Please use knowledge of junior high school physics. Middle School Physics Quiz Questions

Back to Nantian"I don't understand what it means, but there are water droplets on the walls of the floor because the air is humid and the air temperature is higher than the temperature of the walls and floor. As a result, water vapor releases heat when it hits walls and floors, turning into liquid water and sticking to floors and wall surfaces.

The reason why there are water droplets outside the window is because the temperature inside the house is lower than the temperature outside the house, so the temperature of the glass is lower than the temperature of the air outside. The water vapor in the air outside the house gets into the glass, releases heat, turns into liquid water, and adheres to the glass.

Turning on the air conditioner in summer is different from spring, because before turning on the air conditioner in summer, the temperature in the house is high and the humidity is high, but the temperature outside the house is also high, so there will be no gods on the glass. After turning on the air conditioner, the indoor air temperature becomes low, and it should not be condensed inside the glass window, but on the heat exchanger of the air conditioner. Because the hot air in the house passes through the heat exchanger of the air conditioner, and the temperature of the heat exchanger of the air conditioner is lower than the indoor air temperature, the water vapor condenses on the heat exchanger.

The temperature of the air going out has been lowered and is still higher than the heat exchanger when it is mixed with the unexchanged air, so the air will not condense the water vapor into the glass window... You're not right about this summer's question ...

Because the air humidity is high (the air contains a lot of water vapor), the water vapor will liquefy into small water droplets when it encounters cold floors and walls, and the difference between inside and outside is only the liquefaction of water vapor inside or the liquefaction of water vapor outside.

This has something to do with temperature, because there is a temperature difference between indoor and outdoor. Hot air has a high content of water vapor and is converted into water droplets when cold.

Because the temperature inside the window is high, the temperature outside the window is low

The power supply voltage is 6V, which means that the voltage at both ends of the two parallel bulbs is U = 6V [the voltage of each branch in the parallel circuit is equal].

then L1 emits normal light, P1=3W, U=6V

1. The current of l1 i1 = p1 u =3 6 =

2. The power consumed by the bulb L1 per minute [60 seconds] w = p1*t =3 * 60 = 180j

3. The resistance of L2 r = UE2 PE = 144 3 = 48 ohms [the square of the rated voltage and the rated power].

The actual power of L2 P2 = U2 R = 36 48 = [the square resistance of the actual voltage].

Because the actual power is the factor that determines the brightness of the bulb, P1 >P2, so the bulb L1 is brighter

, [2009 Chongqing] Lin Gang of Aisi used the circuit shown in Figure 21 to conduct experiments, and when the switch S was closed, it was found that the brightness of the lamp L1 marked with "6V 3W" and the lamp marked with "12V 3W" L2 were not the same. The supply voltage is known to be 6V, regardless of the change in filament resistance value with temperature. Seeking:

1) What is the current through the bulb L1?

2) How much electricity does the light L2 consume per minute?

3) What is the actual power of the bulb L2? Which is brighter L1 or L2 in the circuit?

Solution: (1) The rated voltage of lamp L1 is U1 = 6V, the rated power P1 = 3 W, and the power supply voltage is 6V

Lamp L1 emits normal light, current through bulb: 2 minutes) (2) When lamp L1 emits normal light, it consumes electrical energy per minute:

w1 = u1i1t = 6v (3 points).

3) The rated current of lamp i2.

Resistance of lamp L2.

The actual current through the lamp L2.

The actual power of lamp L2 P2 = U1I2 =

The actual power of lamp L2 is less than the power of lamp L1.

The lights are brighter.

This is a physics contest question, and I hope it helps you

p=ui=u*u/r=i*i*r

L1, nominal 6V3W, P=3, U=6, R=6*6 3=12 L2, nominal 12V3W, P=3, U=12, R=12*12 3=48 As shown in the figure, L1 and L2 are connected in parallel, the voltages at both ends of L1 and L2 are equal, both are 6V1, the current flowing through L1 I1=U R1=6 12=2, the electric energy consumed by L1 per minute W1=P1*T=6W*(1 60)=kWh) 3, the current flowing through L2 I2=U R2=6 48=P2=ui2=6*

The actual power of L1 and L2 is P1=6W, P2=

p1>p2

So L1 is brighter than L2.

The power supply voltage is 6V

1) According to the formula p=iu, you can get i=p u

L1 is 3W, the voltage is 6V, so the current is 3W, 6V=(2), the power is 3W, the time is one minute, that is, 60S, according to the power formula W=PT, the resistance of 3W*60S=180J (3) L2 can be inversely solved by the formula P=U*U R, R=12V*12V*12V, 3W=48

The input voltage is 6V, so if you bring 6V into that formula, you can calculate its actual power 6V*6V 48=

So L1 is a bit brighter.

