How to use the electric energy meter to calculate the electric power The watt hour meter shows "2500r, kW. H, 220 V, 5A". A bulb is connected to the circuit. The meter turns 5 circles in 3 min to find out the energy of the bulb?

How to use the electric energy meter to calculate the electric power The watt hour meter shows "2500r, kW. H, 220 V, 5A". A bulb is connected to the circuit. The meter turns 5 circles in 3 min to find out the energy of the bulb?

An electric energy meter is marked with "2000r / kW. H". If the turntable of the electric energy meter rotates 1000 times, what is the electric power?

Setting: the electric energy is xkw. H
1000/X=2000/1
The solution is: x = 0.5KW. H
Conversion: 0.5KW. H = 0.5 * 3.6 * 10 ^ 6J = 1.8 * 10 ^ 6J

The watt hour meter is marked with 3000r / kW · h. If a light bulb is connected under the watt hour meter, the power of the light bulb can be calculated by turning the turntable around 300r after being electrified for 1H

Because 300 revolutions have been turned, the work w = 300 / 3000 = 0.1kw · h, then the power is: P = w / T = 0.1 / 1 = 0.1kw = 100W
Don't know to ask me, hope to help you

If the area of a trapezoid is divided into two parts by a diagonal line, what is the area ratio of the two parts of the trapezoid divided by the median line?
It's the process that matters

If the area of the trapezoid is divided by a diagonal line into two parts of 1:2
The height of the two parts is the same, and the ratio of top to bottom is 1:2
Upper bottom: median line: lower bottom = 2:3:4
The ratio of the area of two parts of a trapezoid divided by a median line
=1 / 2 (upper bottom + median line) height: 1 / 2 (median line + lower bottom) height
=5/7

Problems of electric power in series and parallel circuits
In parallel circuit, the light bulb with higher electric power is brighter, and the resistance is higher? In series circuit?

Parallel electric power on the bright series opposite
Because the series connection lies in the voltage division, the resistance with high rated power is small and the voltage division is small, with P = U2 / R and the voltage as the dominant position, so this is the case

If the median length of a trapezoid is 3 and the height is 2, what is the area of the trapezoid?

3*2=6
Trapezoid area = (upper bottom + lower bottom) XH / 2
Midline = (upper bottom + lower bottom) / 2

In series circuit or parallel circuit, is the total electric power or the total electric power equal to the electric power or the sum of electric power of each consumer?
Elder brothers and sisters, please be more detailed. I didn't write such a question in senior high school entrance examination

No matter what circuit, the total electric power must be equal to the sum of the electric power of each consumer
Of course, to be precise, we also need to add the loss of lines (including switches, etc.)
LZ: sorry, I can't explain this in more detail. Think about the conservation of energy

If the median line length of the trapezoid is 5 and the height is 3, the area of the trapezoid is ()

The median line of trapezoid = (upper bottom + lower bottom) △ 2, the area of trapezoid = (upper bottom + lower bottom) × height △ 2, so the area of trapezoid = 15

Calculation formula of junior high school physical watt hour meter

Electric work W: the work done by current is called electric work. The process of electric work is the conversion of electric energy into other forms of energy. Formula: w = UQ, w = uit = u2t / r = i2rt, w = Pt unit: W Joule, u volt, I ampere, t second, Q library, P Watt, 2 electric power P: the electric work done by current in unit time, indicating the speed of electric work done by current

If the trapezoid has an area of 6 and a height of 3, then the median line length of the trapezoid is

(upper bottom + lower bottom) * height / 2 = area
Now that we know the area and the height, we can find out: upper bottom + lower bottom = 4
So the median = (bottom + top) / 2 = 2