Ask for a physics question in the second year of junior high school, please talk about the ideas, es

1) It is known that the bulb is working normally, i.e. U light = 8V, P light = 4W has P=U*i i=U R

We know that r=u is squared p

Bring in to get r=16 ohms.

2) Do you know the partial pressure formula?

u1/r1=u2/r2

It is known that r variable = 4 ohms, r lamp = 16 ohms, u lamp = 8v substitution to obtain u variable = 2v

U total = u lamp + u variable = 10V

3) Still use the partial pressure formula.

It is known that u total = 10 V, r variable = 8 ohms, r light = 16 ohms.

Substituting to obtain u lamp = 20 3V

There is the formula p=u square r lamp.

Substitution yields p=25 9

1,r=u1*u1 w=64 4=16 ohm The algorithm of the resistive circuit r.

2,8v/16 ohm=u2/4 ohm

U2=2V loop currents are equal.

u=u1+u2=8+2=10v

3,i=u/16+8 ohm=10/24=5/12ap=i*i*r=5/12*5/12*16=25/9w=

It is very easy to learn electricity in junior high school, and the key is to familiarize yourself with the formulas and each law and use the formulas to make it more convenient for you to do the questions. Let's take this question as an example:

1.Resistance of the bulb: The rated U, P. can be known by the indication of the bulb

Do you know r=u*u p? This is calculated by a combination of p=ui and i=u r. Replace i=u r with p=ui to get p=u*, that is, p=u*u r is r=u*u p.

In this way, the formula can be directly substituted. r=8*8 4=16 ohms.

Slip = 4 ohms, right. R light = 16 ohms. According to the law of partial pressure in series or U ratio = R ratio.

Do you know any of these? If you don't know, you can also deduce it from the tandem i law and i=u r. This is the case with physics and electricity, where you have to keep reading or thinking about it, familiarize yourself with and master each formula and law, and then prompt other laws.

It's like you know i=u r and you can prompt u=ir and so on. Of course, this is only a small calculation, and the combination of formulas and rules can prompt a lot of convenient questions for you. >

U ratio = r ratio, i.e., 16 ohms 4 ohms = 8V slip uSlip U=2VThen there is the series, U total = U1 + U2can obtain u total = 8 + 2 = 10v

3.To find p, we think that the formula p=ui is fine, but if you are familiar with the formula and often use the formula, you can use the calculated p=u*u r. here"Because it is very troublesome to find i in this p=ui, it is more convenient to use the latter" This is what the electricity needs to do as mentioned earlier, and the familiar and urging of formulas and laws!

In question 2There is a mention of u ratio = r ratio. Got to know the slip r and the light rYou can know that you slide light u = 8 16 = 1 2

To know the total u=10vGet u lamp = 10v 3 * 2 = 20 3Then from p=u*u r, p=<20 3>*<20 3> 16=<25 9>w

Hehe, my other subjects are in the upper middle, and this physics is at the top of the whole school. Just because I had a beautiful dream in junior high school - to be a physicist. Although this dream is no longer there, the foundation of that effort is still there.

I am 16 years old, and I want to exchange friends with junior high school physics or ask questions +++ 1357764330Hehe, in the end, I wish you to learn every day.

1.The resistance can be found according to 8V, 4W.

2.The voltage of the power supply is calculated depending on whether your circuit is connected in series or parallel, but what can be determined is that the voltage on the bulb is 8V when P is in the middle

3.The answer to the second question is foreshadowing.

The first question r=u1 u1 w=64 4=16 the second question i=8v 16 =

u2=i×4ω=

u=u1+u2=8+2=10v

The third question is i=u 16+8 =10 24=5 12ap=i i r=(5 12) (5 12) 16=25 9w=

Complete combustion of 1t household waste exothermic is:

q＝mq，m＝1000kg

The electric energy of 240kw·h is:

e=240kw·h＝240×

The power generation efficiency is: e q base transport.

1 = 1500 steps.

