There is an environment-friendly flashlight. There is no battery in the flashlight. When in use, as long as the flashlight is shaken back and forth to make the permanent magnet move back and forth through the coil between the two rubber pads in the flashlight, the bulb can emit light. This kind of flashlight can generate electricity______ Principle. The method to increase the brightness of the bulb is______ (just write one)

There is an environment-friendly flashlight. There is no battery in the flashlight. When in use, as long as the flashlight is shaken back and forth to make the permanent magnet move back and forth through the coil between the two rubber pads in the flashlight, the bulb can emit light. This kind of flashlight can generate electricity______ Principle. The method to increase the brightness of the bulb is______ (just write one)

When the permanent magnet moves in the coil, the coil cuts the magnetic induction line to produce induced current, so it is manufactured according to the principle of electromagnetic induction. In order to increase the brightness of the bulb, that is, the current becomes larger, the cutting can be accelerated, the number of coil turns can be increased, or the permanent magnet with stronger magnetism can be replaced
So the answer is: electromagnetic induction; Speed up the shaking, increase the number of coil turns or replace the permanent magnet with stronger magnetism (choose any one)

In the circuit, the power supply voltage remains unchanged, and R is a constant resistance. If a bulb L1 marked 3V 2W is connected between a and B, it can emit light normally; If a bulb L2 marked 3v3w is replaced between a and B, L2 A normally emits light and consumes electric power equal to 2W. B normally emits light and consumes electric power equal to 3W C. the electric power consumed by abnormal lighting is less than 3W. D. the electric power consumed by abnormal lighting is greater than 3W Note: the bulb between AB in the circuit is connected in series with the constant resistance. In the circuit, there is only one power supply, one custom resistance and one AB terminal (connected to the bulb)

Upstairs
What do you mean?
And P = UI
I think it should be c
The resistance can be calculated from the data given by the bulb
The calculated resistance shows that R1 is greater than R2
According to the principle of partial pressure
Therefore, the voltage at both ends of L2 becomes smaller
Therefore, it is less than 3V of rated voltage
It must be abnormal lighting
By P = u ²/ R know
U ² Smaller P smaller
Therefore, the electric power is less than 3W
Just give you a way to calculate. You have to do it yourself to make a deep impact

Calculation questions (5 points) The mass of the hollow copper ball with a volume of 30cm3 is m = 178g. After the hollow part is filled with a certain liquid, the total mass is m = 314g. What is the density of the injected liquid? （ ρ Copper = 8.9 × 103㎏/m3）

from ρ Copper = m copper
V copper = m copper
ρ Copper = 178g
8.9g/cm3=20cm3，
Volume of hollow part V liquid = 30cm3-20cm3 = 10cm3,
ρ Liquid = m liquid
Vliquid = 314g − 178g
10cm3=13.6g/cm3．
A: the density of the injected liquid is 13.6g/cm3

Cfxb "220V, 1100W" rice cooker has a shift switch, as shown in the figure, gear 2 is high-temperature cooking; Gear 1 is the heat preservation of stewed rice. When connected to gear 2, the power of the circuit is 1100 watts; When connected to gear 1, the total power of the circuit is 22 watts (the dotted box shows the simple schematic diagram of the rice cooker) Requirements: (1) Resistance value of series resistance R; (2) Power consumed by R in gear 1

(1) When the switch is connected to 2, the resistance R0 = u2p, high temperature = (220V) 21100w = 44 Ω; when the switch is connected to 1, the total resistance R = u2p, insulation = (220V) 222w = 2200 Ω, and the resistance value of resistance R = R total - R0 = 2200 Ω - 44 Ω = 2156 Ω; A: the resistance value of series resistance R is 2156 Ω. (2) in gear 1, the circuit current I = ur, total = 220v22

Connect the bulb L1 marked "8V, 16W" and the bulb L2 marked "12V, 36W" in series at both ends of the power supply. One bulb emits light normally and the other does not reach its rated power, then A. the power supply voltage is 20V The current in circuit B is 3a C the total power of the two bulbs is 32W I know that the L1 current is 2a and the L2 current is 3a. But I also need to ensure that the bulbs are safely connected in series and the current is equal in the circuit. Because the L1 current is less than L2, L1 emits light normally. That is, the actual electric power of L2 is 2 * 12 = 24W, so the total electric power is 16W + 24W = 40W. But the answer is 32W. It's best to write your own answer clearly~

The current of L2 is not 3a, so the voltage is not 12V
The actual electric power of L2 is 2 * 2 * (12 / 3) = 16W,
So 16W + 16W = 32W

Ask a physics question, junior high school, how to do it There are two bulbs marked with 12V 36W and 6V 6W respectively. Then: 1. Connect them in parallel to a circuit. If it is known that the power consumed by lamp a is 0.75w, what is the power consumed by lamp B? 2 what is the maximum current allowed for the main road when the lights of Party A and Party B are used in parallel? (then question 2) what is the maximum power consumed by the current?

Through the rated voltage and rated power of the two bulbs, the resistance of the two bulbs can be calculated as
R a = 122 / 36 = 4 Ω, R B = 62 / 6 = 6 Ω
1. According to the actual power of a, the actual voltage of a can be calculated as: U2 = (0.75 * 4) = 3
Because they are connected in parallel, their actual voltage is the same, so the actual power of B = U2 / R, B = 3 / 6 = 0.5W
2. The parallel voltage shall not exceed the rated voltage of the two, so the maximum parallel voltage can only be 6V
At this time, the actual current of B is equal to its rated current, which is 6 / 6 = 1a
Actual current of a = 6 / 4 = 1.5A
Therefore, the maximum current allowed for the trunk road is 1 + 1.5 = 2.5A