x-x) / (2k) = π / 2, so Lim π / 2 / k = Lim π / (2k) = DT
So the original form
=1/π*(0,π)∫sinx[(0,π/2)∫f(t)dt]dx
=1/π*(0,π)∫sinxdx(0,π/2)∫f(t)dt
=2/π*(0,π/2)∫f(t)dt
=2/π*(0,π/2)∫f(x)dx
It's over