Questions about a junior high school physics problem? Not asking the answer is questioning .

Hello, there is nothing wrong with this question.

Don't be fooled by illusions.

This question is not so cheap as it seems, he gave a diagram, then this is the point.

Pay attention to the conditions given by the question, there will be no conditions that have no effect, which is also a basic method of doing physics.

This is a junior high school question, so the power loss is not considered.

Through the image (this image you can know in high school, which is also correct), we know that when the voltage is 6V, the current is this regardless of him, because this circuit diagram takes into account the loss of the power supply, with 36 is not a resistor), but the next question condition is that 3 identical bulbs are connected in series, so what is the law of the series circuit? Partial voltage, and this problem is equal partial voltage, that is, each bulb is 4V, the current is, then the resistance is 20! The actual power is it!

Do you understand?

From the given image, the resistance of the bulb increases with its voltage, that is, the resistance of the lamp will change, so the correct answer is required, the actual voltage and the corresponding actual current must be found, and then the two are multiplied to be the actual power of the lamp.

Note: Because the three lamps are the same, the total voltage is 12V, the voltage of each lamp is 4V, and 4V and the corresponding current are found in the image, so the actual power of a lamp is 4*W, and then the voltage and current at this time are used to find the lamp resistance is r 4 ohms.

Hello! The resistance of the bulb can be calculated by the amount of u and p when the voltage is rated or the question does not require it, but this question gives an image, and the image should prevail.

The book says that the resistance value is constant in the ideal state, and the linear law is a straight line, but this question is not a straight line.

This question gives a graph, which should be calculated based on the graph data.

The 3 bulbs have the same size, the same resistance, and the same linearity.

Add 12V The average light bulb has a voltage of 4V

Therefore, it is calculated according to the data at 4V.

You are counting the resistance of the bulb at the rated voltage. The actual resistance of a bulb is related to its power, the higher the power, the higher the filament temperature, the greater the filament resistance. The resistance at 6V you calculate is different from the resistance when the bulb actually gets 4V voltage, so you should find the current corresponding to 4V from the diagram and substitute it into the formula.

The image given in the question is not proportional, indicating that the resistance of the lamp should be considered to increase with increasing temperature. And the filament resistance calculated by the formula u and p is the resistance of the lamp when it works at the rated voltage. The three bulbs of the same size in this question all operate at 4V, so 24 cannot be used to calculate.

The voltage of 12V is applied to three identical bulbs, so the voltage at both ends of each bulb is equal (4V), and according to the image, the current is at this time, so p=u*i=4* r=ui=4 ohms.

Summary. It is not right to convert electrical energy into internal energy, it is mainly converted into light energy, which is the role of lighting, and secondly, it will generate heat in the process of its conversion, which is heat energy.

Ask a junior high school physics question.

Does the light bulb convert all the electrical energy into internal energy, and then part of the internal energy into light energy?

Does the light bulb convert all the electrical energy into internal energy, and then part of the internal energy into light energy?

Most of the electrical energy of the light bulb is converted into heat.

There is also light energy, light energy is the main one, and heat energy is secondary.

It is not right to convert electrical energy into internal energy, it is mainly converted into light energy, which is the role of lighting, and secondly, it will generate heat in the process of its conversion, which is heat energy.

Then why can light bulbs use Joule's law.

It is not all converted into internal energy.

Why? The reason for kissing is that he first converts all of it into internal energy, and then continues to convert it outward into light and heat.

So he can use Joule's law.

Good. The light bulb is a pure electrical resistor that emits light by generating heat, while joules are used to calculate heat, so Joule's law is used.

When the two glasses are at the same height, the water stops flowing.

When the water surface of cup A is lower than the water surface of cup B, cup B water flows to cup A.

This is the effect of atmospheric pressure, which causes a pressure difference inside the hose liquid.

Take a liquid sheet at the top of the hose, the pressure of the liquid sheet in the direction of A is greater than the pressure in the direction of Party B, the height of the water column in the hose above the cup of water is small, and the downward pressure is small, and the liquid sheet at the top of the hose is pressured in the direction of A.

This pressure is equal to the atmospheric pressure minus the pressure generated by the water column in the hose above the cup of water).

The height of the water column in the hose above the first glass of water is large, and the downward pressure is strong, and the pressure on the top of the hose is small in the direction of Party B.

This pressure is equal to the atmospheric pressure minus the pressure generated by the water column in the hose above the first cup of water).

The stress area on both sides of the liquid sheet at the top of the hose is equal, because there is a pressure difference between the two sides of the liquid sheet, so there is a pressure difference.

The pressure on the top of the hose is greater than that on Party B. So the water flows in the direction of cup A.

The correct option is: a

Because a small black dot is stained on the convex lens, because the small black dot is located within the collection distance of the convex lens, it will not form a real image on the light screen. So there won't be a small black dot on the light screen. If there is no small black dot, the image of the candle flame will be slightly darker, because the small black dot will block some of the light coming to the light screen.

