Given power and time in physics, how to calculate how many kilowatts of electricity are used?

Given power and time in physics, how to calculate how many kilowatts of electricity are used?

W = Pt, the calculated electric energy unit is Joule. 1 degree of electricity = 3.6x10 ^ 6 Joule, so after calculating the electric energy, divide it by 3.6x10 ^ 6 Joule / degree of electricity

Physics power calculation
As shown in the figure, under the action of horizontal tension F, object a moves in a straight line at a speed of 2m / s along the horizontal plane through the moving pulley. At this time, the friction force on object a is 10N. If the gravity on pulley and rope, the friction between rope and pulley, and between pulley and shaft are ignored, the size of tension F and the work power of tension f are
The power is: P = w / T = FV = F &; 2V = 5N × 2 × 2m / S = 20W
I want to know why the V is going to take two here?

Because of the moving pulley, the rope moves twice as fast as the object

Motor (motor, motor) power calculation
If you want to make a board weighing 3 or 4 kg (6 kg or 8 kg) up and down angle adjustment movement, how much power does the motor need?
How much power does the motor need to adjust the 180 degree angle of a board weighing up to 1 kg? Do you need to consider the length and width of the board?
How to calculate the power and power consumption rate? It's better to put it in a popular way. I don't know much about electronics

It's enough to have tens of watts of electric vehicles. 750W is one horsepower (about the power that an adult horse can give, such as pulling a mill, pulling a cart, etc.) you can imagine. But the specific choice depends on the speed you want to adjust. In fact, a low-power motor can

Some problems about physical electric power
1、 Under the same voltage, isn't the resistance inversely proportional to the electric power? So the higher the resistance, the smaller the electric power. Why can't we say that the brightness of a light bulb depends on its resistance, but only on its electric power
2、 When an electric appliance turns 3000 turns, it consumes 1 kilowatt hour. He turns 45 turns in one minute. What is his power (use the proportional formula to answer) and explain the reason in detail
3、 (6V 3W) (6V 6W) two bulbs are connected in series to the electric appliance (which one can work normally determines the principle of working normally)
4、 (6V, 6V) (6V, 12W) two lamps are connected in series with constant voltage. Why is the total power of the two lamps 4W
Is the total power in the series circuit not the sum of the actual power? Why divide the total resistance of the two bulbs by the actual voltage
OK, I will add points. Please give a brief answer. It's too long for me to understand

1、 Under the same voltage, isn't the resistance inversely proportional to the electric power? So the higher the resistance, the smaller the electric power. Why can't we say that the brightness of a light bulb depends on its resistance, but only on its electric power
When the voltage is fixed, the power is fixed. P = u ^ 2 / r r is the inherent property of conductor
The brighter the bulb is, the more work is done per unit time, that is to say, the greater the current passing through the conductor
2、 When an electric appliance turns 3000 turns, it consumes 1 kilowatt hour. He turns 45 turns in one minute. What is his power (use the proportional formula to answer) and explain the reason in detail
T = 3000 / 45 = 200 / 3 minutes = 4000 seconds
W = 1 kwh = 1000 * 3600 joules
So p = w / T = 900 watts
3、 (6V 3W) (6V 6W) two bulbs are connected in series to the electric appliance (which one can work normally determines the principle of working normally)
I1=3/6=0.5A
I2=6/6=1A
When the current is greater than 0.5, the 3W bulb will burn out
4、 (6V, 6V) (6V, 12W) two lamps are connected in series with constant voltage. Why is the total power of the two lamps 4W
Is the total power in the series circuit not the sum of the actual power? Why divide the total resistance of the two bulbs by the actual voltage
You can only assume that the voltage is 6V, so the solution is as follows:
R1 = 6 Ω
R2 = 3 Euro
R = R1 + R2 = 9 Ω
P=U^2/R=36/9=4W