Does static friction do work or not depends on the selected reference frame?

Does static friction do work or not depends on the selected reference frame?

All work, kinetic energy and momentum are related to the choice of reference frame

Direction of static friction
Why does the body have to be opposite to the movement trend?
What about the circular motion on the disk?
The movement trend is tangent direction of the circle, but static friction provides centripetal force to point to the center of the circle?

In your example, if an object moving in a circle without friction will move along a tangent line, the tangential motion minus the actual circular motion is the motion trend

There are two points a (- 1,0), B (1,0) on the plane, and point P is on the circumference (x-3) 2 + (y-4) 2 = 4. Find the coordinates of point P when ap2 + bp2 is the minimum

According to the meaning of the title, if we make the symmetric point Q of point P about the origin, then the quadrilateral paqb is a parallelogram. According to the properties of parallelogram, there are ap2 + bp2 = 12 (4op2 + AB2), that is, when OP is the minimum, ap2 + bp2 takes the minimum, opmin = 5-2 = 3, PX = 3 × 35 = 95, py = 3 × 45 = 125, P (95125)

Factoring in real numbers: 3x * X-5

3x*x-5
=(√3x-√5)(√3x+√5)

In the cube ABCD --- a1b1c1d1, M is the midpoint of dd1, O is the midpoint of AC, ab = 2

In triangle bdd1,
Connecting Mo, O is the midpoint of AC, and also the midpoint of BD (diagonals bisect each other)
Then Mo is the median line of triangle bdd1,
MO//BD1,
Mo ∈ plane AMC,
So BD1 / / plane ACM (if the line is parallel to a line on the plane, then the line is parallel to the plane)

The problem of n-th power sequence similar to an + 1 = pan + Q
For example, the general formula of a (n + 1) = 2A (n) + 5 ^ n (the brackets are subscripts)

Chenaweiji, there are many such problems. You haven't found out the general form of this series problem. The general form is pa (n + 1) = QaN + R. if you have learned the characteristic equation, it will be easier. The characteristic equation is PX = QX + R (this is a first-order) and then solve the root of the equation. Then there must be p (an + 1 - x) = q (an-x), so

PA.PB.PC What is the cosine of the plane angle of dihedral angle a-pc-b
A.1/2
B.1/3
C. (radical 2) / 2
D. (radical 3) / 2

B)

5 / 6 [6 / 5 (1 / 2-1 / 4) - 7 and 1 / 5] = 3x + 1 / 2

5 / 6 [6 / 5 (1 / 2-1 / 4) - 7 and 1 / 5] = 3x + 1 / 2
(5/6)【(6/5)(2X-1)/4-36/5】=(3X+1)/2
(5/6)【(6X-3)/10-72/10】=(3X+1)/2
(5/6)(6X-75)/10=(3X+1)/2
(6X-75)/12=(3X+1)/2
6X-75=18X+6
12X=-69
X=-23/4
I don't quite understand the topic. I don't know if I understand it wrong

If the diagonal length of a parallelogram is x, y and one side is 12, then the value of X, y may be ()
A. 8 and 14b. 10 and 14C. 18 and 20d. 10 and 34

A. 82 + 142 = 4 + 7 = 11 < 12, so impossible; B, 102 + 142 = 5 + 7 = 12 = 12, so impossible; D, 34-10 = 24, so impossible; so choose C

Solution equation: X: 1.8 = 12-x: 0.3

X:1.8=12-x:0.3
0.3x=1.8×（12-x）
0.3x=21.6-1.8x
0.3x+1.8x=21.6
2.1x=21.6
X = 72 / 7