Hope it helps.

The answer is a

However, the brightness of the image is lower than that of the convex lens without black spots, that is, the image is slightly darker.

The reason for this is that the light that should be reflected behind the lens is blocked.

Choose a for this question. The reason is that in the image of the candle flame, it is crossed through the convex lens, not horizontally through the convex lens.

a Because it does not affect the refractive imaging of light in other positions, even if part of the lens is obscured.

The D-ball sinks to the bottom and has pressure on the bottom of the container So the pressure received at the bottom of the container where the D-ball is located is not only the pressure of the water but also the pressure of the ball.

The pressure received at the bottom of the container where the ABC ball is located is only the pressure of water, so fa=fb=fc< FD option b hope to adopt.

Why is it wrong. It is because of c, the pressure at the bottom of the four containers, which is the total pressure and refers to the pressure of water on the bottom of the container.

If you refer to the pressure of water on the bottom of the container alone, according to p=hpg, both h and p are the same, and p is equal.

Whereas find the pressure of the container on the horizontal plane. If used.

Pressure = water weight in the vessel + ball weight + weight.

Both water weight and ball weight are different.

But the buoyancy of the ball in containers A, B, and C = the weight of the ball = the gravitational force of the water being discharged, so the (water weight + ball weight) in these three containers is equal, but the buoyancy force in container D is "ball weight.

The gravity of the water is displaced.

So the water weight in the container + the ball weight of the other three containers, so b is correct.

Option A: When the ball is put into the water, if the ball is in a state of floating or suspension, then the weight of the ball is equal to the weight of the ball in the water, that is, the volume of the ball immersed in the water multiplied by the density of the water, which is also the principle of Cao Chong's image, then it is not difficult to get 1< 2< 3, and the fourth ball sinks to the bottom, its weight is greater than the weight of the same volume of water as it, so its density is the largest.

Option B: If you want to consider both the weight of the water and the weight of the ball, then obviously you will feel confused, since the A selection can equate the weight of the ball to a part of the water, then we can convert the ball into water, and then compare, obviously the three options A, B, C, when the ball is converted into the same weight of water, the height of the water does not change, and the D option, because the weight of the ball is greater than the weight of the same volume of water, so after the conversion, the height of the water in the cup will rise, so a comparison, it is not difficult to findWhether it is gravity, pressure, or pressure on the bottom of the cup, ABC is equal for all three cups, and D cup is the largest.

PS: In fact, this question is not rigorous enough, although the D ball sinks to the bottom, but it is not said that the D ball has pressure on the bottom of the cup, so the D ball may also be suspended.

It seems that the answer is wrong, and what you said is right. Because there is no conversion between units, it should be counted directly.

Yours is right. The answer is correct, the calculation is wrong.

The method is correct, and the final calculation result is wrong.

The end result was 163333, and the answer was wrong.

Answer: The correct one is A.

a.The nature of microwaves is closer to that of light waves, and they probably propagate in a straight line (at the beginning of the second paragraph on page 108 of the second volume of the second volume of the Physics Textbook).

b.By evenly deploying 3 synchronous satellites around the earth, global communication can be realized, (false!) Because there are many satellites, not all satellites can do, and they must be communication satellites, so it is changed to "3 synchronous [communication] satellites" (the fourth line of the last paragraph of page 108 of the second volume of the physics textbook).

c.Optical fiber communication is to use ordinary light sources to transmit information in optical cables (false!). If the ordinary light source is not good, it must be a single frequency.

1. Lasers with a high degree of concentration in the direction) (the penultimate line of the first paragraph on page 109 of the second volume of the physics textbook of the People's Education Edition).

d, the most commonly used form of network communication is e-mail (false!). It's just often used.,It's not "the most commonly used".,It's not the same with one more word.,Hehe)(The beginning of the second paragraph on page 112 of the second volume of the physics textbook of the People's Education Edition)

I'm a teacher, I hope I can help you, thank you!

Well, I'm in the second year of junior high school, but we learned power in the second year of junior high school. However, I can tell you that microwaves are electromagnetic waves, and the speed of electromagnetic waves is as fast as the speed of light traveling in a vacuum, so microwaves are close to light.

c is not using an ordinary light source, it is a laser beam.

Here's leverage.

20r1=30r2

r1+r2=

Get. r1=r2=That's right. Hehe.

If Xiao Ming closes the switch without checking the circuit, the phenomenon that will occur is a short circuit.

The current in the two lamps is, in addition to the error in (1), there is also a deficiency or error in the range selection.

Burn out ammeter A1, A2. The range of ammeter A1 and A2 is too large, and the gear should be selected. Now the circuit is in series, not in parallel.

1. There is a short circuit and the power supply is burned.

2. The range of the A1 meter is selected incorrectly, and the application, the ammeter should not be connected in parallel with the electrical appliance, and should be replaced with a voltmeter.