A physics problem

It's a hassle to explain, so let's put it simply:

There are two commonly used formulas for q, in addition to q=u 2 rt, there is also q=i 2rt First of all, the wire and the electric heater are connected in series, it can be considered that when switching to wires of different thicknesses, the sum of the currents on the wire and the current of the electric heater are equal, so that the two currents are equal, the formula q=i 2rt should be used, it can be seen that the smaller r is, the smaller q is. This is the control variable method, and the control currents are equal to be comparable.

And the Q=U 2 RT you use can only be used when the voltage on the wire is equal, according to the principle of series voltage division, when you switch to different wires, the voltage of the wire is definitely not equal, so your conclusion is wrong.

The formula for calculating the resistance is r= *l s, which is the resistivity, l is the length, and s is the cross-sectional area.

If the wire is too thin, that is, if S is too small, the resistance will be large, and the heat generated Q=i·i·r, so Q is too large, and the heat will be more.

There is also a formula: q = i 2rt, the larger r is, the greater q is. Because the voltage on r changes before and after r changes.

So it's inaccurate to take advantage of u. The larger the r, the greater the voltage on r, and the smaller the total power, but the power distributed on r in the whole circuit becomes larger, and the heat generated becomes larger.

Compare Hypothesis R1

Hello this student, to discuss the issue of endothermy and exothermy, we must have heat transfer in the process, and only in the process of heat transfer will there be endothermic objects and exothermic objects. According to the heat absorption and release formula q suction = c*m* (t-t0), it can be seen that the amount of heat absorbed or emitted by an object is related to the specific heat capacity, mass and temperature change of the substance, not only the mass and temperature. So the above statement is incorrect.

Regarding the relationship between temperature, heat and internal energy, please note that when the temperature of the substance increases, the internal energy must increase, and the internal energy of the substance increases but the temperature does not necessarily increase (for example, when the crystal melts, the internal energy increases but the temperature does not change). The increase in the internal energy of a substance is not necessarily due to the absorption of heat. Please refer to the attached diagram for the relationship between the three.

Hope it helps.

Objects with high temperatures are heat-absorbing if they are exposed to higher temperatures, such as the sun.

Others, please elaborate on the topic.

Because it is connected in series, the current through the two resistors is the same.

Because the maximum current through B is A so the maximum current that this circuit is allowed to pass through is.

Maximum voltage u=ir=volts.

The maximum allowable current is calculated on the basis of B.

The series circuit current is the same so the maximum allowable current is.

u=(10+15)*

The rated current of B resistor is small, because the current of the series circuit is certain, and the total current can be obtained, so the total U = (r A + r B) is multiplied to get 15V

In series, the total resistance micror = 25 ohms and B's rated current microcircuit has a maximum current of Otherwise B will burn out ......

The maximum voltage is v=

Because they are series circuits with the same current, the maximum current is.

From the formula, we get u=(10+15)*

If the inclined plane is smooth, the extra work is 0, the efficiency is 100%, and the total W = W useful FS = gh

f*5m=800n*

f=240n

I feel like that.

The conditions are sufficient.

When the bevel is smooth:

Total work fl = useful work gh

Thrust f=gh l=800*

800* Ox. Breaking down gravity, the size parallel to the direction of the inclined plane is 240 N. The thrust is equal to the gravitational component, so it is 240 N.

The conditions are sufficient, see the answer:

The ratio of the length of the side is 1: 2:3 earlier than Zheng Shu, the volume = the cube of the side length, then the volume ratio is 1:8:27, the mass = volume * density, the same material is made of the same material, the density is the same, and the density ratio is 1:8:27 if it is solid

3g:24g:36g=1:8:12, the mass of C is small, so C is a hollow square.

If the side length of A in the trapped cavity is l, then the side lengths of B and C are 2L and 3L respectively, and their circular combustion volumes are respectively.

V1 = L 3, V2 = 8L 3, V3 = 27L 3 average densities, respectively.

p1=3/l^3

p2=(24)/(8l^3)=3/l^3

p3=36/(27l^3)

p1=p2> Wang Shan p3

Therefore, C is hollow.

