Bonus points for good physics questions in the second year of junior high school are 50

As the plane flew overhead, the sound began to relay downward. After 10 seconds, the sound has passed 3000m from that point to the human ear, and the plane is still flying forward, and by the time the sound is transmitted to the human ear, the plane has already flown 4000mSo the distance is.

Root number (4000 2 + 3000 2) = 5000

Can it be negative?

The steps to solve the problem are; (Speed of sound - speed of the aircraft) * Time = distance.

Substituting your question is (300m s - 400m s)*10s=-1000m

Maybe you've made a mistake, so you can substitute the correct value and find it.

Let the distance be s,s (400m s —300m s) = 10 seconds ......Chase the problem.

s = 1000 meters.

This topic is not complete, what is the observation point? If the sound source and the plane are in a straight line, the first floor is the correct solution, but the problem is not complete, because the sound source can be moved in a semicircle centered on the plane.

The aircraft flies 4000 meters (400*10) in 10 seconds

The sound traveled 3000 meters (300*10) in 10 seconds

If you deviate, it should be 7,000 meters, no map? The airplane flew towards the ???

According to what you said, it is impossible to hear, it is not possible to use a triangle, it is not possible to use a straight line, and the speed of the plane must be less than the sound - -

If the sound source and the plane are in a straight line.

4000-3000）*10＝1000m

Solution: (1) First of all, this is a rod OA with a uniform texture, so its gravity can be regarded as completely on the center of gravity, that is, the midpoint of the rod.

Look at the A figure, E is the midpoint of OA, that is, the center of gravity, because OA=OB=10M, so at this time OAB is an isosceles right triangle, A=45°, through O as OM AB in M, then OAM is an isosceles right triangle, so OA=10M is obtained OM=5 2, because of the lever equilibrium condition, F1*L1=F2*L2 (depending on gravity F2, then OE is L2, OM is L1), that is, F1*5 2=3000N*5M, F1=1500 2

n and then look at the b figure, aob=90°-30°=60°, e is the midpoint of oa, that is, the center of gravity, because oa=ob=10m, so equilateral aob, so a=60°, through o for om ab in m, at this time, om=5 3, through e to en on, because aon=30°, easy to find on=because of the lever equilibrium condition, f1*l1=f2*l2 (depending on gravity is f2, then on is l2, om is l1), that is, f1'*5 3 = 3000n *, f1 can be obtained'=1500n, i.e., f2 required in the problem so f1:f2=1:2

2) It is easy to find en= from the b diagram, and the gravity is 3000n, so w=fs= because p=w t=7500j 30s=250w

(1) From the table, when the wind speed is 10 m s, the energy obtained in 1s is 5 10 2j, that is, the power is 500j s=500w

At this time, it happens to be the rated wind speed, and the output power is just the rated output power, which is 300W power generation efficiency: =300W 500W=60%.

wxsx15%=45w

Size: S sqm.

mah=18ah

t=w p=25vx18ah 45w=10h---18000mah is the current multiplied by time, i.e. it, and then multiplied by u equals w, w uit).

It can be used for 10 hours of luminaire work.

I don't know what to ask about.

1. Water vapor can be produced no matter what temperature it is, and it should be produced by the thermal movement of substances in the vapor type, and all substances have a certain temperature. There is a definition of absolute zero, i.e. above which there will be steam.

2, isn't this very difficult, choose b. The unflipped thermometer cannot drop below this temperature, but only rises. Because the structure of the thermometer is that there is a bayonet at the end of the meter, if you don't throw it off, the mercury column will not be able to fall.

3. The overall temperature must be guaranteed to be below zero, -2 degrees and 0 degrees combined, the temperature will be between -2 degrees and 0 degrees, so the temperature is absolutely below 0 degrees, and they will freeze. The same can be said for the rest.

4. It is related to pressure and whether there are impurities.

At 1,0 degrees, the ice is directly sublimated to 0 degrees of water vapor.

2,b Because the mercury in the thermometer can only flow out, and it will not go back if you don't shake it off, it will show 38 degrees when it is below 38 degrees, and it can show an accurate value when it is higher than or equal to 38 degrees.

3, B after putting 1 degree of water ice will absorb heat higher than 0 degrees, so there will be some melting, C is the same, and B ice cubes absorb heat, but the temperature will not exceed 0 degrees, water 0 degrees continue to exothermic, so part of the water will freeze.

4. Air pressure.

1. Water below the boiling point can also be evaporated.

2. The thermometer has a thin bend, which does not allow it to fall arbitrarily and is convenient for observation, so it is B.

c is the temperature of the ice-water mixture, so it is b.

4. It is also related to "pressure".

The first question is the definition, at 0 degrees water exists in the form of ice-water mixture, and water vapor exists at any temperature, the second question chooses b, use an unshaken thermometer to measure the body temperature, lower than its degree, the thermometer shows the same number, compared to the higher degree, the correct degree. If you choose B, the water will become -1 degree and the mass of the ice will increase. Four is related to pressure.

You can ask these questions to the teacher.

