AB is two identical ammeters. Now use your hand to turn pointer a clockwise suddenly. At this time, pointer B will turn clockwise. Why? The ammeter is positively connected

AB is two identical ammeters. Now use your hand to turn pointer a clockwise suddenly. At this time, pointer B will turn clockwise. Why? The ammeter is positively connected

Move the coil in a and a to cut the magnetic induction line to generate the induced current, which hinders the clockwise movement of needle A. suppose that the generated current flows from the positive pole to the negative pole, then a moves clockwise when the current flows from the negative pole to the positive pole. Similarly, the current in B flows from the negative pole to the positive pole
Therefore, B also moves clockwise. If the current direction generated by a starts to flow from the negative to the positive, this conclusion can also be deduced

Does the clockwise direction of the time adjustment of the automatic mechanical watch refer to the pointer or knob of the watch? Is the clockwise direction of the calendar adjustment a knob?

Generally, you should let the watch needle run clockwise (regardless of the rotation direction of the watch handle). Calendar adjustment can only make the calendar carry (regardless of the direction of dialing the calendar)

It is known that in the hyperbola 3x ^ 2-4y ^ 2 = 12, a chord AB is bisected by the point P (4,1), and the equation of the straight line AB is obtained

y-1=k(x-4)
y=kx+(1-4k)
Substituting
(3-4k²)x²-8k(1-4k)x+4(1-4k)²-12=0
Midpoint abscissa = (x1 + x2) / 2 = 4K (1-4k) / (3-4k & sup2;) = 4
k=3
So 3x-y-11 = 0

If the slope of a straight line is - 2 and the chord length is 2 cut by the square of circle x + the square of y = 4, the equation of the straight line can be obtained

Because the slope of the straight line is - 2, let the chord length of the straight line y = - 2x + B cut by X * + y * = 4 be 2, we can see that the circle is a circle with x0.y0.as the center and 2 as the radius. If B is greater than 0, the straight line passes through 1,2,4 quadrants. If B is less than 0, the straight line passes through 2,3,4 quadrants

Given the hyperbola 3x ^ 2-y ^ 2 = 3, find the linear equation of the chord with a (2,1) as the midpoint
Please give me the process, thank you!

Let m (x1, Y1). N (X2, Y2) be the intersection of a straight line and a hyperbola, and substitute the point coordinates into the curve equation to get 3x1 ^ 2-y1 ^ 2 = 33x2 ^ 2-y2 ^ 2 = 3. Subtracting the two equations to get 3 (x1 ^ 2-x2 ^ 2) = Y1 ^ 2-y2 ^ 2, and deforming to get (y1-y2) / (x1-x2) = 3 (x1 + x2) / (Y1 + Y2). The left side of the equation is the slope of the straight line, and the right side is the midpoint

The asymptote equation of hyperbola 3x2-y2 = 3 is ()
A. y=±3xB. y=±13xC. y=±3xD. y=±33x

The standard form of hyperbola 3x2-y2 = 3 is x2 − Y23 = 1, and its asymptote equation is x2 − Y23 = 0, so y = ± 3x

Asymptote equation of hyperbolic y square-3x square = 1

Formula
x²/1-y²/(1/3)=1
a=1
b=√3/3
Asymptote equation y = ± A / b x
Asymptote equation y = ± √ 3x

Urgent: A is equal to 2 root sign 3, and when the square of hyperbola x divided by 16 minus y square divided by 4 equals 1, there is a common focus, find the hyperbolic standard equation? Urgent

c=√(16+4)=2√5 b=√(c²-a²)=√(20-8)=2√3
∵ focus on X axis ∵ equation: X & # 178 / A & # 178; - Y & # 178 / B & # 178; = 1 = > X & # 178 / 8-y & # 178 / 12 = 1

The center angle of the inferior arc is ()
A. 30°B. 45°C. 60°D. 90°

Let o be OC ⊥ AB, and the perpendicular foot be point C. according to the equation x2 + y2 = 4 of the circle, the coordinates of the center O of the circle are (0, 0), the radius r = 2, the distance d from the center of the circle to the straight line 3x + y-23 = 0 = | OC | = 232 = 3, the chord | ab | = 2r2 − D2 = 2, and the angle of the center of the circle opposite to the inferior arc AB of the straight line cut by the circle is 60 degrees

If two foci F1, F2 and chord ab of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 pass through F1 and | F1 | + | F2 | = 2 | ab |, then | ab | equals
The answer is 4a. Can you explain it for me? How did you get it?

It is proved that: (F1F2) ^ 2 = (Pf1) ^ 2 + (PF2) ^ 2-2 (Pf1) (PF2) cos ∈ f1pf2 (2C) ^ 2 = (pf1-pf2) ^ 2 + 2 (Pf1) (PF2) (1-cos} f1pf2) 4C ^ 2 = 4A ^ 2 + 2 (Pf1) (PF2) (1-cos} f1pf2) 2B ^ 2 = (Pf1) (PF2) (1-cos} f1pf2)