As shown in figure a is the circuit diagram of "measuring the rated power of small bulb", the power supply is two dry batteries; the small bulb is marked with the word "2.5V", the resistance is about 10 ohm, and it is slippery

As shown in figure a is the circuit diagram of "measuring the rated power of small bulb", the power supply is two dry batteries; the small bulb is marked with the word "2.5V", the resistance is about 10 ohm, and it is slippery


Because the resistance of the filament increases with the temperature, it does not meet the control variable conditions



The equation of a straight line passing through point a (0,3) and having only one common point with hyperbola x ^ 2 / 4-y ^ 2 / 9 = 1


Follow the site oblique, set to
y=kx+3
Substitute into hyperbolic equation
We get a function of degree 2
If there is only one intersection, then b-4ac = 0
The solution is k = the root of plus or minus 2 / 6
So the equation is y = 2 / root 6x + 3 or y = negative 2 / root 6x + 3
… If I'm right, I'll give it to you~



Copper wire of the same length has different thickness, which has higher resistance


The resistance is directly proportional to the length and inversely proportional to the cross-sectional area



Given that F1 is an ellipse X & # 178;; / 25 + Y & # 178;; / 9 = 1 in the left focus, P is the moving point on the ellipse, and a (1,1) is a certain point, what is the maximum value of PA + Pf1


First, through checking calculation, we know that a (1,1) is inside the ellipse; if the major axis 2a of the ellipse is 10, and the right focus coordinate F2 (4,0), then af2 = √ [(1-4) ^ 2 + (1-0) ^ 2] = √ 10; so the maximum pa + Pf1 = 2A + af2 = 10 + √ 10; as shown in the figure above, the sum of the distances from any point of the ellipse to a and F1 is P & # 39; a + P & # 39; F1 & lt; (P & # 39; F2 +



How many watts is one degree of electricity? Can we change the wattage of our light into degree? Take 25 watts as an example


1 degree of electricity = 1 kWh
The wattage of the light we use multiplied by time can be changed into degrees
Take 25 watts as an example, it costs 1 kilowatt hour to turn on the light continuously for 40 hours
25 * 40 = 1000 watt hours



1. Given that the moving point P (x, y) satisfies | X-1 | + | Y-A | = 1, O is the origin of the coordinate, if the value range of the maximum value of | Po | is [root 13, root 17], then the value range of the real number a is____ .
The answer is [- 3, - 1 / 2] and [1 / 2,3]
2. Given the ellipse x * 2 / 16 + y ^ 2 / 4 = 1, the left and right focal points are F1 and F2 respectively, and the point P is on the straight line L: X - (radical 3) * y + 8 + 2 * (radical 2) = 0. When ∠ f1pf2 takes the maximum value, then the value of | Pf1 | / | PF2 | is____ .
Root three minus one


1. It's easy to see which is the farthest point and which is the nearest point
2. I don't know which is rooted in the second question



The rated voltage is 220 V, and the rated power is 100 W and 250 W respectively. The lamp is connected to the power supply in series. Which lamp is on? Parallel connection? Detailed process ≡ [.] ≡


Series known: 100W bulb resistance is large, 250W bulb resistance is small, so when in series, the current is the same, according to the series voltage, 100W bulb voltage is high and P = UI, so the actual power P 100W bulb power is greater than 250W bulb power, so 100W bright parallel shunt voltage constant is the rated voltage



How to use 23.5 * 4 + 4.77/0.25 for simple calculation
This question is divided by 0.25, not multiplied


0.25 = 1 / 4, so 23.5 * 4 + 4.77 / 0.25 = (23.5 + 4.77) * 4



Why is the resistance calculated by Ohm's law when the motor is working larger than that calculated by Ohm's law when it is not working?
I know that Ohm's law cannot be used in non pure resistance circuits
The power of electricity has also increased


When the motor is working, the electric energy is mainly converted into mechanical energy, which is a non pure resistance circuit. Ohm's law can not be used to calculate the pure resistance, but the impedance value of the coil is large; when the motor is not working (stop rotating), the power consumption is small, and the electric energy is converted into internal energy, which is calculated to calculate the resistance value is small



0.24 * 15 / (3.25-2.35) recursive equation calculation
Sorry! I'll ask you another question!


=0.24×15÷0.9
=3.6÷0.9
=4