After connecting the "220 V 60W" bulb and "220 V 25W" bulb in series in the 220 V circuit, what is the actual power of the two lamps? To calculate the process

After connecting the "220 V 60W" bulb and "220 V 25W" bulb in series in the 220 V circuit, what is the actual power of the two lamps? To calculate the process

According to P = u * U / R, the resistance of the two lamps is calculated as R1 = 220 * 220 / 60 = 2420 / 3, R2 = 220 * 220 / 25 = 1936
The current of this circuit is I = u / (R1 + R2) = 165 / 2057
P1=I*I*R1=5.19
P2=I*I*R2=12.46

If x2-5x + 1 = 0, then X3 + 1 / X3=

If x ^ 2-5x + 1 = 0 is divided by X, then X-5 + 1 / x = O can be obtained by simplifying x ^ 2 + 1 / (x ^ 2) + 2 = 25, that is, x ^ 2 + 1 / (x ^ 2) = 23. If x + 1 / x = 5 is multiplied by x ^ 2 + 1 / (x ^ 2) = 23, then x ^ 3 + 1 / (x ^ 3) + X + 1 / x = 115, that is, x ^ 3 + 1 / (x ^ 3) + 5 = 115

50 square aluminum core cable, length is 600 meters, line voltage drop? Power is 60 kW

The impact is not very big. The positive and negative is about 10, and the voltage will be higher when the transformer comes out. It can withstand

If | a + 1 | + | B + 2 | + | C + 3 | = 0, what is the value of the algebraic formula (a + 1) (B + 2) (c + 3)?

Because | a + 1 | + | B + 2 | + | C + 3 | = 0
And a + 1 ≥ 0, B + 2 ≥ 0, C + 3 ≥ 0
So a + 1 = 0, B + 2 = 0, C + 3 = 0
So a = - 1, B = - 2, C = - 3
Bring it into (a + 1) (B + 2) (c + 3)
The result is 0

The panel of a household electric energy meter is marked with 3000r / kWh. When an electric appliance is connected and the turntable turns 30 turns in 6 minutes, what is the electric power of the electric appliance?

The turntable turns 30 turns in 6 minutes, w = (30 / 3000) kwh = 0.01 kwh
The electric power of electric appliance is p = w / T = 0.01kwh/0.1h = 0.1kw = 100W

First decompose the factor, then calculate and evaluate (1) (a + B / 2) &# 178; - (a-b / 2) &# 178;, a = - 1 / 8, B = 2
(2) X & # 178; - 16 / 8-2x, where x = - 2, x + y = 1, find the value of 1 / 2x & # 178; + XY + 1 / 2Y & # 178

(a+b/2)²-(a-b/2)²,
=(a+b/2+a-b/2)(a+b/2-a+b/2)
=2axb
=2ab
=2x(-1/8)x2
=-1/2
x²-16/8-2x
=(x+4)(x-4)/2(4-x)
=-(x+4)/2
=-(-2+4)/2
=-1
1/2x²+xy+1/2y²
=1/2(x+y)²
=1/2x1²
=1/2

One cubic decimeter of iron weighs 7.8 kg. There is a rectangular iron block with a square bottom, which weighs 93.6 kg. It is known that the side length of a square is 20 cm
How many centimeters is the length of this piece of iron

Volume = 93.6 △ 7.8 = 12 cubic decimeter
20 cm = 2 decimeters
Length = 12 ﹣ 2 ﹣ 2 = 3DM = 30cm

Four numbers are 3, - 5,2, - 13, 24 points. What's the formula

-5+3=-2 2*（-13）=-26 -2-（-26）=24

When studying "the relationship between current and resistance under a certain voltage", the circuit is shown in the figure. The power supply voltage is constant 3V, and the sliding rheostat is marked with "15 Ω & nbsp; When the resistance of 20 Ω is connected between a and B, the voltage indication can not reach 1.5V. The reason may be ()
A. The resistance of sliding rheostat is too small B. the power supply voltage is too high 3V C. the resistance of 20 Ω is too small D. the control voltage is too high 1.5V

According to Ohm's law, RPR = upur can be obtained. When a resistance of 20 Ω is connected between a and B, the voltage indication is always unable to reach 1.5V, because the maximum resistance of sliding rheostat is 15 Ω, and the resistance value of 20 Ω is greater than 15 Ω, so the indication of voltmeter will only be greater than 1.5V, so the resistance of sliding rheostat is too small. At the same time, if the power supply voltage is less than 3V, the voltage of sliding rheostat is too small The indication of the voltmeter may also reach 1.5V, so the power supply voltage 3V is too high. Then AB is correct. If 20 Ω is too large, so there are too many partial voltages, the indication of the voltmeter will always be greater than 1.5V, so C is wrong. If the control voltage 1.5V is not high, because the indication of the voltmeter in this experiment is always greater than 1.5V, rather than less than 1.5V, so D is wrong

Help me see if {n = 2 ^ (2n) + 7} is a prime number (n is a positive integer), find out a counterexample, or use the computer to prove that it is true within one million
Give a good answer

2 ^ 12 + 7 = 4096 + 7 = 4103 = 11 × 373; so when n = 6, 2 ^ (2n) + 7 is a composite number