Xiao Hong went to the laboratory to do the experiment of measuring the rated power of the small bulb by voltammetry. The rated voltage of the measured small bulb is 2.5V, and the resistance is about 10 Ω The laboratory has the following equipment: power supply (voltage is 6V), ammeter (0 ~ 0.6A 3a), voltmeter (0 ~ 3V 15V), one switch, several wires, two sliding rheostat: R1 (10 Ω, 0.5A), R2 (50 Ω, 0.5A) (1) Sliding rheostat should be ()? (2) There is a fault in the circuit. The phenomenon is: the bulb is not on, the voltage indication is large, and the ammeter has no indication. Detection: remove the bulb, and the indication of the two meters does not change. Question: what is the cause of the fault?

Xiao Hong went to the laboratory to do the experiment of measuring the rated power of the small bulb by voltammetry. The rated voltage of the measured small bulb is 2.5V, and the resistance is about 10 Ω The laboratory has the following equipment: power supply (voltage is 6V), ammeter (0 ~ 0.6A 3a), voltmeter (0 ~ 3V 15V), one switch, several wires, two sliding rheostat: R1 (10 Ω, 0.5A), R2 (50 Ω, 0.5A) (1) Sliding rheostat should be ()? (2) There is a fault in the circuit. The phenomenon is: the bulb is not on, the voltage indication is large, and the ammeter has no indication. Detection: remove the bulb, and the indication of the two meters does not change. Question: what is the cause of the fault?

(1) Sliding rheostat should be ()? R2 (50 Ω, 0.5A)
(2) Circuit failure, bulb open circuit

In the experiment of measuring the rated power of the small bulb, the power supply is 4V, the rated voltage of the small bulb is 2.5V, and the resistance is about 10 Ω. Which sliding rheostat should be selected
A 20 euro 1A
B 10 euro 0.5A
C 10 euro 0.1A

B

As shown in the figure, the chord AB divides the circumference into 1; 2. Given that the radius is 1, find ab

Because the chord AB divides the circle into 1:2, so the center angle AOB is 120 ° and OP is perpendicular to ab through O. because the radii of the circles are equal, so AP = BP, so AOP = angle BOP = 60 ° and OAP = 30 °, so PA = Pb = two thirds root sign 3, so AB = root sign 3
That "because, so" because the computer is unable to type out, so it should be replaced by the three dots, and the mathematical names in it should be replaced by the symbol language you use, such as Jiao ah
Finally, I venture to ask, which grade of student are you?

The chord AB divides the circle into two parts of 1:2. Given that the radius of the circle O is 1, the length of the chord AB is calculated

AB divides the circle into two parts of 1:2
Then the smaller center angle of AB is 120 degrees
Do NN ⊥ AB through o
A = 30 ° in △ OAN
AN=√3/2,AB=√3

As shown in the figure, the trapezoid ABCD, ab ‖ CD, ∠ B = 90 °, BC = 6cm, CD = 12cm, ab = 20cm is known. The moving point P starts from point a and moves uniformly along the ad direction to D, with a speed of 1cm ／ S; the moving point Q starts from B and moves uniformly along the Ba direction to a, with a speed of 2cm ／ S; when one of them reaches the end point, the two points stop moving at the same time. If two points start at the same time, the moving time is t (s) (T > 0), △ CPQ (1) find the distance from point P to AB; (2) when t is the value, △ Apq is an isosceles triangle with AQ as the base; (3) find the functional relationship between Y and t

(1) Let de ⊥ AB be at point E, PF ⊥ AB be at point F, ∵ ABCD, ab ∥ CD, ∠ B = 90 °, rectangular DEBC, ∵ BC = 6cm, CD = 12cm, ab = 20cm, ∵ de = BC = 6cm, AE = ab-eb = 20-12 = 8cm, ∵ ad = 10cm, ∵ moving point P starts from point a and moves to D at a uniform speed of 1cm ／ S; moving point Q starts from B and moves to a at a uniform speed of 2cm ／ S; AP = TCC, ∵ When △ Apq is an isosceles triangle with AQ as the base, AP = PQ, AF = FQ = 12aq = 12 (ab-bq) = 12 (20-2t) = (10-T) cm, in RT △ AFP, ap2 = af2 + PF2, ｜ (10-T) 2 + (35t) 2 = T2, the solution is t = 50 (rounding off) or T = 509 ｜ when t = 509, △ Apq is AQ based Isosceles triangle; (3) in RT △ APF, ∵ AP = t, PF = 35t ∵ AF = 45t, ∵ y = s trapezoid pfbc-s △ pfq-s △ BCQ = 12 (PF + BC) · fb-12pf · fq-12bc · BQ = 12 [(35t + 6) (20-45t) - 35t (10-T) - 6 × 2T] = 925t2-125t + 60

Given that the lengths of the two parallel strings of the circle are 6 and 2 root sign 6 respectively, and the distance between the two strings is 3, find the radius of the circle (process)

When the center of the circle is on both sides of the two strings, OM + on = radical (R ^ 2-3 ^ 2) + radical (R ^ 2-6) = 3, then r = radical 10
When the center of the circle is on the same side of two strings, on-om = radical (R ^ 2-6) - radical (R ^ 2-9) = 3, r = & nbsp; radical 10

It is known that the lengths of the two parallel strings of the circle are 6 and 2 root sign 6 respectively, and the distance between the two strings is 3, and the radius of the circle is 3

Connect the center of the circle and the intersection of the two chords and the circle, make the vertical line from the center of the circle to the two chords, and the length is x, 3-x respectively. Then use the two right angle triangle row Pythagorean theorem to list two equations, including X and R unknowns, to solve the equation. Note that there are two solutions, namely, the two chords are on the same side and the other side of the center of the circle

In ⊙ the chord Mn ⊥ PQ, OE ⊥ NQ is in E

I don't understand

E is a point in a circle O with a radius of 5cm, and OE is equal to 3cm. Of all the strings passing through e, what are the lengths of the longest string and the shortest string?

The longest is diameter 2x5 = 10 (CM) and the shortest is chord 2x4 = 8 (CM) perpendicular to OE

In the circle O, AB and CB are two equal chords perpendicular to each other. Od is perpendicular to AB and D, OE is perpendicular to AC and e. it is proved that the quadrilateral adoe is a square

AB and CB are two mutually perpendicular and equal strings. The wrong number should be. AB and AC are mutually .
Adoe is a rectangle (∵ a = ∵ d = ∵ e = 80 & ordm;)
And ab = AC, OD = OE
The adoe is a square