Junior 3 Science Contest Questions about electricity

Hehe, I have participated in the national physics competition before, and I won the first place, this kind of problem is actually not difficult, it is troublesome, and the following is the process of solving the problem.

1 kWh = 1kWh w=pt=[(u 2) r]t=(i 2)rt=uit (memorize this formula, extremely useful) Electricity consumption is 11 hours a day.

The resistance of a 220V100W lamp is p=(U2) r, and the value is substituted for 100=220*220 r, and r=22*22=484 is obtained

1) Before not changing, it is ten lamps in parallel, and the total resistance r is total = (1 10) * r = (1 10) * 484 =

Daily electricity consumption.

w does not change = [(u 2) r]t = [(220 * 220) i.e. 11 kWh).

2) After the change, the first 4 hours of each day are parallel, and the last 7 hours are in series, and the first 4 hours.

w change to 1=[(u 2) r]t=[(220*220)

The most important thing is how to calculate the seven hours after the change, and the following is calculated:

After changing to series, the resistance of a lamp is.

r string = (2 3) * 484 euros (don't calculate the specific result, with the score downward, you can naturally make an appointment.) )

In groups of five, the total resistance of each group is.

1 5) *r string.

The resistors of the two groups in series are .

1 5) * r string * 2 = (1 5) * (2 3) * 484 * 2 = (4 15) * 484 euros.

The power consumption for the first 7 hours after series connection is:

w change 2==[(u 2) r]t=220*220*7*(1 [(4 15)*484])=

The answer to the first question came out.

After the change, the daily power consumption is.

W change = W change 1 + W change 2 = 4 + that is, the daily electricity consumption after the change is kWh.

The answer to the second question is:

The daily power saving is w saving = w does not change - w changes =

Electricity savings for a month.

w month = , that is, the monthly power saving is kWh.

If the poster can participate in the contest, he should be able to understand the skills of this algorithm. (Try to substitute the equation as invariant, generally speaking, the resistance and power supply voltage are invariant, and the series is mostly calculated with current, and the parallel is calculated with voltage.) ）

It is known that the series connection is 2 3"

There is something wrong with this statement.

Use 2NH3+H2SO4=(NH4)2SO4 to calculate the mass of ammonia gas as, and then calculate the mass of NH4+ in the compound as, the mass fraction of X is, so the relative atomic mass of X is 18, so the relative molecular mass of 18, and the relative atomic (molecular) mass of X is about 28), because the mass ratio of NH4+ and SO42- is 3:16, so A:C=1 (calculated by relative atomic mass) so B=1 or 3

Because the relative atomic mass of the metal elements in the earth's crust is greater than 10, b=1. Then you can find out what x is based on the relative atomic mass. I don't have your papers, I don't know the reference data given, you can check it yourself.

Also, do you have a standard answer? Can you tell me if it's right or wrong? Can you tell me the correct answer? Thank you

I've done the math for you, and the answer is right.

As for F's request, I agree with what the above person said.

The center of gravity is balanced according to the moment, whether you take the center of gravity as the point or the point B, I will explain the center of gravity as the point below.

First of all, you divide the gravitational force and the static friction force and the support force into two forces, perpendicular to the rod and horizontal to the rod.

Then about the moment balance of the center of gravity, the moment parallel to the rod is 0, you should know this, I will give the formula below.

The neutralization is the component of the frictional force and the supporting force perpendicular to the direction of the rod.

To find x=solve this problem, you must first conduct a force analysis, and figure out the magnitude and direction of each known force, and if you don't understand, you can ask me, I hope to help you draw a diagram for you, and in order to avoid overlapping, I drew a little off.

As shown in the figure, the magnitude of the static friction force acting on the rod in the problem is actually the component fcd used for the equilibrium force f in the horizontal direction, and it is easy to prove that dac=37°, that is, fcd f=sin37°= , that is, f=6n

It can also be found that the component of force f in the vertical direction fac=f*cos37°=, set the center at point e, then the force analysis is done with point b as the fulcrum, and there are 20*be=be=

Yours is the wrong answer.

The two triangles in the diagram are congruent, so the force arms are equal.

According to the leverage equilibrium condition: f1l1=f2l2, because l1=l2, so f1=f2.

This actually has nothing to do with the inclined plane, don't think too complicated.

1 all choose B, the reason is that the vernal equinox is the day when the sun's morning and dusk line coincides with the earth's meridian, at this time the position of the sun's morning line must be six o'clock in the morning, and the position of Shaoxing is just close to the 120e meridian, and the clock in the picture indicates that the time is 6 o'clock in the morning, and six o'clock is just the sunrise, so it is the spring equinox day.

b, the spring equinox or autumn equinox, because this day and night are of equal length.

--b The rotation of the earth is one day, so this day is ---

1,t1*(√2/2*l)=mg*l/2

2, t2*( 3 2*l)=mg* 3 4 The ratio of the two formulas t1 t2 = root number 2:1

It is the power * power arm = resistance * resistance arm.

Select A, let the mass of magnesium be x and aluminum be y. According to the characteristics of their respective reactions with dilute sulfuric acid, the hydrogen gas displaced by magnesium is x 12 and aluminum is y 9. We get the equation x-x 12=y-y 9. After calculation, a is obtained.

5 minutes. Let the resistance value of each section be r

Series resistance: 2r

Series power: U square 2r

Parallel resistance: the reciprocal of (1 r+1 r), i.e. r2 parallel power: u square (r 2).

The ratio is equal to 1 4