First, identify whether it is a parallel circuit or a series circuit.

The parallel branches have equal voltages.

The current of the connected branches is equal.

Then, the basic property of the bulb is that the resistance of each lamp does not change. When solving the problem, first calculate the resistance of each lamp:

r1=(6^2）/3=12;

r2=(12^2)/3=48;

Then solve the problem.

1）、 i1=

2) 3w*60s=180joules.

3) i2 = 6 48 = 6V* = and the actual power of L1 lamp is 6V*

So the L1 light is brighter.

It's good to learn Physics is very rare, and when I took the college entrance examination, I basically scored physics at most, and now it's useless to do questions.

The important thing is the formula, can you tell what the formula is for, the physics question will automatically be, I have more than 100 in the physics college entrance examination, 120 full score, listen to me, you summarize the formulas yourself, quickly say what the formulas are for, how to use them, what are the restrictions, and it is not a problem to understand these questions.

The brightness of the bulb is determined by the current flowing, because the two bulbs are in parallel, the voltage at both ends is the same, but because the resistance of the two bulbs is different, the current flowing through the solid is different, and the brightness is different, 1) because the battery voltage is 6V, so the bulb L1 emits light normally, P=UI, solid I = (2) P=3W, and the electric energy consumed in one minute = PT=180J (3 The voltage at both ends of the bulb 2 is also 6 volts, which should be rated at 12 volts, so it cannot emit light normally, according to P=U2 R, It can be concluded that the bulb resistance is 48 ohms, so the current flowing is 1 8, so the bulb with power = ui = 3 4= is brighter, and solid 1 is brighter.

Analysis: (1) Since the power supply voltage is 6V, the actual voltage of the bulb L1 is equal to the rated voltage, and the actual power is equal to the rated power, so the current of the bulb L1 can be calculated according to the rated power and rated voltage of the bulb L1;

2) First according to the formula r=

U2P calculates the resistance of the bulb L2 and then uses the formula P=

U2R calculates the actual power of the bulb L2; The brightness of the bulb is determined by the actual power of the bulb, and the bulb with the highest actual power is bright

3) Knowing the actual power of the two bulbs, the power consumption of each minute of the bulb can be calculated according to the variation of the electric power formula w=pt

Solution: (1) The rated voltage of lamp L1 is U1=6V, the rated power P1=3W, and the power supply voltage is U=6V, so the lamp L1 emits light normally, and the current through the bulb L1: I1=PU1=3W6V=;

2) Resistance of lamp i2: r2 = u22p2 = (12v) 23w = 48, the actual power of lamp l2 p2 = u2 2r2 = (6v) 248 =

Because the actual power of lamp L2 is less than the actual power of lamp L1, the lamp ll is brighter

3) When the lamp L1 emits light normally, the electric energy consumed per minute: W1 = P1T = 3W 60s = 180J;

When the lamp L2 emits light normally, the electrical energy consumed per minute: W2=P2T=

Answer: (1) The current through the bulb ll is.

2) The actual power of the bulb L2 is, the light ll be brighter

3) The bulb LL consumes 180J of electricity per minute; The bulb L2 consumes 45J of electricity per minute

1.Because the circuit is connected in parallel, the voltage is the same, L1=6V 3W, because the formula P=UI so IL1=P U, IL1=3W 6V=

Minutes = 1/60 of an hour Because w=pt so wl1=3w*1/60=3kw·h of 60 =

3.Because r=u squared p, rl2=12v*12v 3w=48ohm because i=u r i=6v 48ohm= because p=ui so p actual=6v* (2) l1 is brighter because l1 is rated at 6v and the actual voltage is also 6v.

It's actually very simple, remember that the resistance of the bulb is constant, first calculate the resistance of each bulb according to the known conditions, and the others are calculated with the resistance (the voltage and current are variable with the outside world).

Keep this in mind and the rest of the questions will be done.

Little brother, don't say that everyone is short-sighted, you just take a visionary who can see clearly.

(1) is the time of sound propagation, ignoring the propagation time of radio waves s=

2) by p=ui=20000*100000000=2 10 kilowatts to the tenth power.

It can be seen as the time for sound propagation, s=

The power obtained by p=ui is 2 times 10 to the power of 13.

