Xiao Wang's house is decorated. He goes to the store to buy lamps. There are 1000 watt incandescent lamps and 40 Watt energy-saving lamps in the counter. What's the unit price of them 2 yuan and 32 yuan. It is known that the lighting effect and service life of these two kinds of lamps are the same. The electricity price of a little Wang's home is 0.5 yuan per hour. When the service life of these two kinds of lamps exceeds how long, it is worthwhile to choose energy-saving lamps for small lamps? 2+0.5*(100/1000)x>32+0.5x(40/1000)x What is 1000 here?

Xiao Wang's house is decorated. He goes to the store to buy lamps. There are 1000 watt incandescent lamps and 40 Watt energy-saving lamps in the counter. What's the unit price of them 2 yuan and 32 yuan. It is known that the lighting effect and service life of these two kinds of lamps are the same. The electricity price of a little Wang's home is 0.5 yuan per hour. When the service life of these two kinds of lamps exceeds how long, it is worthwhile to choose energy-saving lamps for small lamps? 2+0.5*(100/1000)x>32+0.5x(40/1000)x What is 1000 here?

Here, dividing by 1000 is to change the power from watt to kilowatt, and then multiply it by time hour to get kilowatt hour, which is what we call degree

Factorization of (a + 4b) - 16a and B

Why are lamps of the same voltage and power connected in parallel in the home circuit? The new bulb is brighter than the old one

Due to the sublimation of the filament, the filament of the old bulb becomes thinner and the resistance becomes larger. According to P = u & sup2 / R, the voltage remains unchanged, the resistance increases and the actual power decreases. Therefore, the lamps with the same voltage and power are connected in parallel in the home circuit, and the new bulb is brighter than the old bulb (the brightness of the lamp bulb is determined by its actual power)

First simplify, then evaluate (3m-5m + 4Mn) - (2m-4n + 6mm)~
What is the result when M-N = 7 and Mn = - 5

(3M-5M+4MN)-(2M-4N+6MM)
=3M-5M+4MN-2M+4N-6MN
=4N-2MN

Physics electricity, electric power part
1. Connect a small bulb marked "6V 3W" to a certain circuit, and use an ammeter to measure the current passing through the filament is 450mA. Then what is the electric power consumed by the lamp?
2. Two bulbs A and B marked "6V 2W" and "12V 2W" can work safely after they are connected into the circuit, then their actual power P A and P B are compared ()
A. In series, p a > P B
B. In series, P A

1. Bulb resistance is 6 * 6 / 3 = 12 ohm, power is 0.45 * 0.45 * 12 = 2.43w
2. The resistance of a is 6 * 6 / 2 = 18 Ω, and the resistance of B is 12 * 12 / 2 = 72. Therefore, in the series circuit, the partial voltage of B is greater than that of a, and the currents of the two are the same. According to P = UI, the power of a is less than that of B

As shown in the figure, the object a is placed on the horizontal ground, GA = 100N, GB = 40n, the stiffness coefficient of the spring is 500N / m, regardless of the rope weight and friction, calculate the pressure of the object a on the supporting surface and the elongation of the spring

A is subject to gravity GA, ground supporting force N, spring tension F. from the figure, we can see that: F = GB, x = 0.08m from F = KX, for a: F + n = GA, n = 60N, according to Newton's third law, the ground pressure of a is 60N. Answer: the pressure of object a on supporting surface is 60N, and the elongation of spring is 0.08m

In the circuit with inductor and capacitor, the reason, understanding and image description of lead lag phenomenon of voltage and current?
The book says: "in inductive circuits, the voltage is 90 degrees ahead of the current; in capacitive circuits, the current is 90 degrees ahead of the voltage."
After many searches, the following correlation may be useful
1 said: "it's the advance of waveform, not the advance of time!"
"If there is no charge on the capacitor, when the charging circuit is just connected, the voltage on the capacitor is 0, and the charging current of the capacitor is I = (e-uc) / R, e is the power supply voltage, UC is the voltage on the capacitor, and R is the resistance of the charging circuit. It can be seen that the smaller UC is, the greater I is, With the increase of capacitor UC, the charging current decreases gradually. When UC is close to e, I is close to 0
I don't know whether the two theories are correct or not
First of all, I don't understand under what conditions this lead lag is studied? (voltage at both ends of inductor and capacitor and current in the line? Where is the current? Total current?)
If it is the lead and lag on the waveform, how can we get it?
The more I think, the more confused I am. I hope my good friend will explain it in detail. Thank you again
Brush points of friends please note brush points here also thank you

I invented an explanation method: apply the concept of inertia in mechanics to explain
"Magnetic field inertia" leads to inductance current lag: the physical meaning of inductance current lagging behind voltage is that when the current passes through the inductance, a new magnetic field will be formed. When the new magnetic field is established, the magnetic inertia of the old magnetic field will hinder the establishment of the new magnetic field, that is, hinder the current flow. Therefore, when the voltage is added, the current can not be formed immediately, and it needs a period of time to overcome the magnetic inertia, So there is a lag phenomenon. The 90 degree lag is because the voltage begins to flip at 90 degrees, that is, the direction of voltage changes, resulting in the change of current lagging 90 degrees. It is not convenient to draw, otherwise it is easy to understand
Similarly, the current on the capacitor is 90 degrees ahead of the voltage, which is caused by the "electric field inertia": when the capacitor is initially charged, the voltage of the old electric field is 0. When the current flows into the capacitor to accumulate charge, a new electric field (voltage) can be formed, so the establishment of the new electric field will certainly lag behind the current inflow. After the current flows for a period of time, a certain amount of charge is accumulated on the electrode plate of the capacitor, The new electric field can be formed slowly, so there is a voltage lag phenomenon. The 90 degree relationship between them is the same as inductance

The car B is parked on the smooth horizontal ground. There is a small slider a (which can be regarded as a particle) on the left end of the car. The dynamic friction coefficient between a and B is μ = 0.6, if l is small
=1m, Ma = 1kg, MB = 3kg
(1) If the trolley is pulled to the left with a horizontal force of F = 10, the displacement of the block within 2S can be calculated
(2) If the horizontal force of F = 10 is used to pull the block to the right, the displacement of the trolley within 2S can be calculated

The key to solve the problem is to judge whether the wooden block and the small workshop will slide relative to each other. (1) when the horizontal force of F = 10 is used to pull the car to the left, assuming that the wooden block and the car are relatively stationary, the overall acceleration is A1 = f / (MA + MB) = 10 / (1 + 3) = 2.5 m / S ^ 2. The maximum acceleration of the wooden block caused by the friction force F is a wooden block

It is known that the area of a right triangle is 4, and the lengths of two right sides are 2x and Y respectively, then the functional relationship between Y and X is obtained

I think:
The area of a triangle is 1 / 2 times that of a right angle
That is: 1 / 2 (2x * y) = 4
XY=4
Y=4/x

The upper end of a light spring is fixed, and the lower end is hung with a weight. When balancing, the spring is extended by 4cm, and then the weight is pulled down by 1cm, and then let go,
What is the acceleration of the weight at the moment of release?

Suppose that the stiffness coefficient of the spring is k, the mass of the weight is m, and the spring is extended by X1 = 4cm during balance,
kx1=mg
Then pull down the weight for 1cm, and the elongation of the spring is x2 = 5cm
The elastic force F2 = kx2 can get f = 5mg / 4
The acceleration a = (f-mg) / M = g / 4 at the moment when the weight is just released, and the direction is upward