The school has a round flower bed. In a week of the flower bed, 32 chrysanthemums were planted, and a rose was planted between each two chrysanthemums. How many roses were planted in all?

The school has a round flower bed. In a week of the flower bed, 32 chrysanthemums were planted, and a rose was planted between each two chrysanthemums. How many roses were planted in all?


In fact, the formula is very simple: 32 △ 2 = 16
If you plant 32 chrysanthemums in a circle, there will be 32 empty ones. If you plant a rose in two empty spaces, there will be no front and back ends. You might as well find a few small stones to swing



There are 540 pots of rose, chrysanthemum and tulip in the park. If you remove 2 pots of chrysanthemum and 4 pots of tulip, there will be the same number of three kinds of flowers. How many pots of these three kinds of flowers


After 6 pots of flowers were removed, 534 pots were left. Among the three kinds of flowers, rose was 534 / 3 = 178, chrysanthemum 178 + 2, tulip 178 + 4



There are 60 chrysanthemums in the flower garden. The number of rose flowers is three times that of chrysanthemums, 24 less. How many rose flowers are there


60X3=180
180—24=156
A: there are 156 roses



The maximum current that a resistance wire can pass is 2A. When a voltage of 36V is applied to it, the current passing through is 0.5A______ (fill in "can" or "can't") used in 220 V circuit


The resistance value of resistor wire is r = u1i1 = 36v0.5a = 72 Ω. In the circuit of 220 V, the circuit current is I2 = u2r = 220 V 72 Ω≈ 3a, ∫ 3a > 2A. It can not be used in 220 V circuit



As shown in the figure, a, P, B and C are the four points on the center O, ∠ APC = ∠ CPB = 60 °, judge the shape of △ ABC and prove your conclusion


An equilateral triangle
Because the angles of the same chord to the circumference are equal,
therefore
∠CAB=∠CPB=60°
∠ABC=∠APC=60°
So ABC triangle is 60 degrees



The electric energy meter is an instrument for measuring the electric energy consumption of electrical appliances. Xiaohua wants to verify whether the revolutions per kilowatt hour marked on the dial of his electric energy meter are accurate. Therefore, he will connect the electrical appliance marked with "220V 3A" into the circuit separately. When the electrical appliance works normally for 5 minutes, the turntable of the electric energy meter just turns 110R, then the actual revolutions per kilowatt hour of the electric energy meter should be ()
A. 2400rB. 2000rC. 1500rD. 1200r


The electric energy consumed by the electric appliance in 5 min is: w = Pt = uit = 220 V × 3a × 5 × 60 s = 1.98 × 105 J = 0.055 kW · h, NW = 110 r0.055 kW · H = 2000 R / kW · h, then the actual revolution per kW · h of the electric energy meter is 2000 R



What is (- 81) / (- 2.25) × (- 4 / 9) / - 16,


(- 81) / (- 2.25) × (- 4 / 9) / - 16
=-81×4/9÷(2.25×16)
=-36÷36
=-1



The resistance is a 12 ohm bulb, which is connected to a 12V circuit to calculate the power consumption per minute. It's better to explain it, and it's better to be fast,


W=P*t=U*I*t
=U*(U/R)*t
=72 joules



What are the characteristics of the multiples of 2.3.4.5.6.7.8.9.11.13
A few answers will do


All even numbers are multiples of 2
The sum of each digit of a number is a multiple of 3. If the sum of each digit is a multiple of 9, then the number is a multiple of 9
If the bits of a number are 0 or 5, then the number is a multiple of 5
If a number is an even number and the sum of each digit is a multiple of 3, the number is a multiple of 6



Does the brightness of a bulb depend on the current or the voltage?
What's the relationship between current and voltage? A variable resistor is connected up and down, and a parallel connection. In this case, the current will flow directly from the circuit connected up and down, and the parallel connection will be short circuited directly? (the resistance of the upper and lower connection is not small) but when I do the problem, it seems that it's not the same as when a resistance is in parallel with a wire?
R who said that the voltage must be 220 drops.


Otherwise, I'll be busy in vain. First of all, it's junior high school physics, so I don't consider the internal resistance of ammeter and voltmeter. If the current, voltage and power of the bulb increase, the brightness will increase. L: brighten, L1: unchanged, L2