Life is full of physics, such as eating, which requires gravity, and it is difficult for people to swallow food when they turn upside down or in space;

People walk with speed;

household electricity, which is converted into other forms of energy;

When a person speaks, it is the vocal cords that vibrate; The air passes through the human ear, causing the eardrum of the human ear to vibrate, and then transmits to the ossicles and the auditory nerve to the brain for perception.

Wet clothes dry out in the sun, and the water evaporates into water vapor. Wait a minute.

1: The essence of a short circuit is that a wire connects the positive and negative poles of the power supply (the so-called essence is that the ammeter and the like are regarded as wires).

The essence of the open circuit is that there is a wire disconnection (the so-called essence is that the bad bulb is the filament is disconnected, and the filament is also a wire) or the resistance in the circuit is too large, and the current is too small to be 0 (the so-called essence is that the voltmeter resistance value is large enough to be regarded as an open circuit).

Second, as for how to connect, the key is to draw the wire according to the direction of the current.

If you see a physical diagram that requires connection, you should write from the positive end of the power supply and observe what the electrical appliance is where the current flows next, then connect from the positive end of the power supply to that appliance, and connect it in turn until it is connected to the negative end of the power supply.

If you see the physical diagram requires the circuit diagram to be drawn, then you must be clear about the symbols of each component in the physical diagram, and then according to the direction of the current, first draw the power supply, then draw the symbol of the electrical appliance through which the next current flows, and finally connect to the negative pole of the power supply.

Basically, that's the above points, if you still have anything unclear welcome hi me.

1.When the positive and negative poles are directly connected when there is a short circuit, there is no load, which will cause excessive current. When the circuit is opened, the current cannot pass through.

2.First of all, it is necessary to understand the structure of the physical drawing, which can be drawn according to the physical drawing first, and then simplified. As long as you understand the simple circuit diagram, you can follow it.

In physics, an electric current that is turned on directly without passing through an electrical appliance is called a short circuit. In the event of a short circuit, the machine is often damaged or caught fire due to excessive current.

Series circuitWhen the circuit does not close the switch, or the wire is not connected properly, or the filament is burned out, that is, the circuit is disconnected somewhere. A circuit in this state is called an open circuit.

A short circuit is when both ends of an electrical appliance are connected by a wire, and this electrical appliance does not work, but the other electrical appliances in this circuit do.

Open circuit means that this circuit or the branch where this circuit is located is disconnected somewhere, no current passes through, and other electrical appliances of this circuit or branch work. Sunday.

1.A short circuit is a situation in which the positive and negative poles of the power supply are connected with a single wire, that is, in the circuit, the external circuit (excluding the resistance of the power supply) is 0. Because of the current, voltage, and resistance, when the resistance is 0, the current in the circuit will be infinite, and the power supply and circuit will be burned.

An open circuit, as the name suggests, is when the circuit is not a connected system, and the positive and negative poles of the power supply are not connected by a complete path to form a current. Therefore, in an open circuit, there is no current. At the same time, because the open circuit is caused by broken or unconnected wires, since air is an insulator, it can be infinitely resistant, resulting in zero current.

Your question is both good and difficult. The reason why it is good is because it is a basis for learning circuits, if you can't understand the circuit diagram, learning circuits is like a book in the sky. But it is difficult because the circuit is the most important part of junior high school electricity, there is no shortcut and trick to learn to read the circuit, only after you really understand the nature of the circuit, can you draw the circuit diagram accurately and quickly.

It is a very comprehensive knowledge and ability. But that's not to say that it doesn't have a way to learn. There are a few steps you can take to learn how to draw a circuit diagram according to the rules.

1) Clearly recite the definition of series circuit and parallel circuit, and be able to draw the most basic series and parallel circuit according to the definition. It can find the relationship between resistance and resistance, and the relationship between resistance and current and voltage in the circuit.

2) Don't panic when you see a complex circuit, there is a trick, that is, you start from the positive pole of the power supply, look for it along the direction of the current, you can first find the main circuit according to the flow direction and distance of the current, and draw the main circuit well, that is, connect the positive and negative poles of the power supply with a path. Then find those complex parallel parts, and find the parallel connection is to see that the two ends of the resistor are connected together, and the current will bifurcate when it passes. Then, after the parallel part is connected, carefully look at the series-parallel relationship between several resistors and devices along the current direction to see if a path from the positive pole of the power supply to the negative pole is completed.

3) Drawing a circuit diagram from the real thing is the same. Starting from the positive pole of the power supply, along the direction of the current, find the series-parallel relationship between each resistor and the device. You can draw the main circuit first, or you can connect all the parallel resistors in parallel first, and then connect them in series to form a path.