Can a 220 V, 100 W incandescent lamp light up normally when connected to a 220 V DC power supply? Why?

Can a 220 V, 100 W incandescent lamp light up normally when connected to a 220 V DC power supply? Why?

It glows,
There is a loop
Analogy between two dry batteries and small light bulbs
Furthermore, luminescence is actually one of the manifestations of the thermal effect of resistance, and the reason for the thermal effect is that when the current flows through the resistance, there is a change in the form of energy, that is, electric energy turns into heat energy. As long as there is a loop, this transformation will continue. In addition, what we usually call 200V AC should be a 200V sinusoidal AC voltage with an effective value of 50 Hz, The so-called effective value means that when the sinusoidal voltage and the 220 V DC voltage are applied to the same resistance, the thermal effect is the same. Therefore, not only can it emit light, but also the brightness is the same in both cases

Simple calculation: how easy is 4 - (4 / 5-11 / 15)

4 - (4 / 5-11 / 15)
=4 - (12 / 15-11 / 15)
=4-1 / 15
=3 and 14 / 15

When a motor is connected to a 220 V power supply, the power consumed by the motor is 4.4 kW, and the coil resistance is 2 Ω?

P=UI
4.4*10^3=220I
I=20A
Power of coil: P '= I ^ 2 * r = 20 ^ 2 * 2 = 800W
Therefore, the normal working power of the motor is: P-P '= 4400-800 = 3600W

[1] 2,8 times [1 / 7 plus 3 / 28]. [2] 101 times 99 / 102 [3] 55000 divided by 25 divided by 4 [4] 23 times 14 / 39 plus 14 times 16 / 39

[1] 2,8 times [1 / 7 plus 3 / 28]
=2.8X1/7+2.8X3/28
=0.4+0.3
=0.7
[2] 101 times 99 / 102
=(102-1)X99/102
=102X99/102-1X99/102
=99-99/102
=98 and 1 / 34
[3] 55000 divided by 25 divided by 4
＝55000X25X4
=55000X(25X4)
=55000X100
=5500000
[4] 23 by 14 out of 39 plus 14 by 16 out of 39
=14 / 23x39 + 14 / 16x39
=14x of 39 (23 + 16)
=14x39 out of 39
=14

Why is 1 kcal equal to 4185 joules?
I know the definition of kcal and Joule, but I don't know why we need to convert these two different units

Kcal is the industrial unit, j is the physical unit. In fact, you know their definitions. It's not difficult to understand the physical definition. The heat required for 1kg water to rise 1 ℃ is q = cm Δ t = 4.2 × 10 ^ 3j / (kg ·℃) × 1kg × 1 ℃. According to the definition of CAL, this value is actually = 1kcal, that is, 1kcal = 4200j

16-40% x = 80 to solve the equation

16-40%x=80
-0.4x=80-16
-0.4x=64
x=-160

Wind turbine is a device that converts wind energy (kinetic energy of air flow) into electrical energy. Its main components include wind turbine, gearbox, generator, etc. (as shown in the figure). When wind power is generated, the blades of wind turbine rotate after being pushed by the wind, thus driving the generator to generate electricity, When the wind speed is 8 m / s, the air mass flowing to the blade rotation area per second is 7 200 kg, and the air density is 1.2 kg / m3; It is known that the air kinetic energy is directly proportional to the air mass, and the kinetic energy (E) of 1kg air at different speeds (V) is shown in the following table. E (J) 28183250 V (M / s) 246810 (1) when the wind speed is 8m / s, what is the volume of air passing through the blade rotation area per second? If the calorific value of coal is 3.2 × 107 J / kg, then the wind energy obtained by such a wind turbine in one hour is equivalent to the internal energy generated by the complete combustion of how many kilograms of coal? (2) When the wind speed is 10 m / s, what is the mass of air flowing to the blade rotating area in 1 s? In this case, if the wind turbine can provide 5.4 × 104j of electric energy, then what is the efficiency of the wind energy into electric energy of the wind turbine? (3) Scientists believe that the ideal air power plant should be built at an altitude of 4600m to 10000m. Why?

(1) According to ρ = MV, v = m, ρ = 7200kg, 1.2kg/m3 = 6000m3, the energy produced by air in one hour is q = 32j / kg × 7200kg / s × 3600s = 8.2944 × 108j, the wind energy obtained in one hour is quite complete combustion of coal: M = QQ = 8.2944 × 108j3.2 × 107j / kg = 25.92kg. (2) suppose that the rotating area of the fan blade is s, the wind speed is V, the air flow through the fan in unit time t is v = SL = s · VT, then the air flow through the fan is v The mass of air is: M = ρ v = ρ s · VT; when the air density, the rotating area of the fan blade and the time are the same, m and V are proportional; that is: m ′ V ′ = MV, m ′ = MV ′ v = 7200kg, 8m / s × 10m / S = 9000kg. According to the table, the energy generated is: e wind = 50J / kg × 9000kg = 4.5 × 105J, and the efficiency of wind energy conversion into electric energy of this wind turbine is: η = e wind × 100% = 5.4 × 104j4 5 × 105J × 100% = 12%. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (3) because the higher the altitude is, the higher the air flow velocity is, the more energy is generated, so as to improve the conversion efficiency of wind turbine, so the ideal air power plant should be built in the air with an altitude of 4600m to 10000m

Factorization: A & # 178; B (a-b) - 2Ab (B-A); (a + 1) &# 178; - (A & # 178; + a) &# 178;

a²b(a-b)-2ab(b-a)
=a²b(a-b)+2ab(a-b)
=(a-b)(a²b+2ab)
=ab(a-b)(a+2)
（a+1）²-（a²+a）²
=[(a+1)+(a²+a)][(a+1)-(a²+a)]
=[a²+2a+1][1-a²]
=(a+1)²(1-a)(1+a)
=-(a+1)²(a-1)(a+1)

Excuse me, how many calories is a large calorie?

Calorie is the unit of calculation of heat. 1 calorie is the amount of heat needed to raise the temperature of 1 gram of water by 1 degree centigrade after the food we eat is burned. Kilocalorie is the amount of heat needed to raise the temperature of 1 liter of water by 1 degree centigrade. Kilocalorie is also called large calorie (one large calorie equals one thousand calories)

x²-13x+40=0
Using factorization to do, find the process

x²-13x+40=0
（x-8)(x-5)=0
X = 8 or x = 5