0 water vapor is also there, evidence, if the temperature is 0 today, is there water vapor in the air? There must be, in the same way, water vapor is present at any temperature, and it can be seen that the evaporation of water can be carried out at any temperature.

Question 1, evaporation can be carried out at any temperature, while water at 0 degrees Celsius will remain in the ice-water mixture, will not freeze, and will not melt.

For question 2, choose B, because when the number of sights of the thermometer is lower than it, it will not change, because there is a necking under the thermometer, and it cannot return to the original point.

Question 3, choose B, because it is placed on the melting point of ice, and it is in water, so the mass increases 4 questions, air pressure, impurities contained in the crystals.

1. Evaporation is no longer carried out all the time. There will also be water vapor below zero. 0° is the melting point of ice, so it is possible for both water and ice to exist.

2、b。If he doesn't, he won't go down himself. Even 38° will display 39°. 40° is higher than 39°, normal display.

3、b。Because the ice absorbs the heat of the water at 0 degrees, some of the water freezes again.

4. Pressure. The melting point of an object is related to the pressure.

At a temperature of 0, ice, water, and water vapor are all possible Evaporation Liquefaction Vaporization can occur in any environment.

39, 40 thermometer does not shake and will not fall.

2 of the ice in the water of 0 Ice-water mixture Water may refreeze air pressure Vacuum and high pressure are different.

1. First find the effective volume of the copper ball, v=m=

2. Since the effective volume of the copper ball v<50cm3, it means that it is hollow.

3. Hollow volume =

The hollow mass here is not counted, then the kilogram is the mass of copper, and the volume of copper = kilogram per cubic meter is about equal to cubic centimeters, which is smaller than a ball, and the hollow volume = the volume of the copper ball - the volume of copper = 5 cubic centimeters.

cubic meters = cubic centimeters, hollow, with a volume of cubic centimeters.

1. When only S2 is closed, the ammeter, R1 and R2 are connected in series, let the current in the circuit at this time be i1, and the indication of the ammeter is, then i1=. The voltage applied to both ends of r1 is u1 = i1 * r1 =. The voltage applied to both ends of R2 is U2=I1*R2=.

The power supply voltage u=u1+u2=6v+.

2. When only S2 is disconnected, R1 and R2 are connected in parallel, and the ammeter is on the main road, then according to the nature of the parallel circuit, the voltages at both ends of R1 and R2 are equal, which is equal to the power supply voltage. At this time, it is known that the current through R2 is I2=1A, then the voltage applied to both ends of R2 is the power supply voltage U=I2*R2=1A*R2.

In summary, we can get the equation 6V+, which can be calculated as R2=12 and U=1A*R2=1A*12=12V.

r1 and r2 in series, u (12 + r2) =

Disconnect only S2 (equivalent to R1 and R2 in parallel, ammeter on the dry road. ), the current through R2 is 1A, U R2 = 1A(2).

Solution (1) and (2) yields u=12v r2=12

When S2 is closed, the total supply voltage is u=

When S2 is disconnected, the voltage across the resistor is equal to the supply voltage, so the voltage of R2 = 1A x R2 = U

The simultaneous equation can be solved to u=12v r2=12 ohms.

r1 and r2 in series: u=i*(r1+r2).

R1 and R2 in parallel: U=I2*R2

r1 i i2 is known to solve the equation yourself.

(12* then r2=12 euros u=12*.)

Do you see it ?? Think for yourself why you're doing it.

Proof: Connect OC, pass the O point to do OD BN, OE AC, OF BM, and pass bn, ac, and BM to the points D, E, and F. respectively

ao,bo is the angular bisector of mac, mbn oe=of,od=of

OE=ODOC divides ACN

That is, point o is on the angular bisector of the ACN.

The maximum distance between the chopsticks and the outside of the cup is: 15 cm - 12 cm = 3 cm.

The minimum distance is (chopsticks are placed obliquely) solution: set the diagonal distance between the mouth of the cup and the bottom of the cup to x, and find x as centimeters, so it is 2 to 3 centimeters.

If you analyze it, it should be like this.

The range of h is greater than or equal to 2 cm and less than or equal to 3 cm.

When upright, the longest is 15-12 = 3 cm, when placed obliquely, by the Pythagorean theorem, the hypotenuse = root number (5 2 + 12 2) = 13, and the shortest oblique is 15-13 = 2 cm, then the value range of h is 2 cm h 3 cm.

2-3 When inserting the chopsticks vertically into the cup, h is the maximum, h=15-12=13;When the chopsticks are placed diagonally into the cup, and the lower end is against the wall of the cup at the bottom of the cup, the h value is the smallest, h=15- 12 12+5 5=2 (the square can't be typed, so it is expressed).

h is greater than or equal to 2 less than or equal to 3

h is between 2 and 3, 2 is the use of the Pythagorean theorem, calculated to the diagonal is the longest is 13, then 15-13 to get 2. The other one is very simple!

The Pythagorean theorem is a minimum of 2 and a straight maximum of 3