The total distance of the body height change in the whole process is: 0500 Work done: w = GH 50 10 150 75000J Note: The mass is 50kg, and the weight of the beam is 500N.

The energy of burning 1 ton of Paihu Zhuchang garbage is q=qm=1*10 3kg*, so the efficiency is *100%=

2.The work done to overcome gravity for each undulation is w=g*h=mgh=50kg*10n kg*

1500m total work 1500 1=1500 times. So the total amount of work done is 50j*1500=

1500 divided by 1 equals the number of ups and downs of the body. 50kg is equal to 500N, and 500N is multiplied by 10 times 1500

If there is no diagram, please upload the diagram.

1) At this time, L is connected in series with the rheostat, and L emits light normally, then the voltage at both ends of it is UL=6 volts, and the current in the circuit can be seen from Figure 24 B=1A, at this time, the voltage at both ends of the rheostat U1=U-UL=15-6=9 volts, and the electrical energy it consumes in 10 seconds w=uit=9 1 10=90 joules.

2) At this time, the rheostat r = 20 ohms and the resistance r1 are connected in series, and the ammeter is fully biased, indicating its maximum range and the current in the circuit i =, the total resistance of the circuit r = u i = 15 ohms, and the resistance r 1 = r - r = 25-20 = 5 ohms.

3) R1 = 5 ohms unchanged, then when the voltage at both ends of it is U1 =, i1 = U1 R1 =, through its current i1 =, according to the characteristics of the series circuit, the voltage at both ends of the rheostat is the voltage representation number U2 = -U1 =, and is exactly the middle of the voltmeter 15 range and the ammeter 3a range, the deflection angle of the two meters is the same, and the voltage represents the number =.

This kind of topic couldn't be simpler.

A is a uniform cylinder, if the volume of the cut part is V, and its mass is 1500V, then the remaining mass of A is 9-1500V, and the pressure on the ground is (9-1500V)*10.

Because the density of a is greater than the density of the liquid, the object sinks. And because the title says that the cut part is immersed in the liquid, so V row = V, then the height of the liquid level rise h'=v row s2=v , the final height of the liquid h=, the pressure on the bottom surface is 1000*10*(

Two pressures are equal, the equation about v is listed, and you can solve the equation yourself.

1) Maximum current imax = pmax u= average current i' = q t = 134mA average power p' i' u=u i'=u q t= , about (2) 1, don't hang the phone on your chest, 2, when you connect **, it's best to take the mobile phone to a distance away from the body to turn on, and then put it to your ear to talk, 3, in addition, try not to use the mobile phone to chat, and pay attention not to put the mobile phone next to the pillow when sleeping. Hope, thank you. Satisfied.

Maximum current imax = pmax u=

The average current i'=q t=134mA

The average power p' i' u=u i'=u q t=, approx.

2) 1, do not match the nuclear width to hang the mobile phone on the chest, 2, when the ** is connected, it is best to take the mobile phone to a distance away from the body, and then put it in the ear to talk, 3, this silver, try not to use the mobile phone to chat, and pay attention not to put the mobile phone on the pillow when sleeping.

Hope, thank you.

That is, you can calculate the answer to question 1 3 10 -3m 2

After putting in the wooden block, the water depth h does not change, and the pressure of the water on the bottom of the cup does not change, just solve it according to f=ps.

In fact, it is the first question and the second question.

What about the figure? Count the whole figure up.

Uh-uh... The first picture is half of the front, the table cares, the table cares about it It's very simple 2) Q=cmδt=, Yinzi here you miscalculated, it's 2L, not 1L, look at the topic to be careful to attack.

Then q=i 2rt t=q i r=

Then (3) p=u r=200 44=909w look.

According to p=u r, the total resistance decreases, and the total power will increase, so it should be connected in parallel.

Because of the parallel connection, the voltage at both ends of the drying machine and the heating wire is 220V, and the normal working power of the drying machine is 1000W, so the power of the heating wire is 1500W-1000W=500W, which is obtained by P=U R, R=U P=(220V) 500W=

Why, I dropped out of school in the second year of junior high school and didn't learn.