Use resistance, sliding rheostat, power supply and ammeter to measure rated power of bulb The resistance is known

Use resistance, sliding rheostat, power supply and ammeter to measure rated power of bulb The resistance is known

In the process of experimental measurement, the ammeter A1 is connected in series with the known resistance R0 to act as a voltmeter, as shown in the figure

Why can't we measure the resistance of a small bulb with Peian method

In the process of measuring resistance by voltammetry, if the measured resistance value is basically unchanged, the average value can be taken from multiple results
In this experiment, as the voltage increases, the current in the filament also increases. The important thing is that the temperature of the filament will be higher and its resistance will be larger. Therefore, the average value can not be used

Given that the point P (2x-1, x + 4) is in the second quadrant, and the distance from the point P to the X and Y axes is equal, then x = ()

Because point P (2x-1, x + 4) is in the second quadrant
Then 2x-10
x-4
Because the distance from point P to x-axis and y-axis is equal
be
|2x-1|=|x+4|
-(2x-1)=x+4
-2x+1=x+4
x=-1

Given that point P is in the second quadrant, and the distance to x-axis is 2, the distance to Y-axis is 3, and it is on the left side of y-axis, then the coordinate of point P is ()

Given that point P is in the second quadrant, the abscissa of point P is 0; if the distance from point P to X axis is 2, the ordinate is 2;
So the coordinates of P are (- 3,2)

If point a is in the second quadrant and its distances to x-axis and y-axis are 3 and 2 respectively, then the coordinate of a is___ ．

Let the coordinates of point a be (x, y), ∵ point a be in the second quadrant, and the distances from it to the x-axis and y-axis are 3 and 2 respectively, | x | = 2, | y | = 3, ∵ point a be in the second quadrant, ∵ x < 0, Y > 0, ∵ a coordinates are (- 2, 3)

Given that P (3-m, 2m-5) is on the bisector of the second and fourth quadrants, then M = () if P (2x + 1, x-1) is on the Y axis, then the coordinate of P is ()
(1) given that the point P (3-m, 2m-5) is on the bisector of two or four quadrants, then M = ()
② If P (2x + 1, x-1) is on the Y axis, then the coordinate of point P is ()

① ∵ point P (3-m, 2m-5) is on the bisector of the second and fourth quadrants
∴(3-m﹚＋﹙2m-5)＝0
∴m＝2
② ∵ P (2x + 1, x-1) on the y-axis
∴2x＋1=0
∴x=-½∴x-1=-½－1=－3/2
The coordinates of point P are (0, - 3 / 2)
100% right,

If point P (x2-1, x + 3) is on the angular bisector of three or four quadrants, find the distance from point P to X axis

P (x2-1, x + 3) is on the bisector of three and four quadrants,
No solution for x2-1 + X + 3 = 0
.
So the question is wrong

1, the distance from point P to X axis is 3, and it is on the bisector of two or four quadrants, then the coordinate of point P is_ 2, passing point (- 3)
1, the distance from point P to X axis is 3, and it is on the bisector of two or four quadrants
2. The intersection coordinates of the line passing through the point (- 3,4) and parallel to the Y axis and the line passing through the point (2, - 3) and parallel to the X axis are ｛

Question 1: P (- 3.3) or (3. - 3)
Question 2: intersection coordinates (- 3. - 3)

Given that the point P is in the fourth quadrant, the distance from X axis is 3, the distance from Y axis is 2, and the P coordinate is?

Correct answer: (2, - 3)
Because of the fourth quadrant, x > 0, y < 0
The absolute value of X is 2 and the absolute value of Y is 3
SO 2, - 3

Point P is in the second quadrant, the distance to the x-axis is 3, and the distance to the y-axis is 2, so the coordinate of point P is ()

(-2,3)