Mass = Volume * Density. And the same material, the density is the same, if the stockings are solid, then the mass ratio = volume ratio. Then the mass ratio is 1:8:27, then it can be seen that C does not match the noisy auspicious, he is lighter, so he is hollow.

Because the ratio of side length is 1:2:3, if it is solid, the ratio of volume should be 1:8:27, and the mass is 3g, 24g, and 36g, so C is hollow.

You're right: the brighter it is when connected in series, the smaller the power rating.

First of all, you have to understand one thing: there is only one parameter that determines the brightness of a bulb - the actual power!

You don't have to ask about its current voltage and current, or its resistance. That's it, so the second problem has been solved first. Right. (Because in the second problem, normal work is normal power.) ）

You also need to understand a concept: the greater the resistance of the electrical appliance, the smaller the rated power.

Let's move on to your first question.

First of all, ask: "Now the first light is brighter than the B light", why? It can only be explained: the actual power of the first lamp is large!

Ask again: Why is the power of the first lamp large? It can only be stated:

A lamp actually acquires a large voltage! (Because the two are connected in series) ask: why does the first lamp get a large voltage, it can only show that its resistance is large!

Because the two are connected in series) conclusion: the resistance of the lamp is large, and the rated power is small (the greater the resistance of the electrical appliance, the smaller the power) Finally, many things do not need to be calculated, as long as you grasp the main point and think a little, you can get the answer.

Think: What would happen if two light bulbs, one 220v15 watt and one 220v100 watt, were connected in series to a 220v circuit?

Think before you look down.

At this point, 15 watts of normal light (almost normal, to be exact), while 100 watt bulbs don't light at all.

Theoretically, the 15 watt bulb obtains 87% of the voltage, but in fact, we know that the resistance is also related to the temperature, the temperature of the 15 watt bulb filament is very high, and the resistance becomes larger, while the temperature of the 100 watt filament is very low, the resistance becomes smaller, so the strong are stronger, the weak are weaker, and the 15 watt bulb gets more than 95% of the voltage, so there is the above phenomenon.

Because the current in the series circuit is the same, you can calculate the resistance of the lamp according to the rated power and rated voltage of the lamp, the resistance is fixed, and the actual power is calculated according to the resistance, and you know that the larger the resistance under the same current, the greater the voltage, p=ui

The first question: for two bulbs with a rated voltage of 220V, it can be seen from P=UI=U*U R that the resistance of the bulb with a large rated power is small. When two bulbs are connected in series, the current passing through them is the same; From p=ui=i*i*r, it can be seen that the bulb with a large resistance actually has more power, that is, it is brighter.

As a result, bulbs with a small power rating are brighter when connected in series.

The second question: I feel that the normal work in this problem means that they are all connected to the circuit of their respective rated voltage, so naturally the higher the power, the brighter it is.

na In the series circuit, the current is equal, and the title shows that lamp A is brighter than lamp B, so it can be seen that the voltage at both ends of lamp A is higher, so the resistance of lamp A is larger. According to the formula p=u2 r, the power of the first lamp is smaller.

It should be that when the series is the brighter the light, the actual power is smaller. ）

Natural gas is widely used in our city, and the calorific value of natural gas is known to be 4 107 J m. The complete combustion of natural gas can release 2*10 6 J of heat, if only 42 of these heat is absorbed by water, the temperature of 5kg of pin tung water at room temperature can not rise 40.

Do it with energy-conserved columns! Ignore the lost heat energy.

I forgot the specific formula when I entered the university, hehe, I will know it until I read the book! Mingchi read the book under the eggplant shirt by himself, it's very simple.

The first question is q1=mq

The second Dan is absolutely frank and asks Q2=42%Q1 Q2=CWater m Hongbi t

Put the number of analog Tong Dai inside, I don't count--

It puts more pressure on the sides. The total pressure of the water is constant, but when the pressure on the side is greater, the pressure on the bottom side is smaller.

The total pressure of the water is constant, but as the height of the side increases, the pressure on the side is greater, and the pressure on the bottom surface is smaller.